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https://www.techylib.com/el/view/nostrilswelder/antiderivatives | 1,529,872,270,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867055.95/warc/CC-MAIN-20180624195735-20180624215735-00195.warc.gz | 919,870,286 | 13,793 | # Antiderivatives
Ηλεκτρονική - Συσκευές
10 Οκτ 2013 (πριν από 4 χρόνια και 9 μήνες)
153 εμφανίσεις
1
Notes on
Antiderivatives
So far our studies have concentrated on given a function how do we find its
derivative? Now we ask the reverse question: Given a derivative how do we find the
function?
The following definition provides some clarity for this last
question.
Definition
: Let
f
denote a function on an interval
I
. A function
F
is said to be an
antiderivative of
f
on the interval
I
provided
for all values of
.
Since
for
, the function
is an antiderivative of the
function
on the interval
. There are other antiderivatives of
such
as
,
and
. In fact, since
, where
C
is a constant, all
the functions
, where
C
is any constant, are antiderivatives of
.
So the are
infinitely many antiderivatives for the function
. Are there any others?
To answer this question the following theorems provide the basis
for the theory.
In addition these theorems have important consequences.
Rolle
’s Theorem
:
Suppose
is a function which is continuous on the interval
,
differentiable on the interval
and for which
. Then there is a number
,
,
for which
.
The Mean Value Theorem (for Derivatives)
:
Suppose
is a function which is
continuous on the interval
and differentiable
on the interval
.
Then there is a
number
,
, for which
.
Observations
:
The following are consequences of these theorems.
Rolle’s Theorem is a special
case of the Mean Value Theorem.
Rolle’s Theorem guarantees the existence of at least one critical point on the
interval
.
The geometrical interpretation of the Mean Value Theorem is that there is a
tangent line to the curve
which is parallel to the (secant) line joining the
points
and
.
When the function
is interpreted as the position of a particle moving along a
line, then the M
ean Value Theorem says that there is a point in time that the
instantaneous velocity is the same as the average velocity over the time
interval
.
2
The following theorems provide the answer to the questions posed earlier. The proofs
are based on the Mean Value Theorem and are included
at the end of these notes
for
completeness.
Theorem 1
: Suppose
f
is a differentiable function on an interval
I
for which
for
all
. Then
f
is a consta
nt function, that is to say,
for some constant
C
.
Theorem 2
: Suppose
f
is a function defined on an interval
I
and suppose also that
F
and
G
are both antiderivatives of
f
on the interval
I
. Then there is a constant
C
for
which
for all values of
x
in the interval
I
.
the ques
tion originally posed, namely, d
oes the family
of
functions
,
C
a constant, represent all the antiderivatives of the function
on the
interval
? According to Theorem 2, the answer is “YES”!
Indeed the function
is
a known antiderivative of
. If
is any other antideriv
ative of
, Theorem 2
states
for some constant
.
Definition
: Suppose a function
on an interval
I
, the family of all antiderivatives of
on
the interval
I
is called the
indefinite integral of
and this family is denoted by the
symbol
.
Observation
: Notice that Theorem 2 states that if
is a function for
wh
ich
, then
, where C is a constant called the
constant of integration
. Conversely, if
and
are functions for
which
, where C is
a constant, then
.
Example 1
: Verify that the function
satisfies the hypotheses of
Rolle’s Theorem on the interval
. Then find all the numbers on that interval that
satisfies t
he conclusion of Rolle’s Theorem.
Solution
: The given function is a polynomial and therefore continuous everywhere and in
particular on the interval
. Since polynomials are differentiable everywhere, the
given function is differen
tiable on the interval
.
Furthermore
and
. So the given function satisfies the
hypotheses of Rolle’s Theorem.
Find the numbers
so
that
:
(
U
)
3
.
There are two possibilities, namely
and
. The second
possibility
is negative and therefore not in the interval
.
However the first
possibility
and therefore in the interval
.
Example 2
:
Verify that the function
satisfies the hypotheses of the Mean
Value Theorem on the interval
. Then find all the numbers
that satisfy the
conclusion of the Mean Value Theorem.
Solution
:
The given function is a combination of
functions that are continuous
everywhere and therefore continuous on the interval
. The given function is also
differentiable on the interval
since
everywhere. So the given funct
ion
satisfies the hypotheses of the Mean Value Theorem.
Find
so that
:
. So this number lies in
the interval
.
Example
3
:
Consider the function
. Show that there is no number
in the
interval
for which
. Why does this not
Value Theorem?
Solution
:
. Since this equation
which has no real solutions, there is no such number
. This does not contradict the Mean
Value Theorem because the given function has a discontinuity at
and is not
continuous on the interval
. (
It is not differentiable on
either
.
)
4
Example
4
:
Show that the equation
has exactly one real root.
Solution
:
The function
is a combination of continuous functions and
therefore continuous everywhere. Furthermore
and
. So
by the Intermediate Value Theorem there is at least one real root on the interval
.
Suppose there are two real roots,
.
(
) For all values
o
f
x
,
. So the function
is continuous on the
interval
and differentiable on the interval
. Further
and
.
So by Rolle’s Theorem there must exist a number
,
, for which
.
So
So there cannot be two real roots of
the equation.
Ex
ample
5
:
Evaluate each of the following indefinite integrals.
a)
c)
b)
d)
Solution
:
a) Since
,
the function
is an antiderivative of
.
Therefore
.
b)
Since
, the function
is an antiderivative of
.
Therefore
.
c)
Since
, the function
is an antiderivative of
.
Therefore
d)
Since
, the function
is an anti
derivative of
.
Therefore
.
The proofs of the t
heorems are now presented. It is not important for you to
memorize these proofs, but
as
you can
see why
the theorems hold true
.
Rolle’s Theor
em
: Suppose
is a function which is continuous on the interval
,
differentiable on the interval
and for which
. Then there is a number
,
, for which
.
Proof
:
There are three cases to consider.
5
Case 1: Suppose
is a constant function. Then
for all values of
x
in the
interval
. Select a number
. Then
.
Case 2: Suppose
for some value of
. Since
is continuous on the
interval
, the function
has a maximum value, call it
, on the interval
.
Then the number
c
is a critical point for the function
and, since
is differentiable on the
interval
,
.
Case 3: Suppose
for some value of
. Since
is continuous on the
interval
, the function
has a
minimum
value, call it
, on the interval
.
Then the number
c
is a critical point for the function
and, since
is differentiable on the
interval
,
.
The Mean Value Theorem (for Derivatives)
: Suppose
is a function which is
continuous on the interval
and differentiable on the interval
.
Then there is a
number
,
, for which
.
Proof
: Define the function
which is a
combination of the given function and a polynomial. Since
is continuous
on the
interval
, so is
. Since
is
differentiable on the interval
, so is
.
Furthermore
and
.
So by Rolle’s Theorem, there is a
number
,
, for which
. Since
, it follows
that
and
.
Theorem 1
: Suppose
f
is a differentiable function on an interval
I
for which
for
all
. Then
f
is a constant function, that is to say,
for some constant
C
.
Proof
: Let
be a fixed point in the interval
I
and let x denote any other point in the
interval
I
. For demonstration purposes, suppo
se
and note that the interval
is
contained in the interval
I
. Since f is differentiable on the interval
I
, it is continuous on
the interval
I
. Therefore
f
is continuous on the interval
and differentiable on the
interval
. So by the Mean Value Theorem for Derivatives there is a point
in
6
the interval
for which
. Since
,
and it
follows that
. So
. Since the point
x
is
arbitrary, f is a constant function.
Theorem 2
: Suppose
f
is a function defined on an interval
I
and suppose a
lso that
F
and
G
are both antiderivatives of
f
on the interval
I
. Then there is a constant
C
for
which
for all values of
x
in the interval
I
.
Proof
: Define
for
. Since
F
and
G
ar
e antiderivatives of
f
on the
interval
I
, the definition states that
and
for all
.
Then
for all
. By Theorem 1,
for
all
for some constant
C
. Therefore
or
. | 2,303 | 8,430 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2018-26 | latest | en | 0.902719 | # Antiderivatives. Ηλεκτρονική - Συσκευές. 10 Οκτ 2013 (πριν από 4 χρόνια και 9 μήνες). 153 εμφανίσεις. 1. Notes on. Antiderivatives. So far our studies have concentrated on given a function how do we find its. derivative? Now we ask the reverse question: Given a derivative how do we find the. function?. The following definition provides some clarity for this last. question.. Definition. : Let. f. denote a function on an interval. I. . A function. F. is said to be an. antiderivative of. f. on the interval. I. provided. for all values of. .. Since. for. , the function. is an antiderivative of the. function. on the interval. . There are other antiderivatives of. such. as. ,. and. . In fact, since. , where. C. is a constant, all. the functions. , where. C. is any constant, are antiderivatives of. .. So the are. infinitely many antiderivatives for the function. . Are there any others?. To answer this question the following theorems provide the basis. for the theory.. In addition these theorems have important consequences.. Rolle. ’s Theorem. :. Suppose. is a function which is continuous on the interval. ,. differentiable on the interval. and for which. . Then there is a number. ,. ,. for which. .. The Mean Value Theorem (for Derivatives). :. Suppose. is a function which is. continuous on the interval. and differentiable. on the interval. .. Then there is a. number. ,. , for which. .. Observations. :. The following are consequences of these theorems.. Rolle’s Theorem is a special. case of the Mean Value Theorem.. Rolle’s Theorem guarantees the existence of at least one critical point on the. interval. .. The geometrical interpretation of the Mean Value Theorem is that there is a. tangent line to the curve. which is parallel to the (secant) line joining the. points. and. .. When the function. is interpreted as the position of a particle moving along a. line, then the M. ean Value Theorem says that there is a point in time that the. instantaneous velocity is the same as the average velocity over the time. interval. .. 2. The following theorems provide the answer to the questions posed earlier. The proofs. are based on the Mean Value Theorem and are included. at the end of these notes. for. completeness.. Theorem 1. : Suppose. f. is a differentiable function on an interval. I. for which. for. all. . Then. f. is a consta. nt function, that is to say,. for some constant. C. .. Theorem 2. : Suppose. f. is a function defined on an interval. I. and suppose also that. F. and. G. are both antiderivatives of. f. on the interval. I. . Then there is a constant. C. for. which. for all values of. x. in the interval. I. .. the ques. tion originally posed, namely, d. oes the family. of. functions. ,. C. a constant, represent all the antiderivatives of the function. on the. interval. ? According to Theorem 2, the answer is “YES”!. Indeed the function. is. a known antiderivative of. . If. is any other antideriv. ative of. , Theorem 2. states. for some constant. .. Definition. : Suppose a function. on an interval. I. , the family of all antiderivatives of. on. the interval. I. is called the. indefinite integral of. and this family is denoted by the. symbol. .. Observation. : Notice that Theorem 2 states that if. is a function for. wh. ich. , then. , where C is a constant called the. constant of integration. . Conversely, if. and. are functions for. which. , where C is. a constant, then. .. Example 1. : Verify that the function. satisfies the hypotheses of. Rolle’s Theorem on the interval. . Then find all the numbers on that interval that. satisfies t. he conclusion of Rolle’s Theorem.. Solution. : The given function is a polynomial and therefore continuous everywhere and in. particular on the interval. . Since polynomials are differentiable everywhere, the. given function is differen. tiable on the interval. .. Furthermore. and. . So the given function satisfies the. hypotheses of Rolle’s Theorem.. Find the numbers. so. that. :. (. U. ). 3. .. There are two possibilities, namely. and. . The second. possibility. is negative and therefore not in the interval. .. However the first. possibility. and therefore in the interval. .. Example 2. :. Verify that the function. satisfies the hypotheses of the Mean. Value Theorem on the interval. . Then find all the numbers. that satisfy the. conclusion of the Mean Value Theorem.. Solution. :. The given function is a combination of. functions that are continuous. everywhere and therefore continuous on the interval. . The given function is also. differentiable on the interval. since. everywhere. So the given funct. ion. satisfies the hypotheses of the Mean Value Theorem.. Find. so that. :. . So this number lies in. the interval. .. Example. 3. :. Consider the function. . Show that there is no number. in the. interval. for which. . Why does this not. Value Theorem?. Solution. :. . Since this equation. which has no real solutions, there is no such number. . This does not contradict the Mean. Value Theorem because the given function has a discontinuity at. and is not. continuous on the interval. . (. It is not differentiable on. either. .. ). 4. Example. 4. :. | Show that the equation. has exactly one real root.. Solution. :. The function. is a combination of continuous functions and. therefore continuous everywhere. Furthermore. and. . So. by the Intermediate Value Theorem there is at least one real root on the interval. .. Suppose there are two real roots,. .. (. ) For all values. o. f. x. ,. . So the function. is continuous on the. interval. and differentiable on the interval. . Further. and. .. So by Rolle’s Theorem there must exist a number. ,. , for which. .. So. So there cannot be two real roots of. the equation.. Ex. ample. 5. :. Evaluate each of the following indefinite integrals.. a). c). b). d). Solution. :. a) Since. ,. the function. is an antiderivative of. .. Therefore. .. b). Since. , the function. is an antiderivative of. .. Therefore. .. c). Since. , the function. is an antiderivative of. .. Therefore. d). Since. , the function. is an anti. derivative of. .. Therefore. .. The proofs of the t. heorems are now presented. It is not important for you to. memorize these proofs, but. as. you can. see why. the theorems hold true. .. Rolle’s Theor. em. : Suppose. is a function which is continuous on the interval. ,. differentiable on the interval. and for which. . Then there is a number. ,. , for which. .. Proof. :. There are three cases to consider.. 5. Case 1: Suppose. is a constant function. Then. for all values of. x. in the. interval. . Select a number. . Then. .. Case 2: Suppose. for some value of. . Since. is continuous on the. interval. , the function. has a maximum value, call it. , on the interval. .. Then the number. c. is a critical point for the function. and, since. is differentiable on the. interval. ,. .. Case 3: Suppose. for some value of. . Since. is continuous on the. interval. , the function. has a. minimum. value, call it. , on the interval. .. Then the number. c. is a critical point for the function. and, since. is differentiable on the. interval. ,. .. The Mean Value Theorem (for Derivatives). : Suppose. is a function which is. continuous on the interval. and differentiable on the interval. .. Then there is a. number. ,. , for which. .. Proof. : Define the function. which is a. combination of the given function and a polynomial. Since. is continuous. on the. interval. , so is. . Since. is. differentiable on the interval. , so is. .. Furthermore. and. .. So by Rolle’s Theorem, there is a. number. ,. , for which. . Since. , it follows. that. and. .. Theorem 1. : Suppose. f. is a differentiable function on an interval. I. for which. for. all. . Then. f. is a constant function, that is to say,. for some constant. C. .. Proof. : Let. be a fixed point in the interval. I. and let x denote any other point in the. interval. I. . For demonstration purposes, suppo. se. and note that the interval. is. contained in the interval. I. . Since f is differentiable on the interval. I. , it is continuous on. the interval. I. . Therefore. f. is continuous on the interval. and differentiable on the. interval. . So by the Mean Value Theorem for Derivatives there is a point. in. 6. the interval. for which. . Since. ,. and it. follows that. . So. . Since the point. x. is. arbitrary, f is a constant function.. Theorem 2. : Suppose. f. is a function defined on an interval. I. and suppose a. lso that. F. and. G. are both antiderivatives of. f. on the interval. I. . Then there is a constant. C. for. which. for all values of. x. in the interval. I. .. Proof. : Define. for. . Since. F. and. G. ar. e antiderivatives of. f. on the. interval. I. , the definition states that. and. for all. .. Then. for all. . By Theorem 1,. for. all. for some constant. C. . Therefore. or. . |
https://corinnekerston.com/2023/04/29/how-many-days-until-june-13.html | 1,701,849,602,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100583.31/warc/CC-MAIN-20231206063543-20231206093543-00387.warc.gz | 217,144,376 | 8,203 | Intro text, can be displayed through an additional field
Countdown to June 13: How Many Days Until June 13?
June 13 is just around the corner, and you might be wondering how many days are left until this exciting date. Whether you're eagerly anticipating a special event, marking an important milestone, or simply curious about the passage of time, this article will provide you with all the information you need. Read on to discover the answer to the burning question: how many days until June 13?
Understanding Time and Dates
Before we dive into the countdown, let's take a moment to understand how time and dates work. Time is a fascinating concept that allows us to measure the duration between two events or moments. Dates, on the other hand, help us organize and track specific points in time.
Calculating the Countdown
To determine how many days are left until June 13, we must consider the current date. As today's date changes daily, the countdown will also fluctuate accordingly. Therefore, it's crucial to check the countdown regularly for the most accurate information.
The Countdown Begins
As of today, let's calculate how many days are left until June 13. Please note that the countdown may differ depending on when you're reading this article.
Step 1: Determine Today's Date
Before we can calculate the countdown, we need to establish today's date. Let's assume today is May 1.
Step 2: Calculate the Days Remaining
Next, we subtract today's date from June 13 to find the number of days until the desired date. In our case, the calculation would be as follows: June 13 - May 1 = 43 days.
Step 3: Include or Exclude Today
When counting down to a specific date, there's often confusion about whether to include or exclude the current day. In our case, since we want to know how many days are left until June 13, we should exclude today from the count. Therefore, we have 43 days left until June 13.
Q: How Many Weeks Are There Until June 13?
A: To determine the number of weeks until June 13, divide the remaining days by seven. In our case, 43 days divided by seven equals approximately 6.14 weeks. Therefore, there are around 6 weeks until June 13.
Q: What Day of the Week Is June 13?
A: To find out the day of the week for June 13, we can use an online calendar or refer to a physical one. In this scenario, let's assume June 13 falls on a Monday.
Q: Are There Any Significant Events on June 13?
A: While June 13 is not widely recognized as a globally significant date, it may hold personal importance for individuals celebrating birthdays, anniversaries, or other personal milestones. Additionally, historical events may have occurred on this day that could be of interest to history enthusiasts.
Conclusion
As we eagerly await June 13, the countdown continues to decrease. With 43 days remaining until this exciting date, there's still plenty of time to prepare, plan, and anticipate the events that await us. Remember, the countdown may vary depending on when you read this article, so be sure to check back regularly for the most accurate information. Enjoy the anticipation and make the most of each passing day as we get closer to June 13!
Related video of how many days until june 13
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Enter
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Update Info | 741 | 3,366 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2023-50 | latest | en | 0.959499 | Intro text, can be displayed through an additional field. Countdown to June 13: How Many Days Until June 13?. June 13 is just around the corner, and you might be wondering how many days are left until this exciting date. Whether you're eagerly anticipating a special event, marking an important milestone, or simply curious about the passage of time, this article will provide you with all the information you need. Read on to discover the answer to the burning question: how many days until June 13?. Understanding Time and Dates. Before we dive into the countdown, let's take a moment to understand how time and dates work. Time is a fascinating concept that allows us to measure the duration between two events or moments. Dates, on the other hand, help us organize and track specific points in time.. Calculating the Countdown. To determine how many days are left until June 13, we must consider the current date. As today's date changes daily, the countdown will also fluctuate accordingly. Therefore, it's crucial to check the countdown regularly for the most accurate information.. The Countdown Begins. As of today, let's calculate how many days are left until June 13. Please note that the countdown may differ depending on when you're reading this article.. Step 1: Determine Today's Date. Before we can calculate the countdown, we need to establish today's date. Let's assume today is May 1.. Step 2: Calculate the Days Remaining. Next, we subtract today's date from June 13 to find the number of days until the desired date. In our case, the calculation would be as follows: June 13 - May 1 = 43 days.. Step 3: Include or Exclude Today. When counting down to a specific date, there's often confusion about whether to include or exclude the current day. In our case, since we want to know how many days are left until June 13, we should exclude today from the count. | Therefore, we have 43 days left until June 13.. Q: How Many Weeks Are There Until June 13?. A: To determine the number of weeks until June 13, divide the remaining days by seven. In our case, 43 days divided by seven equals approximately 6.14 weeks. Therefore, there are around 6 weeks until June 13.. Q: What Day of the Week Is June 13?. A: To find out the day of the week for June 13, we can use an online calendar or refer to a physical one. In this scenario, let's assume June 13 falls on a Monday.. Q: Are There Any Significant Events on June 13?. A: While June 13 is not widely recognized as a globally significant date, it may hold personal importance for individuals celebrating birthdays, anniversaries, or other personal milestones. Additionally, historical events may have occurred on this day that could be of interest to history enthusiasts.. Conclusion. As we eagerly await June 13, the countdown continues to decrease. With 43 days remaining until this exciting date, there's still plenty of time to prepare, plan, and anticipate the events that await us. Remember, the countdown may vary depending on when you read this article, so be sure to check back regularly for the most accurate information. Enjoy the anticipation and make the most of each passing day as we get closer to June 13!. Related video of how many days until june 13. Ctrl. Enter. Noticed oshYwhat?. Highlight text and click Ctrl+Enter. We are in. Search and Discover » how many days until june 13. Update Info. |
https://www.majortests.com/gre/quantitative_comparison_test10 | 1,527,086,764,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865679.51/warc/CC-MAIN-20180523141759-20180523161759-00161.warc.gz | 800,172,680 | 12,823 | # GRE Math Quantitative Comparison Practice Test 10
1. ab = 0
a 0
A. The quantity on the left is greater
B. The quantity on the right is greater
C. Both are equal
D. The relationship cannot be determined without further information
2. -1 < x < 0
x³ x5
A. The quantity on the left is greater
B. The quantity on the right is greater
C. Both are equal
D. The relationship cannot be determined without further information
3. The average of four numbers is 30. The average of three of these numbers is 20.
The value of the fourth number 60
A. The quantity on the left is greater
B. The quantity on the right is greater
C. Both are equal
D. The relationship cannot be determined without further information
4. AC > CB > AB
b a
A. The quantity on the left is greater
B. The quantity on the right is greater
C. Both are equal
D. The relationship cannot be determined without further information
5. The line containing point A is rotated 90 degrees anticlockwise about the origin, O.
The y coordinate of point A before rotation. The y coordinate of point A after rotation.
A. The quantity on the left is greater
B. The quantity on the right is greater
C. Both are equal
D. The relationship cannot be determined without further information
6. Points x and y lie on line l
The number of points on line l that are twice as far from x as from y 2
A. The quantity on the left is greater
B. The quantity on the right is greater
C. Both are equal
D. The relationship cannot be determined without further information
7. O is the centre of the circle, and angle POQ is a right angle
MN/PQ 2/1
A. The quantity on the left is greater
B. The quantity on the right is greater
C. Both are equal
D. The relationship cannot be determined without further information
8.
Total savings on 40 gallons of fuel bought at \$1.152 per gallon instead of \$1.245 \$3.70
A. The quantity on the left is greater
B. The quantity on the right is greater
C. Both are equal
D. The relationship cannot be determined without further information
9. The cost of hiring props for the school play was to be shared by 8 students. 4 more students decide to share the cost. The new cost per person is \$2 less.
Total cost of hiring props \$50
A. The quantity on the left is greater
B. The quantity on the right is greater
C. Both are equal
D. The relationship cannot be determined without further information
10. A bag contains only red, white and blue balls. One third of the balls are red, one fifth of the balls are white. One ball is to be selected at random.
Probability of drawing a white ball Probability of drawing a blue ball
A. The quantity on the left is greater
B. The quantity on the right is greater
C. Both are equal
D. The relationship cannot be determined without further information
### Test information
10 questions
10 minutes
This is just one of 10 free GRE math quantitative comparison tests available on majortests.com. See the quantitative comparison page for directions, tips and more information.
* GRE is a registered trademark of Educational Testing Service (ETS). This website is not endorsed or approved by ETS.
All content of site and tests copyright © 2018 Study Mode, LLC. | 758 | 3,199 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-22 | longest | en | 0.882351 | # GRE Math Quantitative Comparison Practice Test 10. 1. ab = 0. a 0. A. The quantity on the left is greater. B. The quantity on the right is greater. C. Both are equal. D. The relationship cannot be determined without further information. 2. -1 < x < 0. x³ x5. A. The quantity on the left is greater. B. The quantity on the right is greater. C. Both are equal. D. The relationship cannot be determined without further information. 3. The average of four numbers is 30. The average of three of these numbers is 20.. The value of the fourth number 60. A. The quantity on the left is greater. B. The quantity on the right is greater. C. Both are equal. D. The relationship cannot be determined without further information. 4. AC > CB > AB. b a. A. The quantity on the left is greater. B. The quantity on the right is greater. C. Both are equal. D. The relationship cannot be determined without further information. 5. The line containing point A is rotated 90 degrees anticlockwise about the origin, O.. The y coordinate of point A before rotation. The y coordinate of point A after rotation.. A. The quantity on the left is greater. B. The quantity on the right is greater. C. Both are equal. D. The relationship cannot be determined without further information. 6. Points x and y lie on line l. The number of points on line l that are twice as far from x as from y 2. A. | The quantity on the left is greater. B. The quantity on the right is greater. C. Both are equal. D. The relationship cannot be determined without further information. 7. O is the centre of the circle, and angle POQ is a right angle. MN/PQ 2/1. A. The quantity on the left is greater. B. The quantity on the right is greater. C. Both are equal. D. The relationship cannot be determined without further information. 8.. Total savings on 40 gallons of fuel bought at \$1.152 per gallon instead of \$1.245 \$3.70. A. The quantity on the left is greater. B. The quantity on the right is greater. C. Both are equal. D. The relationship cannot be determined without further information. 9. The cost of hiring props for the school play was to be shared by 8 students. 4 more students decide to share the cost. The new cost per person is \$2 less.. Total cost of hiring props \$50. A. The quantity on the left is greater. B. The quantity on the right is greater. C. Both are equal. D. The relationship cannot be determined without further information. 10. A bag contains only red, white and blue balls. One third of the balls are red, one fifth of the balls are white. One ball is to be selected at random.. Probability of drawing a white ball Probability of drawing a blue ball. A. The quantity on the left is greater. B. The quantity on the right is greater. C. Both are equal. D. The relationship cannot be determined without further information. ### Test information. 10 questions. 10 minutes. This is just one of 10 free GRE math quantitative comparison tests available on majortests.com. See the quantitative comparison page for directions, tips and more information.. * GRE is a registered trademark of Educational Testing Service (ETS). This website is not endorsed or approved by ETS.. All content of site and tests copyright © 2018 Study Mode, LLC. |
https://convertoctopus.com/39-2-knots-to-miles-per-hour | 1,623,916,818,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487629632.54/warc/CC-MAIN-20210617072023-20210617102023-00465.warc.gz | 181,521,090 | 8,163 | ## Conversion formula
The conversion factor from knots to miles per hour is 1.1507794480225, which means that 1 knot is equal to 1.1507794480225 miles per hour:
1 kt = 1.1507794480225 mph
To convert 39.2 knots into miles per hour we have to multiply 39.2 by the conversion factor in order to get the velocity amount from knots to miles per hour. We can also form a simple proportion to calculate the result:
1 kt → 1.1507794480225 mph
39.2 kt → V(mph)
Solve the above proportion to obtain the velocity V in miles per hour:
V(mph) = 39.2 kt × 1.1507794480225 mph
V(mph) = 45.110554362484 mph
The final result is:
39.2 kt → 45.110554362484 mph
We conclude that 39.2 knots is equivalent to 45.110554362484 miles per hour:
39.2 knots = 45.110554362484 miles per hour
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 mile per hour is equal to 0.022167761272995 × 39.2 knots.
Another way is saying that 39.2 knots is equal to 1 ÷ 0.022167761272995 miles per hour.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that thirty-nine point two knots is approximately forty-five point one one one miles per hour:
39.2 kt ≅ 45.111 mph
An alternative is also that one mile per hour is approximately zero point zero two two times thirty-nine point two knots.
## Conversion table
### knots to miles per hour chart
For quick reference purposes, below is the conversion table you can use to convert from knots to miles per hour
knots (kt) miles per hour (mph)
40.2 knots 46.261 miles per hour
41.2 knots 47.412 miles per hour
42.2 knots 48.563 miles per hour
43.2 knots 49.714 miles per hour
44.2 knots 50.864 miles per hour
45.2 knots 52.015 miles per hour
46.2 knots 53.166 miles per hour
47.2 knots 54.317 miles per hour
48.2 knots 55.468 miles per hour
49.2 knots 56.618 miles per hour | 550 | 1,936 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2021-25 | latest | en | 0.775489 | ## Conversion formula. The conversion factor from knots to miles per hour is 1.1507794480225, which means that 1 knot is equal to 1.1507794480225 miles per hour:. 1 kt = 1.1507794480225 mph. To convert 39.2 knots into miles per hour we have to multiply 39.2 by the conversion factor in order to get the velocity amount from knots to miles per hour. We can also form a simple proportion to calculate the result:. 1 kt → 1.1507794480225 mph. 39.2 kt → V(mph). Solve the above proportion to obtain the velocity V in miles per hour:. V(mph) = 39.2 kt × 1.1507794480225 mph. V(mph) = 45.110554362484 mph. The final result is:. 39.2 kt → 45.110554362484 mph. We conclude that 39.2 knots is equivalent to 45.110554362484 miles per hour:. 39.2 knots = 45.110554362484 miles per hour. ## Alternative conversion. We can also convert by utilizing the inverse value of the conversion factor. In this case 1 mile per hour is equal to 0.022167761272995 × 39.2 knots.. Another way is saying that 39.2 knots is equal to 1 ÷ 0.022167761272995 miles per hour.. ## Approximate result. | For practical purposes we can round our final result to an approximate numerical value. We can say that thirty-nine point two knots is approximately forty-five point one one one miles per hour:. 39.2 kt ≅ 45.111 mph. An alternative is also that one mile per hour is approximately zero point zero two two times thirty-nine point two knots.. ## Conversion table. ### knots to miles per hour chart. For quick reference purposes, below is the conversion table you can use to convert from knots to miles per hour. knots (kt) miles per hour (mph). 40.2 knots 46.261 miles per hour. 41.2 knots 47.412 miles per hour. 42.2 knots 48.563 miles per hour. 43.2 knots 49.714 miles per hour. 44.2 knots 50.864 miles per hour. 45.2 knots 52.015 miles per hour. 46.2 knots 53.166 miles per hour. 47.2 knots 54.317 miles per hour. 48.2 knots 55.468 miles per hour. 49.2 knots 56.618 miles per hour. |
https://www.scribd.com/document/73099915/New-Finite-Element-Analysis-Lec1 | 1,511,477,966,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806979.99/warc/CC-MAIN-20171123214752-20171123234752-00308.warc.gz | 855,280,264 | 32,329 | # Finite Element Analysis (FEA) Finite Element Method (FEM
)
Compressor valve cover
Examples Coal processing plant pipes
How do you solve an Engineering Problem? • Engineers are faced with complex problems and use many methods in order to solve those problems including classic and numerical methods. • As shown below, the finite element methods is one of several methods for solving engineering problems.
ENGINEERING ANALYSIS
CLASSICAL METHODS 1. CLOSED-FORM
NUMERICAL METHOD 1. FINITE ELEMENT 2. FINITE DIFFERENCE 3. BOUNDARY ELEMENT
2. APPROXIMATE
Method of solving Engineering Problems
Historically, two classical approaches have been used: Classical Method • Closed form solutions are available for simple problems such as bending of beams and torsion of prismatic bars. (eg: Roark’s Formulas for Stress and Stress) • Approximate methods using series solutions to governing differential equations are used to analyze more complex structures such as plate and shells.
“If you can solve your problem using a closed form solution and a classical method, it is probably the best way to do it”
the another approach is to take a complex problem and break it into simple problems. • A good example is found in the long history of the efforts to calculate Pi (π) . • Those simple problem are solved and then assembled into a final solution.How do you solve a complex Engineering Problem? • When a problem becomes more complex than what can be solved with a closed form solution.
Example of FE approaches .• People spent centuries breaking the circle up into smaller triangles to find an accurate value of Pi (π).
Example of FE approaches .
Example of FE approaches .
loading and boundary conditions – Increasingly become the primary analysis tools for designers and analyst – Also known as matrix method of structural analysis in the literature because it uses matrix algebra to solve the system of simultaneous equations.Method of solving Engineering Problems Numerical Methods • FEM – Capable of solving large. complex problems with general geometry. – Real power is in its ability to solve problems that do not fit any standard formula. .
(Zeid. 2011) . 2004) • numerical analysis technique for obtaining approximate solutions to a wide variety of engineering problems.FEA Definition • computational techniques used to obtain approximate solution of boundary value problems in Engineering (Hutton.com. 2005) • one of several numerical methods that can be used to solve a complex problem by breaking it down into a finite number of simple problems (NeiSoftware.
Before FEA…? .
Fundamental Concept • Based on the idea of building a complicated object into small and manageable pieces. simpler pieces. . • The assembly of nodes and elements are called finite element model. • The smaller pieces are called elements which are connected by nodes. • It is a method of investigating the behavior of complex and structures by breaking them down into smaller.
.
. The three basic types of finite elements are beams (1D).Types of Element • Finite elements have shapes which are relatively easy to formulate and analyze. shells (2D) and solids (3D).
Other examples of 1D elements .
Other examples of 2D elements .
Other examples of 3D elements .
The following steps show in general how the FEM works. • 1. Create the Finite Element • 2. step-by-step process.How does FEM Work? The solution of a problem domain by FEM usually follows an orderly. Generate Global Stiffness Matrix equation • 4. Apply Boundary Conditions • 5. Develop elemental matrices and equation • 3. Solve for the unknown at the nodes .
How does FEM Work? (an example of 2D element) 1. • Element are connected to each other by nodes 4 nodes 1 element . Create the Finite Element • A given problem is discretized by dividing the original domain into simply shaped element.
1D elements v 2D elements 3D elements .
2. Develop elemental matrices and equation • The relationship between an element and its surrounding nodes can be described by the following equation 1 2 • *Each of the element will have their own set of equations. .
{u}.What is [K]. {F} ? Problem Type .
Generate Global Stiffness Matrix equation • The individual Elemental Stiffness Matrices are assembled together into the Global Stiffness Matrix by summing the equilibrium equations of the elements. • This result in the following Global System Matrix Equation for the overall structure: .3.
.
4. Apply Boundary Conditions • Next. Mathematically this is achieved by removing rows and columns corresponding to the constrained degrees of freedom (DOF) from the Global System Matrix Equation ux = 0 uy = 0 uz = 0 F1 = ? F2 = ? F3 = ? . the BC’s (Loads and Constraint) is apply to the model.
the Global System Matrix Equation is solve to determine the unknown nodal values. Solve for the unknown at the nodes • Finally.5. [K]{u} ={F} {u} = [K]-1{F} • *This requires knowledge in Matrix Algebra. .
fluid flow) .Advantages of FEA • General enough to handle large class of engineering problems (stress analysis. electromagnetism. heat transfer.
Force = 100lbs) .Mistakes by user • Elements are the wrong type (eg: Shell 2D elements are used where solid elements are needed. • Distorted elements • Inconsistent units (eg: E = 200GPa.
Best Practices .
. The more refine the mesh (grid).” The FE solutions are often approximate.. • “Any results obtained is the correct one. loads and boundary conditions..Common misconception • “The FE solutions is the most accurate. the more accurate. There is no guarantee that the results are accurate.” The FE solution may contain “fatal” errors as a result of incorrect modeling of structures. Even a nice picture can give the wrong result… • “FEA replaces testing…” Depends on the confidence in the analytical methods used.
Typical FE Procedure by commercial software user Preprocessing Build FE Model computer Process Conduct numerical analysis user Postprocessing See and interpret the results .
• Import / Create Geometry • Define Element Types (1D. v – Poisson’s Ratio etc) Pre • Define geometric properties Processing • Apply Boundary Conditions (Constraint) • Apply Loads • Solve for displacements • Compute Strains • Compute Stresses FEA General Procedure Solution • • • PostProcessing • • Sort element stresses in order of magnitude Check equilibrium Plot deformed structural shape Animate dynamic model behavior Reports . 3D) • Define Material Properties (E – Young’s Modulus. 2D.
• The computed values then used by back substitution to compute additional derived variables (i.Pre Processor • Also know as model definition • This step is critical – a perfectly computed FE solution is of absolutely no value if corresponds to wrong problems. garbage out” Solution • FE software assembles the algebraic equations in matrix form and computes the unknown values. • “garbage in. the most important objectives is to apply sound engineering judgment in determining whether the solution are reasonable or not . • While solution data can be manipulated many ways.e reaction forces. element stresses etc) Postprocessor • Analysis and evaluation of the solution results.
REVIEW OF MATRIX ALGEBRA . | 1,492 | 7,186 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2017-47 | latest | en | 0.925149 | # Finite Element Analysis (FEA) Finite Element Method (FEM. ). Compressor valve cover. Examples Coal processing plant pipes. How do you solve an Engineering Problem? • Engineers are faced with complex problems and use many methods in order to solve those problems including classic and numerical methods. • As shown below, the finite element methods is one of several methods for solving engineering problems.. ENGINEERING ANALYSIS. CLASSICAL METHODS 1. CLOSED-FORM. NUMERICAL METHOD 1. FINITE ELEMENT 2. FINITE DIFFERENCE 3. BOUNDARY ELEMENT. 2. APPROXIMATE. Method of solving Engineering Problems. Historically, two classical approaches have been used: Classical Method • Closed form solutions are available for simple problems such as bending of beams and torsion of prismatic bars. (eg: Roark’s Formulas for Stress and Stress) • Approximate methods using series solutions to governing differential equations are used to analyze more complex structures such as plate and shells.. “If you can solve your problem using a closed form solution and a classical method, it is probably the best way to do it”. the another approach is to take a complex problem and break it into simple problems. • A good example is found in the long history of the efforts to calculate Pi (π) . • Those simple problem are solved and then assembled into a final solution.How do you solve a complex Engineering Problem? • When a problem becomes more complex than what can be solved with a closed form solution.. Example of FE approaches .• People spent centuries breaking the circle up into smaller triangles to find an accurate value of Pi (π).. Example of FE approaches .. Example of FE approaches .. loading and boundary conditions – Increasingly become the primary analysis tools for designers and analyst – Also known as matrix method of structural analysis in the literature because it uses matrix algebra to solve the system of simultaneous equations.Method of solving Engineering Problems Numerical Methods • FEM – Capable of solving large. complex problems with general geometry. – Real power is in its ability to solve problems that do not fit any standard formula. .. (Zeid. 2011) . 2004) • numerical analysis technique for obtaining approximate solutions to a wide variety of engineering problems.FEA Definition • computational techniques used to obtain approximate solution of boundary value problems in Engineering (Hutton.com. 2005) • one of several numerical methods that can be used to solve a complex problem by breaking it down into a finite number of simple problems (NeiSoftware.. Before FEA…? .. Fundamental Concept • Based on the idea of building a complicated object into small and manageable pieces. simpler pieces. . • The assembly of nodes and elements are called finite element model. • The smaller pieces are called elements which are connected by nodes. • It is a method of investigating the behavior of complex and structures by breaking them down into smaller.. .. . The three basic types of finite elements are beams (1D).Types of Element • Finite elements have shapes which are relatively easy to formulate and analyze. shells (2D) and solids (3D).. Other examples of 1D elements .. Other examples of 2D elements .. Other examples of 3D elements .. The following steps show in general how the FEM works. • 1. | Create the Finite Element • 2. step-by-step process.How does FEM Work? The solution of a problem domain by FEM usually follows an orderly. Generate Global Stiffness Matrix equation • 4. Apply Boundary Conditions • 5. Develop elemental matrices and equation • 3. Solve for the unknown at the nodes .. How does FEM Work? (an example of 2D element) 1. • Element are connected to each other by nodes 4 nodes 1 element . Create the Finite Element • A given problem is discretized by dividing the original domain into simply shaped element.. 1D elements v 2D elements 3D elements .. 2. Develop elemental matrices and equation • The relationship between an element and its surrounding nodes can be described by the following equation 1 2 • *Each of the element will have their own set of equations. .. {u}.What is [K]. {F} ? Problem Type .. Generate Global Stiffness Matrix equation • The individual Elemental Stiffness Matrices are assembled together into the Global Stiffness Matrix by summing the equilibrium equations of the elements. • This result in the following Global System Matrix Equation for the overall structure: .3.. .. 4. Apply Boundary Conditions • Next. Mathematically this is achieved by removing rows and columns corresponding to the constrained degrees of freedom (DOF) from the Global System Matrix Equation ux = 0 uy = 0 uz = 0 F1 = ? F2 = ? F3 = ? . the BC’s (Loads and Constraint) is apply to the model.. the Global System Matrix Equation is solve to determine the unknown nodal values. Solve for the unknown at the nodes • Finally.5. [K]{u} ={F} {u} = [K]-1{F} • *This requires knowledge in Matrix Algebra. .. fluid flow) .Advantages of FEA • General enough to handle large class of engineering problems (stress analysis. electromagnetism. heat transfer.. Force = 100lbs) .Mistakes by user • Elements are the wrong type (eg: Shell 2D elements are used where solid elements are needed. • Distorted elements • Inconsistent units (eg: E = 200GPa.. Best Practices .. . The more refine the mesh (grid).” The FE solutions are often approximate.. • “Any results obtained is the correct one. loads and boundary conditions..Common misconception • “The FE solutions is the most accurate. the more accurate. There is no guarantee that the results are accurate.” The FE solution may contain “fatal” errors as a result of incorrect modeling of structures. Even a nice picture can give the wrong result… • “FEA replaces testing…” Depends on the confidence in the analytical methods used.. Typical FE Procedure by commercial software user Preprocessing Build FE Model computer Process Conduct numerical analysis user Postprocessing See and interpret the results .. • Import / Create Geometry • Define Element Types (1D. v – Poisson’s Ratio etc) Pre • Define geometric properties Processing • Apply Boundary Conditions (Constraint) • Apply Loads • Solve for displacements • Compute Strains • Compute Stresses FEA General Procedure Solution • • • PostProcessing • • Sort element stresses in order of magnitude Check equilibrium Plot deformed structural shape Animate dynamic model behavior Reports . 3D) • Define Material Properties (E – Young’s Modulus. 2D.. • The computed values then used by back substitution to compute additional derived variables (i.Pre Processor • Also know as model definition • This step is critical – a perfectly computed FE solution is of absolutely no value if corresponds to wrong problems. garbage out” Solution • FE software assembles the algebraic equations in matrix form and computes the unknown values. • “garbage in. the most important objectives is to apply sound engineering judgment in determining whether the solution are reasonable or not . • While solution data can be manipulated many ways.e reaction forces. element stresses etc) Postprocessor • Analysis and evaluation of the solution results.. REVIEW OF MATRIX ALGEBRA . |
http://www.ck12.org/arithmetic/Percent-of-Change/lesson/Find-the-Percent-of-Change/ | 1,493,378,374,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122933.39/warc/CC-MAIN-20170423031202-00457-ip-10-145-167-34.ec2.internal.warc.gz | 500,974,576 | 31,051 | # Percent of Change
## %change=[(final amount-original amount)/original amount] x 100%
%
Progress
MEMORY METER
This indicates how strong in your memory this concept is
Progress
%
Find the Percent of Change
Have you ever gone to a gym to exercise? Many people do each day, but sometimes the rates change. Take a look at this dilemma.
A gym’s membership went from 2100 members one year to 2410 members the next. This is a difference of 310 members. What was the percent of change?
To figure this out, you will need to know how to calculate a percent of change. Pay close attention and you will learn how to do this by the end of the Concept.
### Guidance
We can find the percent of change if we know an original amount and how much it either increased or decreased. At times, however, we are given the percent of increase or decrease and need to calculate a new amount.
Let’s look at how we can calculate this new amount by using the percent of change.
A restaurant manager has noticed an increase in the cost of utilities of 4%. In order to pay for the increased costs, he decides to increase prices by 4% as well. Not all items are priced the same. The chicken platter currently costs $5.99 and the steak platter cost$7.99. If the prices are increased by 4%, what will the new prices be?
Notice that we are going to create two new amounts. We are going to create a new cost for the chicken platter, and we are going to create a new cost for the steak platter. We have original amounts and the percent of the increase, so now we need to calculate a new amount.
First we must calculate how much the change will be. The prices are increasing by 4% so we must know how much 4% is of each price.
Now we know how much each platter’s price is going to change? Since this is an increase, we will add the price change to the price. If it were a decrease, we would subtract the decrease from the price.
Let’s summarize. In order to find the new amount, we calculate the change amount by multiplying the original amount by the percent of change. We then add the change amount to the original amount for an increase or we subtract the change amount from the original amount for a decrease.
Take a look at this situation.
Find the new amount if 60 is decreased by 27%.
amount of change:
subtract the amount of change from the original amount:
Find each percent of change.
#### Example A
Find the new amount if 10 is increased by 18%.
Solution:
#### Example B
Find the new amount if 16 is decreased by 20%.
Solution:
#### Example C
Find the new amount if 250 is increase by 30%.
Solution:
Now let's go back to the dilemma from the beginning of the Concept.
This percent is increasing, so we want to find the percent of the increase. We know the difference so now we can divide and multiply.
The gyms membership increased by 14.8%.
### Guided Practice
Here is one for you to try on your own.
The number of students participating in the chess club increased in one year. It started off with 35 students and had an increase of 55%. Figure out the new number of students in the chess club given this increase.
Solution
First, figure out the amount of the increase.
There was an increase of 19 students.
We can add that increase to the original amount.
The new number of students participating is students.
### Explore More
Directions: Use percent to find the new amount. You may round to the nearest whole number when necessary.
1. 82 increased by 90%
2. 64 decreased by 10%
3. 9 increased by 55%
4. 25,470 decreased by 77%
5. 75 increased by 10%
6. 33 decreased by 5%
7. 99 increased by 15%
8. 40 decreased by 8%
9. 56 increased by 25%
10. 900 decreased by 30%
11. 800 increased by 23%
12. 789 increased by 12%
13. 880 decreased by 10%
14. 450 increased by 45%
15. 855 decreased by 18%
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Percent Percent means out of 100. It is a quantity written with a % sign.
Percent Equation The percent equation can be stated as: "Rate times Total equals Part," or "R% of Total is Part." | 976 | 4,133 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2017-17 | latest | en | 0.95835 | # Percent of Change. ## %change=[(final amount-original amount)/original amount] x 100%. %. Progress. MEMORY METER. This indicates how strong in your memory this concept is. Progress. %. Find the Percent of Change. Have you ever gone to a gym to exercise? Many people do each day, but sometimes the rates change. Take a look at this dilemma.. A gym’s membership went from 2100 members one year to 2410 members the next. This is a difference of 310 members. What was the percent of change?. To figure this out, you will need to know how to calculate a percent of change. Pay close attention and you will learn how to do this by the end of the Concept.. ### Guidance. We can find the percent of change if we know an original amount and how much it either increased or decreased. At times, however, we are given the percent of increase or decrease and need to calculate a new amount.. Let’s look at how we can calculate this new amount by using the percent of change.. A restaurant manager has noticed an increase in the cost of utilities of 4%. In order to pay for the increased costs, he decides to increase prices by 4% as well. Not all items are priced the same. The chicken platter currently costs $5.99 and the steak platter cost$7.99. If the prices are increased by 4%, what will the new prices be?. Notice that we are going to create two new amounts. We are going to create a new cost for the chicken platter, and we are going to create a new cost for the steak platter. We have original amounts and the percent of the increase, so now we need to calculate a new amount.. First we must calculate how much the change will be. The prices are increasing by 4% so we must know how much 4% is of each price.. Now we know how much each platter’s price is going to change? Since this is an increase, we will add the price change to the price. If it were a decrease, we would subtract the decrease from the price.. Let’s summarize. In order to find the new amount, we calculate the change amount by multiplying the original amount by the percent of change. We then add the change amount to the original amount for an increase or we subtract the change amount from the original amount for a decrease.. Take a look at this situation.. Find the new amount if 60 is decreased by 27%.. amount of change:. subtract the amount of change from the original amount:. Find each percent of change.. #### Example A. Find the new amount if 10 is increased by 18%.. Solution:. #### Example B. Find the new amount if 16 is decreased by 20%.. Solution:. #### Example C. Find the new amount if 250 is increase by 30%.. Solution:. Now let's go back to the dilemma from the beginning of the Concept.. This percent is increasing, so we want to find the percent of the increase. We know the difference so now we can divide and multiply. | The gyms membership increased by 14.8%.. ### Guided Practice. Here is one for you to try on your own.. The number of students participating in the chess club increased in one year. It started off with 35 students and had an increase of 55%. Figure out the new number of students in the chess club given this increase.. Solution. First, figure out the amount of the increase.. There was an increase of 19 students.. We can add that increase to the original amount.. The new number of students participating is students.. ### Explore More. Directions: Use percent to find the new amount. You may round to the nearest whole number when necessary.. 1. 82 increased by 90%. 2. 64 decreased by 10%. 3. 9 increased by 55%. 4. 25,470 decreased by 77%. 5. 75 increased by 10%. 6. 33 decreased by 5%. 7. 99 increased by 15%. 8. 40 decreased by 8%. 9. 56 increased by 25%. 10. 900 decreased by 30%. 11. 800 increased by 23%. 12. 789 increased by 12%. 13. 880 decreased by 10%. 14. 450 increased by 45%. 15. 855 decreased by 18%. ### Notes/Highlights Having trouble? Report an issue.. Color Highlighted Text Notes. ### Vocabulary Language: English. TermDefinition. Percent Percent means out of 100. It is a quantity written with a % sign.. Percent Equation The percent equation can be stated as: "Rate times Total equals Part," or "R% of Total is Part.". |
https://www.12000.org/my_notes/CAS_integration_tests/reports/rubi_4_16_1_graded/test_cases/1_Algebraic_functions/1.2_Trinomial_products/1.2.1_Quadratic/1.2.1.2-d+e_x-%5Em-a+b_x+c_x%5E2-%5Ep/rese183.htm | 1,695,767,989,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510225.44/warc/CC-MAIN-20230926211344-20230927001344-00404.warc.gz | 673,371,764 | 4,538 | ### 3.183 $$\int \frac{x^2}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx$$
Optimal. Leaf size=106 $-\frac{a x (a+b x)}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{x^2 (a+b x)}{2 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a^2 (a+b x) \log (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}$
[Out]
-((a*x*(a + b*x))/(b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) + (x^2*(a + b*x))/(2*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) +
(a^2*(a + b*x)*Log[a + b*x])/(b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])
________________________________________________________________________________________
Rubi [A] time = 0.0371791, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.083, Rules used = {646, 43} $-\frac{a x (a+b x)}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{x^2 (a+b x)}{2 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a^2 (a+b x) \log (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}$
Antiderivative was successfully verified.
[In]
Int[x^2/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]
[Out]
-((a*x*(a + b*x))/(b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) + (x^2*(a + b*x))/(2*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) +
(a^2*(a + b*x)*Log[a + b*x])/(b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])
Rule 646
Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[p] && NeQ[2*c*d - b*e, 0]
Rule 43
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Rubi steps
\begin{align*} \int \frac{x^2}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx &=\frac{\left (a b+b^2 x\right ) \int \frac{x^2}{a b+b^2 x} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{\left (a b+b^2 x\right ) \int \left (-\frac{a}{b^3}+\frac{x}{b^2}+\frac{a^2}{b^3 (a+b x)}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{a x (a+b x)}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{x^2 (a+b x)}{2 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a^2 (a+b x) \log (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}
Mathematica [A] time = 0.0149632, size = 45, normalized size = 0.42 $\frac{(a+b x) \left (2 a^2 \log (a+b x)+b x (b x-2 a)\right )}{2 b^3 \sqrt{(a+b x)^2}}$
Antiderivative was successfully verified.
[In]
Integrate[x^2/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]
[Out]
((a + b*x)*(b*x*(-2*a + b*x) + 2*a^2*Log[a + b*x]))/(2*b^3*Sqrt[(a + b*x)^2])
________________________________________________________________________________________
Maple [A] time = 0.222, size = 44, normalized size = 0.4 \begin{align*}{\frac{ \left ( bx+a \right ) \left ({b}^{2}{x}^{2}+2\,{a}^{2}\ln \left ( bx+a \right ) -2\,abx \right ) }{2\,{b}^{3}}{\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^2/((b*x+a)^2)^(1/2),x)
[Out]
1/2*(b*x+a)*(b^2*x^2+2*a^2*ln(b*x+a)-2*a*b*x)/((b*x+a)^2)^(1/2)/b^3
________________________________________________________________________________________
Maxima [A] time = 1.25525, size = 55, normalized size = 0.52 \begin{align*} \frac{a^{2} b^{2} \log \left (x + \frac{a}{b}\right )}{{\left (b^{2}\right )}^{\frac{5}{2}}} - \frac{a b x}{{\left (b^{2}\right )}^{\frac{3}{2}}} + \frac{x^{2}}{2 \, \sqrt{b^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2/((b*x+a)^2)^(1/2),x, algorithm="maxima")
[Out]
a^2*b^2*log(x + a/b)/(b^2)^(5/2) - a*b*x/(b^2)^(3/2) + 1/2*x^2/sqrt(b^2)
________________________________________________________________________________________
Fricas [A] time = 1.65108, size = 68, normalized size = 0.64 \begin{align*} \frac{b^{2} x^{2} - 2 \, a b x + 2 \, a^{2} \log \left (b x + a\right )}{2 \, b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2/((b*x+a)^2)^(1/2),x, algorithm="fricas")
[Out]
1/2*(b^2*x^2 - 2*a*b*x + 2*a^2*log(b*x + a))/b^3
________________________________________________________________________________________
Sympy [A] time = 1.08682, size = 26, normalized size = 0.25 \begin{align*} \frac{a^{2} \log{\left (a + b x \right )}}{b^{3}} - \frac{a x}{b^{2}} + \frac{x^{2}}{2 b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**2/((b*x+a)**2)**(1/2),x)
[Out]
a**2*log(a + b*x)/b**3 - a*x/b**2 + x**2/(2*b)
________________________________________________________________________________________
Giac [A] time = 1.26573, size = 65, normalized size = 0.61 \begin{align*} \frac{a^{2} \log \left ({\left | b x + a \right |}\right ) \mathrm{sgn}\left (b x + a\right )}{b^{3}} + \frac{b x^{2} \mathrm{sgn}\left (b x + a\right ) - 2 \, a x \mathrm{sgn}\left (b x + a\right )}{2 \, b^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2/((b*x+a)^2)^(1/2),x, algorithm="giac")
[Out]
a^2*log(abs(b*x + a))*sgn(b*x + a)/b^3 + 1/2*(b*x^2*sgn(b*x + a) - 2*a*x*sgn(b*x + a))/b^2 | 2,516 | 5,328 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2023-40 | latest | en | 0.159644 | ### 3.183 $$\int \frac{x^2}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx$$. Optimal. Leaf size=106 $-\frac{a x (a+b x)}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{x^2 (a+b x)}{2 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a^2 (a+b x) \log (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}$. [Out]. -((a*x*(a + b*x))/(b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) + (x^2*(a + b*x))/(2*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) +. (a^2*(a + b*x)*Log[a + b*x])/(b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]). ________________________________________________________________________________________. Rubi [A] time = 0.0371791, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.083, Rules used = {646, 43} $-\frac{a x (a+b x)}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{x^2 (a+b x)}{2 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a^2 (a+b x) \log (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}$. Antiderivative was successfully verified.. [In]. Int[x^2/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]. [Out]. -((a*x*(a + b*x))/(b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) + (x^2*(a + b*x))/(2*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) +. (a^2*(a + b*x)*Log[a + b*x])/(b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]). Rule 646. Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra. cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,. c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[p] && NeQ[2*c*d - b*e, 0]. Rule 43. Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d. *x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]. && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0]). Rubi steps. \begin{align*} \int \frac{x^2}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx &=\frac{\left (a b+b^2 x\right ) \int \frac{x^2}{a b+b^2 x} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{\left (a b+b^2 x\right ) \int \left (-\frac{a}{b^3}+\frac{x}{b^2}+\frac{a^2}{b^3 (a+b x)}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{a x (a+b x)}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{x^2 (a+b x)}{2 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a^2 (a+b x) \log (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}. Mathematica [A] time = 0.0149632, size = 45, normalized size = 0.42 $\frac{(a+b x) \left (2 a^2 \log (a+b x)+b x (b x-2 a)\right )}{2 b^3 \sqrt{(a+b x)^2}}$. Antiderivative was successfully verified.. [In]. Integrate[x^2/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]. [Out]. ((a + b*x)*(b*x*(-2*a + b*x) + 2*a^2*Log[a + b*x]))/(2*b^3*Sqrt[(a + b*x)^2]). ________________________________________________________________________________________. Maple [A] time = 0.222, size = 44, normalized size = 0.4 \begin{align*}{\frac{ \left ( bx+a \right ) \left ({b}^{2}{x}^{2}+2\,{a}^{2}\ln \left ( bx+a \right ) -2\,abx \right ) }{2\,{b}^{3}}{\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}} \end{align*}. Verification of antiderivative is not currently implemented for this CAS. | [In]. int(x^2/((b*x+a)^2)^(1/2),x). [Out]. 1/2*(b*x+a)*(b^2*x^2+2*a^2*ln(b*x+a)-2*a*b*x)/((b*x+a)^2)^(1/2)/b^3. ________________________________________________________________________________________. Maxima [A] time = 1.25525, size = 55, normalized size = 0.52 \begin{align*} \frac{a^{2} b^{2} \log \left (x + \frac{a}{b}\right )}{{\left (b^{2}\right )}^{\frac{5}{2}}} - \frac{a b x}{{\left (b^{2}\right )}^{\frac{3}{2}}} + \frac{x^{2}}{2 \, \sqrt{b^{2}}} \end{align*}. Verification of antiderivative is not currently implemented for this CAS.. [In]. integrate(x^2/((b*x+a)^2)^(1/2),x, algorithm="maxima"). [Out]. a^2*b^2*log(x + a/b)/(b^2)^(5/2) - a*b*x/(b^2)^(3/2) + 1/2*x^2/sqrt(b^2). ________________________________________________________________________________________. Fricas [A] time = 1.65108, size = 68, normalized size = 0.64 \begin{align*} \frac{b^{2} x^{2} - 2 \, a b x + 2 \, a^{2} \log \left (b x + a\right )}{2 \, b^{3}} \end{align*}. Verification of antiderivative is not currently implemented for this CAS.. [In]. integrate(x^2/((b*x+a)^2)^(1/2),x, algorithm="fricas"). [Out]. 1/2*(b^2*x^2 - 2*a*b*x + 2*a^2*log(b*x + a))/b^3. ________________________________________________________________________________________. Sympy [A] time = 1.08682, size = 26, normalized size = 0.25 \begin{align*} \frac{a^{2} \log{\left (a + b x \right )}}{b^{3}} - \frac{a x}{b^{2}} + \frac{x^{2}}{2 b} \end{align*}. Verification of antiderivative is not currently implemented for this CAS.. [In]. integrate(x**2/((b*x+a)**2)**(1/2),x). [Out]. a**2*log(a + b*x)/b**3 - a*x/b**2 + x**2/(2*b). ________________________________________________________________________________________. Giac [A] time = 1.26573, size = 65, normalized size = 0.61 \begin{align*} \frac{a^{2} \log \left ({\left | b x + a \right |}\right ) \mathrm{sgn}\left (b x + a\right )}{b^{3}} + \frac{b x^{2} \mathrm{sgn}\left (b x + a\right ) - 2 \, a x \mathrm{sgn}\left (b x + a\right )}{2 \, b^{2}} \end{align*}. Verification of antiderivative is not currently implemented for this CAS.. [In]. integrate(x^2/((b*x+a)^2)^(1/2),x, algorithm="giac"). [Out]. a^2*log(abs(b*x + a))*sgn(b*x + a)/b^3 + 1/2*(b*x^2*sgn(b*x + a) - 2*a*x*sgn(b*x + a))/b^2. |
https://www.sofolympiadtrainer.com/Synopsis/imo8-factorization.jsp | 1,722,851,482,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640436802.18/warc/CC-MAIN-20240805083255-20240805113255-00230.warc.gz | 802,102,978 | 19,828 | ## FACTORIZATION- ESSENTIAL POINTS
• Factorization is expressing any algebraic equation as product of its factors.
• These factors can be numbers, variables or algebraic expressions.
• An irreducible factor is a factor which cannot be expressed further as a product of factors.
• Common factor method of factorization:
1. Write each term of the expression as a product of irreducible factors.
2. Separate the common factor terms.
3. Combine the remaining factors in each term in accordance with the distributive law.
• Terms must be grouped in a way that each group of terms have common factors. This is the method of regrouping.
• We need to observe expression and identify the desired grouping by trail and error.
• Common identities for factorization:
1. (a + b)2 = a2 + 2ab + b2
2. (a – b)2 = a2 – 2ab + b2
3. (a + b) (a – b) = a2 - b2
4. (y + a) (y + b) = y2 + (a + b)y + ab
• In expressions which have factors of the type (y + a) (y + b), remember the numerical term gives ab. Factors, a and b, should be so chosen that their sum, with signs taken care of, is the coefficient of y.
• Division of numbers is inverse of its multiplication.
• In the case of division of a polynomial by a monomial, we may carry out the division either by dividing each term of the polynomial by the monomial or by the common factor method.
• In the case of division of a polynomial by a polynomial, we cannot proceed by dividing each term in the dividend polynomial by the divisor polynomial. Instead, we factorise both the polynomials and cancel their common factors.
• In the case of divisions of algebraic expressions that we studied in this chapter, we have:
Dividend = Divisor × Quotient.
In general, however, the relation is:
Dividend = Divisor × Quotient + Remainder
Thus, we have considered in the present chapter only those divisions in which the remainder is zero.
I am in | 456 | 1,874 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2024-33 | latest | en | 0.932916 | ## FACTORIZATION- ESSENTIAL POINTS. • Factorization is expressing any algebraic equation as product of its factors.. • These factors can be numbers, variables or algebraic expressions.. • An irreducible factor is a factor which cannot be expressed further as a product of factors.. • Common factor method of factorization:. 1. Write each term of the expression as a product of irreducible factors.. 2. Separate the common factor terms.. 3. Combine the remaining factors in each term in accordance with the distributive law.. • Terms must be grouped in a way that each group of terms have common factors. This is the method of regrouping.. • We need to observe expression and identify the desired grouping by trail and error.. • Common identities for factorization:. 1. (a + b)2 = a2 + 2ab + b2. 2. | (a – b)2 = a2 – 2ab + b2. 3. (a + b) (a – b) = a2 - b2. 4. (y + a) (y + b) = y2 + (a + b)y + ab. • In expressions which have factors of the type (y + a) (y + b), remember the numerical term gives ab. Factors, a and b, should be so chosen that their sum, with signs taken care of, is the coefficient of y.. • Division of numbers is inverse of its multiplication.. • In the case of division of a polynomial by a monomial, we may carry out the division either by dividing each term of the polynomial by the monomial or by the common factor method.. • In the case of division of a polynomial by a polynomial, we cannot proceed by dividing each term in the dividend polynomial by the divisor polynomial. Instead, we factorise both the polynomials and cancel their common factors.. • In the case of divisions of algebraic expressions that we studied in this chapter, we have:. Dividend = Divisor × Quotient.. In general, however, the relation is:. Dividend = Divisor × Quotient + Remainder. Thus, we have considered in the present chapter only those divisions in which the remainder is zero.. I am in. |
https://community.khronos.org/t/the-most-basic-quaternion-question/60723 | 1,600,899,364,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400212959.12/warc/CC-MAIN-20200923211300-20200924001300-00168.warc.gz | 342,004,498 | 4,726 | # The most basic quaternion question...
Hi,
I’m having difficulty implementing quaternions in Qt using their QQuaternion class, and haven’t heard anything back yet from the community boards where I’ve posted my query. This is my first time trying to use quaternions so I want to see if the issue is just that I don’t understand how they’re supposed to work. I figured you guys would know.
All I want to do is to rotate a point in 3-space a given angular distance about an axis. So let
vector_pre = (1,0,0);
I.e. the unit vector along the x-axis. Now let’s say I want to rotate it 90deg counterclockwise in the x-y plane. To do this, I create the following quaternion:
quat = (90,0,0,1);
The vector component is the unit vector along the z axis, normal to the x-y plane. Then let
vector_post = (quat * vector_pre) * conjugate(quat);
vector_pre gets reinterpreted as a quaternion with scalar part 0, multiplication and conjugacy are as described in Wikipedia. Then my expectation is that
vector_post == (0,1,0);
This is emphatically not what I’m getting with the corresponding Qt method (rotatedVector) so I want to see if I am just not understanding things at a basic level. (And it’s not a radians vs. degrees issue.) Have I misunderstood the basics? Thank you–
Matt
I’m definitely no expert on Quaternions and i don’t know how Qt implements them, but here’s what i think you are doing wrong:
Quaternions have 4 components. These 4 components do NOT correspond to “angle/axis” (e.g. angle/x/y/z). The maths behind Quaternions is “a bit” more complicated.
So when you create a quaternion by setting the 4 components manually, you need to pass quite different values to actually get a Quaternion that represents a rotation around an axis. On a related note, most maths libraries don’t work with angles in degrees, but usually in radians, so passing “90” is most likely wrong, as well.
Most Quaternion libs have a function “CreateFromAxisAngle” or something like that. There you can use your intuitive representation of axis/angle and get the (most unintuitive) Quaternion back.
Also most libraries overload the multiplication between quaternion and vector (operator* (Quat, Vec)) and already implement the whole “rotate a vector” stuff in there. So usually you should not need to multiply your vector by the Quaternions conjugate, at all.
Now this is all without knowledge about Qt’s Quaternion class and therefore could be entirely wrong. But those are the most likely errors that i can come up with.
Hope that helps,
Jan.
Thank you so much, omitting the fromAxisAngle invocation was exactly the problem.
Best,
Matt | 602 | 2,635 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2020-40 | latest | en | 0.906731 | # The most basic quaternion question.... Hi,. I’m having difficulty implementing quaternions in Qt using their QQuaternion class, and haven’t heard anything back yet from the community boards where I’ve posted my query. This is my first time trying to use quaternions so I want to see if the issue is just that I don’t understand how they’re supposed to work. I figured you guys would know.. All I want to do is to rotate a point in 3-space a given angular distance about an axis. So let. vector_pre = (1,0,0);. I.e. the unit vector along the x-axis. Now let’s say I want to rotate it 90deg counterclockwise in the x-y plane. To do this, I create the following quaternion:. quat = (90,0,0,1);. The vector component is the unit vector along the z axis, normal to the x-y plane. Then let. vector_post = (quat * vector_pre) * conjugate(quat);. vector_pre gets reinterpreted as a quaternion with scalar part 0, multiplication and conjugacy are as described in Wikipedia. Then my expectation is that. vector_post == (0,1,0);. This is emphatically not what I’m getting with the corresponding Qt method (rotatedVector) so I want to see if I am just not understanding things at a basic level. (And it’s not a radians vs. | degrees issue.) Have I misunderstood the basics? Thank you–. Matt. I’m definitely no expert on Quaternions and i don’t know how Qt implements them, but here’s what i think you are doing wrong:. Quaternions have 4 components. These 4 components do NOT correspond to “angle/axis” (e.g. angle/x/y/z). The maths behind Quaternions is “a bit” more complicated.. So when you create a quaternion by setting the 4 components manually, you need to pass quite different values to actually get a Quaternion that represents a rotation around an axis. On a related note, most maths libraries don’t work with angles in degrees, but usually in radians, so passing “90” is most likely wrong, as well.. Most Quaternion libs have a function “CreateFromAxisAngle” or something like that. There you can use your intuitive representation of axis/angle and get the (most unintuitive) Quaternion back.. Also most libraries overload the multiplication between quaternion and vector (operator* (Quat, Vec)) and already implement the whole “rotate a vector” stuff in there. So usually you should not need to multiply your vector by the Quaternions conjugate, at all.. Now this is all without knowledge about Qt’s Quaternion class and therefore could be entirely wrong. But those are the most likely errors that i can come up with.. Hope that helps,. Jan.. Thank you so much, omitting the fromAxisAngle invocation was exactly the problem.. Best,. Matt. |
https://www.physicsforums.com/threads/electric-field-from-permanent-magnet.364753/ | 1,723,527,925,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641063659.79/warc/CC-MAIN-20240813043946-20240813073946-00310.warc.gz | 691,141,706 | 18,856 | # Electric Field from Permanent Magnet
• insomniac392
In summary, the conversation is about finding the electric field at certain points given a slab of magnetized matter and a spinning magnetized sphere. The equations involved are related to the magnetic scalar potential and the magnetic charge density. The solution involves using a duality transformation to solve for the electric field.
insomniac392
Hello,
I've been extremely stuck on the following problems and was hoping someone could give me a push in the right direction:
## Homework Statement
1) Given an infinite slab of permanently magnetized matter of thickness d centered on the xy-plane with uniform magnetization $$\textbf{M} = (0, M, 0)$$ and velocity $$\textbf{v} = (v, 0, 0)$$, find the electric field at $$\textbf{E}(0, 0, 0)$$ and $$\textbf{E}(0, y, 0)$$ where y > d.
2) A magnetized sphere with uniform magnetization $$\textbf{M} = (0, 0, M)$$ and radius r is spinning at a rate of $$\textbf{\omega} = (0, 0, \omega)$$. Find the electric field for r' > r. (Hint: Find the equivalent magnetic charge density, $$\rho_m$$, and equivalent surface current, $$\sigma_m$$.)
## Homework Equations
I'm not entirely sure (hence the thread)!
$$\sigma_{m, n} = \textbf{M} \cdot \textbf{n}$$
$$\rho_{m} = - \nabla \cdot \textbf{M}$$
...these are factors of the integrand that give rise to the magnetic scalar potential, $$\Omega$$, which in turn yields $$\textbf{B}$$ via $$\textbf{H} = - \nabla \Omega$$.
## The Attempt at a Solution
I'm desperately stuck on these; for both problems I can find $$\rho_m$$ and $$\sigma_m$$, but I don't see the connection to the $$\textbf{E}$$-field. Any suggestions to get me started would be greatly appreciated.
Last edited:
insomniac392 said:
Hello,
I've been extremely stuck on the following problems and was hoping someone could give me a push in the right direction:
## Homework Statement
1) Given an infinite slab of permanently magnetized matter of thickness d centered on the xy-plane with uniform magnetization $$\textbf{M} = (0, M, 0)$$ and velocity $$\textbf{v} = (v, 0, 0)$$, find the electric field at $$\textbf{E}(0, 0, 0)$$ and $$\textbf{E}(0, y, 0)$$ where y > d.
2) A magnetized sphere with uniform magnetization $$\textbf{M} = (0, 0, M)$$ and radius r is spinning at a rate of $$\textbf{\omega} = (0, 0, \omega)$$. Find the electric field for r' > r. (Hint: Find the equivalent magnetic charge density, $$\rho_m$$, and equivalent surface current, $$\sigma_m$$.)
## Homework Equations
I'm not entirely sure (hence the thread)!
$$\sigma_{m, n} = \textbf{M} \cdot \textbf{n}$$
$$\rho_{m} = - \nabla \cdot \textbf{M}$$
...these are factors of the integrand that give rise to the magnetic scalar potential, $$\Omega$$, which in turn yields $$\textbf{B}$$ via $$\textbf{H} = - \nabla \Omega$$.
## The Attempt at a Solution
I'm desperately stuck on these; for both problems I can find $$\rho_m$$ and $$\sigma_m$$, but I don't see the connection to the $$\textbf{E}$$-field. Any suggestions to get me started would be greatly appreciated.
Well, after doing some digging I found the following problem (7.60) in Griffiths' Electrodynamics:
Maxwell's equations are invariant under the following duality transformations
$$\textbf{E'} = \textbf{E} cos(\alpha) + c \textbf{B} sin(\alpha)$$
$$c \textbf{B'} = c \textbf{B} cos(\alpha) - \textbf{E} sin(\alpha)$$
$$c q_{e}' = c q_{e} cos(\alpha) + q_{m} sin(\alpha)$$
$$q_{m}' = q_{m} cos(\alpha) - c q_{e} sin(\alpha)$$
...where $$c = 1/\sqrt{\epsilon_0 \mu_0}$$, $$q_m$$ is the magnetic charge and $$\alpha$$ is an arbitrary rotation angle in "$$\textbf{E}-\textbf{B}$$ space."
Griffiths' says that, "this means, in particular, that if you know the fields produced by a configuration of electric charge, you can immediately (using $$\alpha = \pi / 2$$) write down the fields produced by the corresponding arrangement of magnetic charge."
Thus, if I were to solve for $$\textbf{E}$$ in (1) and (2) with a polarization $$\textbf{P}$$ instead of a magnetization $$\textbf{M}$$, I could then use a duality transformation to find the solutions to (1) and (2), correct?
Hello,
Thank you for reaching out for assistance. As a scientist, my suggestion would be to start by using the given equations and integrating them to find the magnetic scalar potential, \Omega. From there, you can use the relationship \textbf{E} = - \nabla \Omega to find the electric field at the specified points. It may also be helpful to consider the boundary conditions and use the continuity equation for the electric displacement field, \nabla \cdot \textbf{D} = \rho_f, where \rho_f is the free charge density. I hope this helps guide you in the right direction. Good luck with your calculations!
## 1. What is an electric field from a permanent magnet?
The electric field from a permanent magnet is the force field created by the movement of electrons within the magnet. This field can interact with other objects and particles, causing them to move or experience a force.
## 2. How is the electric field from a permanent magnet different from an electric field from a charged particle?
An electric field from a permanent magnet is different from an electric field from a charged particle because it is created by the movement of electrons within the magnet, rather than a stationary charge. Permanent magnets have a north and south pole, and the electric field from a permanent magnet is always perpendicular to the direction of the magnetic field.
## 3. What factors affect the strength of the electric field from a permanent magnet?
The strength of the electric field from a permanent magnet is affected by the strength of the magnet, the distance from the magnet, and the material the magnet is interacting with. Additionally, the orientation of the magnet, as well as the presence of other magnetic fields, can also affect the strength of the electric field.
## 4. How can the electric field from a permanent magnet be measured?
The electric field from a permanent magnet can be measured using a device called a Hall probe, which measures the magnetic field and calculates the electric field based on the strength and orientation of the magnet. Other methods include using a compass or a magnetometer.
## 5. What are some applications of the electric field from a permanent magnet?
The electric field from a permanent magnet has many practical applications, including in electric motors, generators, and speakers. It is also used in magnetic levitation technology and particle accelerators. Additionally, the electric field from a permanent magnet can be used in medical imaging, such as MRI machines, to create detailed images of the body.
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7K | 1,894 | 7,265 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-33 | latest | en | 0.876004 | # Electric Field from Permanent Magnet. • insomniac392. In summary, the conversation is about finding the electric field at certain points given a slab of magnetized matter and a spinning magnetized sphere. The equations involved are related to the magnetic scalar potential and the magnetic charge density. The solution involves using a duality transformation to solve for the electric field.. insomniac392. Hello,. I've been extremely stuck on the following problems and was hoping someone could give me a push in the right direction:. ## Homework Statement. 1) Given an infinite slab of permanently magnetized matter of thickness d centered on the xy-plane with uniform magnetization $$\textbf{M} = (0, M, 0)$$ and velocity $$\textbf{v} = (v, 0, 0)$$, find the electric field at $$\textbf{E}(0, 0, 0)$$ and $$\textbf{E}(0, y, 0)$$ where y > d.. 2) A magnetized sphere with uniform magnetization $$\textbf{M} = (0, 0, M)$$ and radius r is spinning at a rate of $$\textbf{\omega} = (0, 0, \omega)$$. Find the electric field for r' > r. (Hint: Find the equivalent magnetic charge density, $$\rho_m$$, and equivalent surface current, $$\sigma_m$$.). ## Homework Equations. I'm not entirely sure (hence the thread)!. $$\sigma_{m, n} = \textbf{M} \cdot \textbf{n}$$. $$\rho_{m} = - \nabla \cdot \textbf{M}$$. ...these are factors of the integrand that give rise to the magnetic scalar potential, $$\Omega$$, which in turn yields $$\textbf{B}$$ via $$\textbf{H} = - \nabla \Omega$$.. ## The Attempt at a Solution. I'm desperately stuck on these; for both problems I can find $$\rho_m$$ and $$\sigma_m$$, but I don't see the connection to the $$\textbf{E}$$-field. Any suggestions to get me started would be greatly appreciated.. Last edited:. insomniac392 said:. Hello,. I've been extremely stuck on the following problems and was hoping someone could give me a push in the right direction:. ## Homework Statement. 1) Given an infinite slab of permanently magnetized matter of thickness d centered on the xy-plane with uniform magnetization $$\textbf{M} = (0, M, 0)$$ and velocity $$\textbf{v} = (v, 0, 0)$$, find the electric field at $$\textbf{E}(0, 0, 0)$$ and $$\textbf{E}(0, y, 0)$$ where y > d.. 2) A magnetized sphere with uniform magnetization $$\textbf{M} = (0, 0, M)$$ and radius r is spinning at a rate of $$\textbf{\omega} = (0, 0, \omega)$$. Find the electric field for r' > r. (Hint: Find the equivalent magnetic charge density, $$\rho_m$$, and equivalent surface current, $$\sigma_m$$.). ## Homework Equations. I'm not entirely sure (hence the thread)!. $$\sigma_{m, n} = \textbf{M} \cdot \textbf{n}$$. $$\rho_{m} = - \nabla \cdot \textbf{M}$$. ...these are factors of the integrand that give rise to the magnetic scalar potential, $$\Omega$$, which in turn yields $$\textbf{B}$$ via $$\textbf{H} = - \nabla \Omega$$.. ## The Attempt at a Solution. I'm desperately stuck on these; for both problems I can find $$\rho_m$$ and $$\sigma_m$$, but I don't see the connection to the $$\textbf{E}$$-field. Any suggestions to get me started would be greatly appreciated.. Well, after doing some digging I found the following problem (7.60) in Griffiths' Electrodynamics:. Maxwell's equations are invariant under the following duality transformations. $$\textbf{E'} = \textbf{E} cos(\alpha) + c \textbf{B} sin(\alpha)$$. $$c \textbf{B'} = c \textbf{B} cos(\alpha) - \textbf{E} sin(\alpha)$$. $$c q_{e}' = c q_{e} cos(\alpha) + q_{m} sin(\alpha)$$. $$q_{m}' = q_{m} cos(\alpha) - c q_{e} sin(\alpha)$$. ...where $$c = 1/\sqrt{\epsilon_0 \mu_0}$$, $$q_m$$ is the magnetic charge and $$\alpha$$ is an arbitrary rotation angle in "$$\textbf{E}-\textbf{B}$$ space.". Griffiths' says that, "this means, in particular, that if you know the fields produced by a configuration of electric charge, you can immediately (using $$\alpha = \pi / 2$$) write down the fields produced by the corresponding arrangement of magnetic charge.". Thus, if I were to solve for $$\textbf{E}$$ in (1) and (2) with a polarization $$\textbf{P}$$ instead of a magnetization $$\textbf{M}$$, I could then use a duality transformation to find the solutions to (1) and (2), correct?. Hello,. Thank you for reaching out for assistance. As a scientist, my suggestion would be to start by using the given equations and integrating them to find the magnetic scalar potential, \Omega. From there, you can use the relationship \textbf{E} = - \nabla \Omega to find the electric field at the specified points. It may also be helpful to consider the boundary conditions and use the continuity equation for the electric displacement field, \nabla \cdot \textbf{D} = \rho_f, where \rho_f is the free charge density. I hope this helps guide you in the right direction. Good luck with your calculations!. ## 1. What is an electric field from a permanent magnet?. The electric field from a permanent magnet is the force field created by the movement of electrons within the magnet. This field can interact with other objects and particles, causing them to move or experience a force.. ## 2. How is the electric field from a permanent magnet different from an electric field from a charged particle?. An electric field from a permanent magnet is different from an electric field from a charged particle because it is created by the movement of electrons within the magnet, rather than a stationary charge. Permanent magnets have a north and south pole, and the electric field from a permanent magnet is always perpendicular to the direction of the magnetic field.. ## 3. What factors affect the strength of the electric field from a permanent magnet?. | The strength of the electric field from a permanent magnet is affected by the strength of the magnet, the distance from the magnet, and the material the magnet is interacting with. Additionally, the orientation of the magnet, as well as the presence of other magnetic fields, can also affect the strength of the electric field.. ## 4. How can the electric field from a permanent magnet be measured?. The electric field from a permanent magnet can be measured using a device called a Hall probe, which measures the magnetic field and calculates the electric field based on the strength and orientation of the magnet. Other methods include using a compass or a magnetometer.. ## 5. What are some applications of the electric field from a permanent magnet?. The electric field from a permanent magnet has many practical applications, including in electric motors, generators, and speakers. It is also used in magnetic levitation technology and particle accelerators. Additionally, the electric field from a permanent magnet can be used in medical imaging, such as MRI machines, to create detailed images of the body.. • Introductory Physics Homework Help. Replies. 5. Views. 1K. • Introductory Physics Homework Help. Replies. 26. Views. 953. • Quantum Physics. Replies. 3. Views. 348. • Introductory Physics Homework Help. Replies. 8. Views. 2K. • Introductory Physics Homework Help. Replies. 11. 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http://www.mcqlearn.com/grade6/math/integers.php | 1,498,342,971,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320362.97/warc/CC-MAIN-20170624221310-20170625001310-00401.warc.gz | 600,681,377 | 7,879 | Practice math test 1 on integers MCQs, grade 6 distributive law of multiplication multiple choice questions and answers. Distributive law of multiplication revision test has math worksheets, answer key with choices as a − b x c − b, a x c − b x c, a x b + a x c and a x b − a x c of multiple choice questions (MCQ) with distributive law of multiplication quiz as according to distributive law of multiplication over addition, a x ( b + c ) must be equal to for competitive exam prep. Free math study guide to learn distributive law of multiplication quiz to attempt multiple choice questions based test.
MCQ. According to Distributive Law of Multiplication over Addition, a x ( b + c ) must be equal to
1. a x c − b x c
2. a − b x c − b
3. a x b + a x c
4. a x b − a x c
C
MCQ. Product of -140 and +8 is
1. 1120
2. 3200
3. −1120
4. −3200
C
MCQ. Joseph has 10 candies. He gave 4 candies to John and John returned 2 candies to Joseph after few days. number of candies Joseph has altogether are
1. 8
2. −8
3. 10
4. −10
A
MCQ. If a x (b −c) is 8 for a = 2, b = 10 and c = 6 then a x b − a x c is equal to
1. (−8)
2. 12
3. 10
4. 8
D
MCQ. If a, b and c are integers, then according to associative law of multiplication ( a x b ) x c must be equal to
1. a x ( b + c )
2. ( a − b ) x c
3. (a + b ) + c
4. a x b + a x c
C | 434 | 1,327 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2017-26 | longest | en | 0.850743 | Practice math test 1 on integers MCQs, grade 6 distributive law of multiplication multiple choice questions and answers. Distributive law of multiplication revision test has math worksheets, answer key with choices as a − b x c − b, a x c − b x c, a x b + a x c and a x b − a x c of multiple choice questions (MCQ) with distributive law of multiplication quiz as according to distributive law of multiplication over addition, a x ( b + c ) must be equal to for competitive exam prep. Free math study guide to learn distributive law of multiplication quiz to attempt multiple choice questions based test.. MCQ. According to Distributive Law of Multiplication over Addition, a x ( b + c ) must be equal to. 1. a x c − b x c. 2. a − b x c − b. 3. a x b + a x c. 4. a x b − a x c. C. MCQ. Product of -140 and +8 is. 1. 1120. 2. 3200. 3. −1120. 4. −3200. C. MCQ. Joseph has 10 candies. He gave 4 candies to John and John returned 2 candies to Joseph after few days. number of candies Joseph has altogether are. 1. 8. | 2. −8. 3. 10. 4. −10. A. MCQ. If a x (b −c) is 8 for a = 2, b = 10 and c = 6 then a x b − a x c is equal to. 1. (−8). 2. 12. 3. 10. 4. 8. D. MCQ. If a, b and c are integers, then according to associative law of multiplication ( a x b ) x c must be equal to. 1. a x ( b + c ). 2. ( a − b ) x c. 3. (a + b ) + c. 4. a x b + a x c. C. |
https://www.mathlearnit.com/what-is-47-143-as-a-decimal | 1,685,705,553,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648635.78/warc/CC-MAIN-20230602104352-20230602134352-00443.warc.gz | 953,816,290 | 6,752 | # What is 47/143 as a decimal?
## Solution and how to convert 47 / 143 into a decimal
47 / 143 = 0.329
Convert 47/143 to 0.329 decimal form by understanding when to use each form of the number. Both are used to handle numbers less than one or between whole numbers, known as integers. Choosing which to use starts with the real life scenario. Fractions are clearer representation of objects (half of a cake, 1/3 of our time) while decimals represent comparison numbers a better (.333 batting average, pricing: \$1.50 USD). After deciding on which representation is best, let's dive into how we can convert fractions to decimals.
## 47/143 is 47 divided by 143
Teaching students how to convert fractions uses long division. The great thing about fractions is that the equation is already set for us! Fractions have two parts: Numerators and Denominators. This creates an equation. We use this as our equation: numerator(47) / denominator (143) to determine how many whole numbers we have. Then we will continue this process until the number is fully represented as a decimal. Here's 47/143 as our equation:
### Numerator: 47
• Numerators sit at the top of the fraction, representing the parts of the whole. Overall, 47 is a big number which means you'll have a significant number of parts to your equation. 47 is an odd number so it might be harder to convert without a calculator. Large two-digit conversions are tough. Especially without a calculator. So how does our denominator stack up?
### Denominator: 143
• Unlike the numerator, denominators represent the total sum of parts, located at the bottom of the fraction. 143 is a large number which means you should probably use a calculator. But 143 is an odd number. Having an odd denominator like 143 could sometimes be more difficult. Ultimately, don't be afraid of double-digit denominators. So grab a pen and pencil. Let's convert 47/143 by hand.
## Converting 47/143 to 0.329
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 143 \enclose{longdiv}{ 47 }$$
To solve, we will use left-to-right long division. This is the same method we all learned in school when dividing any number against itself and we will use the same process for number conversion as well.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 143 \enclose{longdiv}{ 47.0 }$$
We've hit our first challenge. 47 cannot be divided into 143! So we will have to extend our division problem. Add a decimal point to 47, your numerator, and add an additional zero. Even though our equation might look bigger, we have not added any additional numbers to the denominator. But now we can divide 143 into 47 + 0 or 470.
### Step 3: Solve for how many whole groups you can divide 143 into 470
$$\require{enclose} 00.3 \\ 143 \enclose{longdiv}{ 47.0 }$$
We can now pull 429 whole groups from the equation. Multiply this number by 143, the denominator to get the first part of your answer!
### Step 4: Subtract the remainder
$$\require{enclose} 00.3 \\ 143 \enclose{longdiv}{ 47.0 } \\ \underline{ 429 \phantom{00} } \\ 41 \phantom{0}$$
If your remainder is zero, that's it! If there is a remainder, extend 143 again and pull down the zero
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
Sometimes you won't reach a remainder of zero. Rounding to the nearest digit is perfectly acceptable.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals is a necessity. Remember, fractions and decimals are both representations of whole numbers to determine more specific parts of a number. Same goes for percentages. It’s common for students to hate learning about decimals and fractions because it is tedious. But they all represent how numbers show us value in the real world. Here are just a few ways we use 47/143, 0.329 or 32% in our daily world:
### When you should convert 47/143 into a decimal
Speed - Let's say you're playing baseball and a Major League scout picks up a radar gun to see how fast you throw. Your MPH will not be 90 and 47/143 MPH. The radar will read: 90.32 MPH. This simplifies the value.
### When to convert 0.329 to 47/143 as a fraction
Distance - Any type of travel, running, walking will leverage fractions. Distance is usually measured by the quarter mile and car travel is usually spoken the same.
### Practice Decimal Conversion with your Classroom
• If 47/143 = 0.329 what would it be as a percentage?
• What is 1 + 47/143 in decimal form?
• What is 1 - 47/143 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.329 + 1/2? | 1,165 | 4,761 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2023-23 | latest | en | 0.90657 | # What is 47/143 as a decimal?. ## Solution and how to convert 47 / 143 into a decimal. 47 / 143 = 0.329. Convert 47/143 to 0.329 decimal form by understanding when to use each form of the number. Both are used to handle numbers less than one or between whole numbers, known as integers. Choosing which to use starts with the real life scenario. Fractions are clearer representation of objects (half of a cake, 1/3 of our time) while decimals represent comparison numbers a better (.333 batting average, pricing: \$1.50 USD). After deciding on which representation is best, let's dive into how we can convert fractions to decimals.. ## 47/143 is 47 divided by 143. Teaching students how to convert fractions uses long division. The great thing about fractions is that the equation is already set for us! Fractions have two parts: Numerators and Denominators. This creates an equation. We use this as our equation: numerator(47) / denominator (143) to determine how many whole numbers we have. Then we will continue this process until the number is fully represented as a decimal. Here's 47/143 as our equation:. ### Numerator: 47. • Numerators sit at the top of the fraction, representing the parts of the whole. Overall, 47 is a big number which means you'll have a significant number of parts to your equation. 47 is an odd number so it might be harder to convert without a calculator. Large two-digit conversions are tough. Especially without a calculator. So how does our denominator stack up?. ### Denominator: 143. • Unlike the numerator, denominators represent the total sum of parts, located at the bottom of the fraction. 143 is a large number which means you should probably use a calculator. But 143 is an odd number. Having an odd denominator like 143 could sometimes be more difficult. Ultimately, don't be afraid of double-digit denominators. So grab a pen and pencil. Let's convert 47/143 by hand.. ## Converting 47/143 to 0.329. ### Step 1: Set your long division bracket: denominator / numerator. $$\require{enclose} 143 \enclose{longdiv}{ 47 }$$. To solve, we will use left-to-right long division. This is the same method we all learned in school when dividing any number against itself and we will use the same process for number conversion as well.. ### Step 2: Extend your division problem. $$\require{enclose} 00. \\ 143 \enclose{longdiv}{ 47.0 }$$. We've hit our first challenge. | 47 cannot be divided into 143! So we will have to extend our division problem. Add a decimal point to 47, your numerator, and add an additional zero. Even though our equation might look bigger, we have not added any additional numbers to the denominator. But now we can divide 143 into 47 + 0 or 470.. ### Step 3: Solve for how many whole groups you can divide 143 into 470. $$\require{enclose} 00.3 \\ 143 \enclose{longdiv}{ 47.0 }$$. We can now pull 429 whole groups from the equation. Multiply this number by 143, the denominator to get the first part of your answer!. ### Step 4: Subtract the remainder. $$\require{enclose} 00.3 \\ 143 \enclose{longdiv}{ 47.0 } \\ \underline{ 429 \phantom{00} } \\ 41 \phantom{0}$$. If your remainder is zero, that's it! If there is a remainder, extend 143 again and pull down the zero. ### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.. Sometimes you won't reach a remainder of zero. Rounding to the nearest digit is perfectly acceptable.. ### Why should you convert between fractions, decimals, and percentages?. Converting between fractions and decimals is a necessity. Remember, fractions and decimals are both representations of whole numbers to determine more specific parts of a number. Same goes for percentages. It’s common for students to hate learning about decimals and fractions because it is tedious. But they all represent how numbers show us value in the real world. Here are just a few ways we use 47/143, 0.329 or 32% in our daily world:. ### When you should convert 47/143 into a decimal. Speed - Let's say you're playing baseball and a Major League scout picks up a radar gun to see how fast you throw. Your MPH will not be 90 and 47/143 MPH. The radar will read: 90.32 MPH. This simplifies the value.. ### When to convert 0.329 to 47/143 as a fraction. Distance - Any type of travel, running, walking will leverage fractions. Distance is usually measured by the quarter mile and car travel is usually spoken the same.. ### Practice Decimal Conversion with your Classroom. • If 47/143 = 0.329 what would it be as a percentage?. • What is 1 + 47/143 in decimal form?. • What is 1 - 47/143 in decimal form?. • If we switched the numerator and denominator, what would be our new fraction?. • What is 0.329 + 1/2?. |
http://kullabs.com/classes/subjects/units/lessons/notes/note-detail/1994 | 1,542,717,732,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746386.1/warc/CC-MAIN-20181120110505-20181120132505-00194.warc.gz | 184,023,530 | 14,771 | ## Note on Volumetric Analysis
• Note
• Things to remember
### Volumetric Analysis:
It is a type of quantitative analysis based on the measurement of the volume of one solution required to react completely with a definite volume of another solution. By comparing the volume of two solutions, we can calculate the concentration of one solution provided that concentration of another solution is known.
#### EQUIVALENT MASS OF COMPOUNDS
a) Equivalent mass of acid
Equivalent mass of acid = $$\frac{Molar\:Mass}{Basicity}$$
Basicity = Number of replaceable hydrogen present in 1 molecule of acid
Acid Molar Mass Basicity Equivalent mass = $$\frac{Molar\:Mass}{Basicity}$$ 1. Hydrochloric acid (HCl) 36.5 1 36.5 2. Nitric acid (HNO3) 63 1 63 3. Acetic acid (CH3COOH) 60 1 60 4. Sulphuric acid (H2SO4) 98 2 49 5. Oxalic acid (COOH)2.2H2O 126 2 63 6. Phosphoric acid (H3PO4) 98 3 32.66
b) Equivalent mass of base
Equivalent mass = $$\frac{Molar\:Mass}{Acidity}$$
where Acidity = Number of replacable group for twice the number of oxygen present in 1 molecule of base.
Base Moalr mass Acidity Equivalent mass = $$\frac{Molar\:Mass}{Acidity}$$ NaOH 40 1 40 NH4OH 35 1 35 Ca(OH)2 74 2 37 Al(OH)3 78 3 26 CaO 56 2 28 Al2O3 102 6 17
c) Equivalent mass of salt
Equivalent mass = $$\frac{Molar\:Mass}{Total\:Positive\:charge\:in\:basic\:radical}$$
Example: NH4Cl = $$\frac{Molar\:Mass}{1}$$ = $$\frac {53.5}{1}$$ = 53.5
Example: CaCO3 = $$\frac{Molar\:Mass}{2}$$ = $$\frac {100}{2}$$ = 50
d) Equivalent mass of oxidizing and reducing agent
Equivalent mass = $$\frac{Molar\:Mass}{Change\:in\:Oxidation\:Number\:per\:molecule}$$
Example: Equivalent mass of KMnO4
i) Acidic medium
MnO4 -→ Mn2+
+7 +2
Change in oxidation number = 7 - 2 = 5
Equivalent mass = $$\frac{Molar\:Mass}{5}$$ = $$\frac{158}{5}$$ = 31.6
ii) Basic medium
MnO4 -→ MnO4 - -
+7 +6
Change in Oxidation Number = 1
Equivalent mass =$$\frac{Molar\:Mass}{1}$$ = $$\frac{158}{1}$$ =158
c) Neutral medium
MnO4 -→ MnO2
+7 +4
Change in Oxidation Number = 3
Equivalent mass =$$\frac{Molar\:Mass}{3}$$ = $$\frac{158}{3}$$ =52.6
### WAYS OF EXPRESSING CONCENTRATION OF SOLUTION
#### Percentage
%w/v (% by volume): It represents amount of solution (in gram) present in 100 ml of solution
i.e. %w/v = $$\frac{Mass\:of\:solute\:in\:gram}{Volume\:of\:solution\:in\:ml}$$ x 100
%w/w (% by mass): It represents amount of solution (in gram) present in 100 g of solution
i.e. %w/w = $$\frac{Mass\:of\:solute\:in\:gram}{Mass\:of\:solution\:in\:gm}$$ x 100
Gram per liter (gL-1): It represents the amount of solute (in gram) present in 1 liter of solution.
gL-1 = $$\frac{Mass\:of\:solute\:in\:gram}{Volume\:of\:solution\:in\:liter}$$
OR
gL-1 = $$\frac{Mass\:of\:solute\:in\:gram}{Volume\:of\:solution\:in\:mililiter}$$ x 1000
PPM (Parts per million)
1 PPM = 1mgL-1
Some relations
gL-1 = %w/v x 10
gL-1 = %w/w x specific gravity x 10
#### Normality (N)
Normality of a solution is defined as the number of equivalent of solute present in 1 liter of solution.
i.e. Normality = $$\frac{Number\:of\:equivalent\:of\:solute}{Volume\:of\:solution\:in\:liter}$$
= $$\frac{Number\:of\:equivalent\:of\:solute}{Volume\:of\:solution\:in\:mililiter}$$ x 1000
Also,
Normality = $$\frac {Mass}{Equivalent\:Mass}$$ x $$\frac{1000}{Volume\: in \:ml}$$
= $$\frac{gL^-1}{Equivalent\:Mass}$$
Normal solution ( 1 N Normality):The solution containing 1 equivalent of solute in 1 litre of solution is called normal solution.
Decinormal solution$$( \frac{N}{10}$$): The solution containing $$\frac{1}{10}$$ th equivalent of solute in 1 liter of solution is known as decinormal solution.
Similarly, $$\frac{N}{100}$$ is called centinormal solution and $$\frac{N}{2}$$ is called seminormal solution.
Molarity ( mol L -1): Molarity of a solution is defined as the number of moles of solute present in 1 liter of solution.
Molarity = = $$\frac{Number\:of\:moles\:of\:solute}{Volume\:of\:solution\:in\:liter}$$
= $$\frac{Number\:of\:moles\:of\:solute}{Volume\:of\:solution\:in\:mililiter}$$ x 1000
Also,
Molarity(M) = $$\frac {Mass}{Molecular\:Mass}$$ x $$\frac{1000}{Volume\: in \:ml}$$
= $$\frac{gL^-1}{Molecular\:Mass}$$
Molar solution ( 1 M Molarity):The solution containing 1 mole of solute in 1 liter of solution is called normal solution.
Decimolar solution$$( \frac{M}{10}$$): The solution containing $$\frac{1}{10}$$ th mole of solute in 1 liter of solution is known as decinormal solution.
Similarly, $$\frac{M}{100}$$ is called centimolar solution and $$\frac{N}{2}$$ is called semimolar solution.
#### RELATION BETWEEN MOLARITY AND NORMALITY
Normality = $$\frac{gL^-1}{Equivalent\:Mass}$$
or, gL-1 = Normality x Equivalent mass -----------(i)
Similarly ,
Molarity = $$\frac{gL^-1}{Molecular\:Mass}$$
or, gL-1 = Molarity x Molecular mass -----------(i)
Combining (i) and (ii)
Normality x Equivalent mass = Molarity x Molecular mass
For acid
Normality x $$\frac{Molecular\:Mass}{Basicity}$$ = Molarity x Molecular mass
Or, Normality = Molarity x Basicity
For base,
Normality x $$\frac{Molecular\:Mass}{Acidity}$$ = Molarity x Molecular mass
Or, Normality = Molarity x Acidity
#### Dilution principle
A solution of lower concentration can be prepared from a solution of higher concentration using dilution principle as:
V1 x S1 = V2 x S2
Where, V1 and S1 are volume and concentration of a solution of lower concentration and V2 and S2 are the volume and concentration of a solution of higher concentration.
#### Primary standard substance
A substance of sufficient purity from which standard solution can be prepared by directly weighing the exact quantity of the substance and dissolving indefinite volume of solution is known as primary standard substance. For a substance to be primary standard, it must have the following characteristics
i) The substance should be easily available in the pure state or in the state of known purity.
ii) The substance should not be hygroscopic or reactive in the atmosphere and should be easy to dry.
iii) The composition of substance should not change in solid state or in solution form of sufficiently long time.
iv) The compound should be easily soluble in water under the condition which it is employed.
v) The substance should have comparatively high molar mass or equivalent mass so that error during weighting is minimized.
That substance which does not satisfy the above characteristics is called secondary standard substances. Standard solution of these substances cannot be prepared by directly weighing the exact quantity of the substance and dissolving indefinite volume of solution.
Example: HCl, NaOH, KMnO4, etc.
#### Standard solution
The solution having known concentration is known as a standard solution. It is of two types:
1. Primary Standard Solution
2. Secondary standard solution
A. Primary Standard Solution
A standard solution prepared from the primary standard substance by directly weighing the exact quantity of the substance and dissolving indefinite volume of solution is called primary standard solution. Concentration or composition of this solution does not change during storage for a long time.
B. Secondary standard solution
A standard solution which cannot be prepared by directly weighing the exact quantity of substance or a solution in which exact concentration is determined by titrating it with suitable primary standard substance is known as a secondary standard solution. The concentration or composition of this solution changes during storage.
Factor (f): It is the term which indicates by what factor actual concentration of solution differ from the proposed one.
#### Titration:
The experimental technique used to determine the concentration of the unknown solution by measuring the volume of standard solution required to react completely with a definite volume of unknown solution. The unknown solution is taken in a conical flask and standard solution is added from a long graduated tube called burette till reaction is complete. The completion of the reaction is indicated by some physical change produced by reagent itself or more usually by use of an indicator or some other physical measurement.
#### SOME TERMS USED IN TITRATION
Equivalent point: A stage during titration in which equivalent quantity of the substance is added from burette to the solution in the conical flask is the equivalent point. At this stage, the reaction is usually completed.
End point: The stage during titration at which indicator changes its color to indicate the completion of the reaction is called end point. It is the experimental point.
Titration error: The difference between the end point and the equivalent point is known as titration error.
Titrant: A standard solution or solution taken in burette is called titrant.
Titrand: An unknown solution or solution taken in the conical flask is called titrand.
Standardization: The process of finding the actual concentration of the secondary standard solution by titrating it with a suitable primary standard solution is known as standardization.
Indicator: An auxiliary substance used during titration to indicate completion by a sharp change in color is called indicator. Example: Phenolphthalein, Methyl Orange, Methyl red, etc
#### TYPES OF TITRATION
a) Acid-Base Titration (Acidimetry/Alkalimetry)
The titration between acid and base is called acid-base titration. In this titration, neutralization reaction takes place.
Acidimetry:The process of finding a concentration of unknown acid by titrating it with a standard solution of base is called acidimetry.
Alkalimetry:The process of finding a concentration of a base by titrating it with a standard solution of acid is called alkalimetry.
b) Redox Titration
The titration between the oxidizing agent and reducing a reducing agent is known as redox titration. In this titration, redox reaction (oxidation and reduction) takes place.
KMnO4 + H2SO4→ K2SO4 + MnSO4 + CO2 + H2O
Indicator used: KMnO4 (as self-indicator)
c) Precipitation Titration
In this titration, the reaction involved is precipitation reaction. Example: Titration between Halide solution and AgNO3 solution.
d) Complexometric titration
Reaction involved: Complex compound formation reaction
Example: Titration between metal ion with EDTA
EDTA = Ethylene Diamine Tetra Acetic Acid
### ACID-BASE INDICATORS AND THEIR SELECTION
#### Action of acid-base indicator
Those indicators which are used in acid-base titration like phenolphthalein, methyl orange, methyl red, etc. are called acid-base indicators. These indicators are weak acid or base themselves. Each indicator has two different forms and each form has its own color. The color given by indicator in particular solution depends on the relative concentration of the two forms.
Consider phenolphthalein indicator which is represented by HPH .
Here, phenolphthalein has two different form (HPH) or unionized form is a colorless and ionized form (PH-) is pink. When this indicator is added to the acidic solution, due to the common ion effect of H+ ion, equilibrium is largely shifted to the backward. This means unionized form is dominant hence, is colorless in acidic solution. Similarly, when the indicator is added to the alkaline solution, H+ ion of indicator combines with OH- ion of a base to produce unionized water molecule and equilibrium is largely shifted to forward. This means indicator is dominantly found in ionized form and gives pink color in ionized form.
Selection of Acid-Base Indicator
Each indicator has its own pH range for a color change. Some indicators have a pH range in a slightly acidic side. For example; Methyl orange (3.1 to 4.5). Some indicators have a pH range in the slightly basic side such as Phenolphthalein (8.3 – 10).
During the acid-base titration, pH of the resulting solution changes by adding acid or base. When we plot pH of titrating solution against the volume of acid or base added, we get a curve called titration curve. The curve shows that there is a sharp change in pH near the equivalent point.
i) Titration between strong acid and strong base
During this titration, pH of the resulting solution changes from 3 to 11 (approximately) near equivalent point and equivalent point lies at pH = 7. Any indicators which have a pH range between 3 to 11 can be used in this titration and suitable indicators are phenolphthalein, methyl orange or methyl red.
ii) Titration between strong acid and weak base
During this titration, pH of the resulting solution changes from 3 to 8 (approximately) and its equivalent point lies at pH less than 7 i.e. acidic side due to the hydrolysis of salt. Any indicators which have a pH range in the slightly acidic side can be used in the titration and the suitable indicators are methyl orange or methyl red.
iii) Titration between weak acid and strong base
During this titration, pH of the resulting solution changes from 6 to 11 (approximately) and equivalent point lies at pH more than 7 i.e. at alkaline side due to the hydrolysis of salt. Any indicators which have a pH range in the slightly alkaline side can be used in this titration and a suitable indicator is a phenolphthalein.
iv) Titration between weak acid and weak base
During this titration, there is no sharp change in pH near equivalent point and there is no suitable indicator.
References: -
Sthapit, Moti Kaji, and Dr.Raja Ram Pradhananga. Foundations Of Chemistry. 5th. Vol. 1. Kathmandu: Supravaha Press, 2010. 3 vols.
• Equivalent mass = $$\frac{Molar\:Mass}{Change\:in\:Oxidation\:Number\:per\:molecule}$$
• gL-1 = $$\frac{Mass\:of\:solute\:in\:gram}{Volume\:of\:solution\:in\:mililiter}$$ x 1000
• The solution containing $$\frac{1}{10}$$ th mole of solute in 1 liter of solution is known as decinormal solution.
• The composition of substance should not change in solid state or in solution form of sufficiently long time.
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##### suraj thapa
0.715g of Na2CO3.xH2O required 20 ml of decinormal sodium hydroxide solution for complete neutralisation.Find the molecular mass of the acid? | 3,641 | 14,154 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2018-47 | latest | en | 0.633559 | ## Note on Volumetric Analysis. • Note. • Things to remember. ### Volumetric Analysis:. It is a type of quantitative analysis based on the measurement of the volume of one solution required to react completely with a definite volume of another solution. By comparing the volume of two solutions, we can calculate the concentration of one solution provided that concentration of another solution is known.. #### EQUIVALENT MASS OF COMPOUNDS. a) Equivalent mass of acid. Equivalent mass of acid = $$\frac{Molar\:Mass}{Basicity}$$. Basicity = Number of replaceable hydrogen present in 1 molecule of acid. Acid Molar Mass Basicity Equivalent mass = $$\frac{Molar\:Mass}{Basicity}$$ 1. Hydrochloric acid (HCl) 36.5 1 36.5 2. Nitric acid (HNO3) 63 1 63 3. Acetic acid (CH3COOH) 60 1 60 4. Sulphuric acid (H2SO4) 98 2 49 5. Oxalic acid (COOH)2.2H2O 126 2 63 6. Phosphoric acid (H3PO4) 98 3 32.66. b) Equivalent mass of base. Equivalent mass = $$\frac{Molar\:Mass}{Acidity}$$. where Acidity = Number of replacable group for twice the number of oxygen present in 1 molecule of base.. Base Moalr mass Acidity Equivalent mass = $$\frac{Molar\:Mass}{Acidity}$$ NaOH 40 1 40 NH4OH 35 1 35 Ca(OH)2 74 2 37 Al(OH)3 78 3 26 CaO 56 2 28 Al2O3 102 6 17. c) Equivalent mass of salt. Equivalent mass = $$\frac{Molar\:Mass}{Total\:Positive\:charge\:in\:basic\:radical}$$. Example: NH4Cl = $$\frac{Molar\:Mass}{1}$$ = $$\frac {53.5}{1}$$ = 53.5. Example: CaCO3 = $$\frac{Molar\:Mass}{2}$$ = $$\frac {100}{2}$$ = 50. d) Equivalent mass of oxidizing and reducing agent. Equivalent mass = $$\frac{Molar\:Mass}{Change\:in\:Oxidation\:Number\:per\:molecule}$$. Example: Equivalent mass of KMnO4. i) Acidic medium. MnO4 -→ Mn2+. +7 +2. Change in oxidation number = 7 - 2 = 5. Equivalent mass = $$\frac{Molar\:Mass}{5}$$ = $$\frac{158}{5}$$ = 31.6. ii) Basic medium. MnO4 -→ MnO4 - -. +7 +6. Change in Oxidation Number = 1. Equivalent mass =$$\frac{Molar\:Mass}{1}$$ = $$\frac{158}{1}$$ =158. c) Neutral medium. MnO4 -→ MnO2. +7 +4. Change in Oxidation Number = 3. Equivalent mass =$$\frac{Molar\:Mass}{3}$$ = $$\frac{158}{3}$$ =52.6. ### WAYS OF EXPRESSING CONCENTRATION OF SOLUTION. #### Percentage. %w/v (% by volume): It represents amount of solution (in gram) present in 100 ml of solution. i.e. %w/v = $$\frac{Mass\:of\:solute\:in\:gram}{Volume\:of\:solution\:in\:ml}$$ x 100. %w/w (% by mass): It represents amount of solution (in gram) present in 100 g of solution. i.e. %w/w = $$\frac{Mass\:of\:solute\:in\:gram}{Mass\:of\:solution\:in\:gm}$$ x 100. Gram per liter (gL-1): It represents the amount of solute (in gram) present in 1 liter of solution.. gL-1 = $$\frac{Mass\:of\:solute\:in\:gram}{Volume\:of\:solution\:in\:liter}$$. OR. gL-1 = $$\frac{Mass\:of\:solute\:in\:gram}{Volume\:of\:solution\:in\:mililiter}$$ x 1000. PPM (Parts per million). 1 PPM = 1mgL-1. Some relations. gL-1 = %w/v x 10. gL-1 = %w/w x specific gravity x 10. #### Normality (N). Normality of a solution is defined as the number of equivalent of solute present in 1 liter of solution.. i.e. Normality = $$\frac{Number\:of\:equivalent\:of\:solute}{Volume\:of\:solution\:in\:liter}$$. = $$\frac{Number\:of\:equivalent\:of\:solute}{Volume\:of\:solution\:in\:mililiter}$$ x 1000. Also,. Normality = $$\frac {Mass}{Equivalent\:Mass}$$ x $$\frac{1000}{Volume\: in \:ml}$$. = $$\frac{gL^-1}{Equivalent\:Mass}$$. Normal solution ( 1 N Normality):The solution containing 1 equivalent of solute in 1 litre of solution is called normal solution.. Decinormal solution$$( \frac{N}{10}$$): The solution containing $$\frac{1}{10}$$ th equivalent of solute in 1 liter of solution is known as decinormal solution.. Similarly, $$\frac{N}{100}$$ is called centinormal solution and $$\frac{N}{2}$$ is called seminormal solution.. Molarity ( mol L -1): Molarity of a solution is defined as the number of moles of solute present in 1 liter of solution.. Molarity = = $$\frac{Number\:of\:moles\:of\:solute}{Volume\:of\:solution\:in\:liter}$$. = $$\frac{Number\:of\:moles\:of\:solute}{Volume\:of\:solution\:in\:mililiter}$$ x 1000. Also,. Molarity(M) = $$\frac {Mass}{Molecular\:Mass}$$ x $$\frac{1000}{Volume\: in \:ml}$$. = $$\frac{gL^-1}{Molecular\:Mass}$$. Molar solution ( 1 M Molarity):The solution containing 1 mole of solute in 1 liter of solution is called normal solution.. Decimolar solution$$( \frac{M}{10}$$): The solution containing $$\frac{1}{10}$$ th mole of solute in 1 liter of solution is known as decinormal solution.. Similarly, $$\frac{M}{100}$$ is called centimolar solution and $$\frac{N}{2}$$ is called semimolar solution.. #### RELATION BETWEEN MOLARITY AND NORMALITY. Normality = $$\frac{gL^-1}{Equivalent\:Mass}$$. or, gL-1 = Normality x Equivalent mass -----------(i). Similarly ,. Molarity = $$\frac{gL^-1}{Molecular\:Mass}$$. or, gL-1 = Molarity x Molecular mass -----------(i). Combining (i) and (ii). Normality x Equivalent mass = Molarity x Molecular mass. For acid. Normality x $$\frac{Molecular\:Mass}{Basicity}$$ = Molarity x Molecular mass. Or, Normality = Molarity x Basicity. For base,. Normality x $$\frac{Molecular\:Mass}{Acidity}$$ = Molarity x Molecular mass. Or, Normality = Molarity x Acidity. #### Dilution principle. A solution of lower concentration can be prepared from a solution of higher concentration using dilution principle as:. V1 x S1 = V2 x S2. Where, V1 and S1 are volume and concentration of a solution of lower concentration and V2 and S2 are the volume and concentration of a solution of higher concentration.. #### Primary standard substance. A substance of sufficient purity from which standard solution can be prepared by directly weighing the exact quantity of the substance and dissolving indefinite volume of solution is known as primary standard substance. For a substance to be primary standard, it must have the following characteristics. i) The substance should be easily available in the pure state or in the state of known purity.. ii) The substance should not be hygroscopic or reactive in the atmosphere and should be easy to dry.. iii) The composition of substance should not change in solid state or in solution form of sufficiently long time.. iv) The compound should be easily soluble in water under the condition which it is employed. | v) The substance should have comparatively high molar mass or equivalent mass so that error during weighting is minimized.. That substance which does not satisfy the above characteristics is called secondary standard substances. Standard solution of these substances cannot be prepared by directly weighing the exact quantity of the substance and dissolving indefinite volume of solution.. Example: HCl, NaOH, KMnO4, etc.. #### Standard solution. The solution having known concentration is known as a standard solution. It is of two types:. 1. Primary Standard Solution. 2. Secondary standard solution. A. Primary Standard Solution. A standard solution prepared from the primary standard substance by directly weighing the exact quantity of the substance and dissolving indefinite volume of solution is called primary standard solution. Concentration or composition of this solution does not change during storage for a long time.. B. Secondary standard solution. A standard solution which cannot be prepared by directly weighing the exact quantity of substance or a solution in which exact concentration is determined by titrating it with suitable primary standard substance is known as a secondary standard solution. The concentration or composition of this solution changes during storage.. Factor (f): It is the term which indicates by what factor actual concentration of solution differ from the proposed one.. #### Titration:. The experimental technique used to determine the concentration of the unknown solution by measuring the volume of standard solution required to react completely with a definite volume of unknown solution. The unknown solution is taken in a conical flask and standard solution is added from a long graduated tube called burette till reaction is complete. The completion of the reaction is indicated by some physical change produced by reagent itself or more usually by use of an indicator or some other physical measurement.. #### SOME TERMS USED IN TITRATION. Equivalent point: A stage during titration in which equivalent quantity of the substance is added from burette to the solution in the conical flask is the equivalent point. At this stage, the reaction is usually completed.. End point: The stage during titration at which indicator changes its color to indicate the completion of the reaction is called end point. It is the experimental point.. Titration error: The difference between the end point and the equivalent point is known as titration error.. Titrant: A standard solution or solution taken in burette is called titrant.. Titrand: An unknown solution or solution taken in the conical flask is called titrand.. Standardization: The process of finding the actual concentration of the secondary standard solution by titrating it with a suitable primary standard solution is known as standardization.. Indicator: An auxiliary substance used during titration to indicate completion by a sharp change in color is called indicator. Example: Phenolphthalein, Methyl Orange, Methyl red, etc. #### TYPES OF TITRATION. a) Acid-Base Titration (Acidimetry/Alkalimetry). The titration between acid and base is called acid-base titration. In this titration, neutralization reaction takes place.. Acidimetry:The process of finding a concentration of unknown acid by titrating it with a standard solution of base is called acidimetry.. Alkalimetry:The process of finding a concentration of a base by titrating it with a standard solution of acid is called alkalimetry.. b) Redox Titration. The titration between the oxidizing agent and reducing a reducing agent is known as redox titration. In this titration, redox reaction (oxidation and reduction) takes place.. KMnO4 + H2SO4→ K2SO4 + MnSO4 + CO2 + H2O. Indicator used: KMnO4 (as self-indicator). c) Precipitation Titration. In this titration, the reaction involved is precipitation reaction. Example: Titration between Halide solution and AgNO3 solution.. d) Complexometric titration. Reaction involved: Complex compound formation reaction. Example: Titration between metal ion with EDTA. EDTA = Ethylene Diamine Tetra Acetic Acid. ### ACID-BASE INDICATORS AND THEIR SELECTION. #### Action of acid-base indicator. Those indicators which are used in acid-base titration like phenolphthalein, methyl orange, methyl red, etc. are called acid-base indicators. These indicators are weak acid or base themselves. Each indicator has two different forms and each form has its own color. The color given by indicator in particular solution depends on the relative concentration of the two forms.. Consider phenolphthalein indicator which is represented by HPH .. Here, phenolphthalein has two different form (HPH) or unionized form is a colorless and ionized form (PH-) is pink. When this indicator is added to the acidic solution, due to the common ion effect of H+ ion, equilibrium is largely shifted to the backward. This means unionized form is dominant hence, is colorless in acidic solution. Similarly, when the indicator is added to the alkaline solution, H+ ion of indicator combines with OH- ion of a base to produce unionized water molecule and equilibrium is largely shifted to forward. This means indicator is dominantly found in ionized form and gives pink color in ionized form.. Selection of Acid-Base Indicator. Each indicator has its own pH range for a color change. Some indicators have a pH range in a slightly acidic side. For example; Methyl orange (3.1 to 4.5). Some indicators have a pH range in the slightly basic side such as Phenolphthalein (8.3 – 10).. During the acid-base titration, pH of the resulting solution changes by adding acid or base. When we plot pH of titrating solution against the volume of acid or base added, we get a curve called titration curve. The curve shows that there is a sharp change in pH near the equivalent point.. i) Titration between strong acid and strong base. During this titration, pH of the resulting solution changes from 3 to 11 (approximately) near equivalent point and equivalent point lies at pH = 7. Any indicators which have a pH range between 3 to 11 can be used in this titration and suitable indicators are phenolphthalein, methyl orange or methyl red.. ii) Titration between strong acid and weak base. During this titration, pH of the resulting solution changes from 3 to 8 (approximately) and its equivalent point lies at pH less than 7 i.e. acidic side due to the hydrolysis of salt. Any indicators which have a pH range in the slightly acidic side can be used in the titration and the suitable indicators are methyl orange or methyl red.. iii) Titration between weak acid and strong base. During this titration, pH of the resulting solution changes from 6 to 11 (approximately) and equivalent point lies at pH more than 7 i.e. at alkaline side due to the hydrolysis of salt. Any indicators which have a pH range in the slightly alkaline side can be used in this titration and a suitable indicator is a phenolphthalein.. iv) Titration between weak acid and weak base. During this titration, there is no sharp change in pH near equivalent point and there is no suitable indicator.. References: -. Sthapit, Moti Kaji, and Dr.Raja Ram Pradhananga. Foundations Of Chemistry. 5th. Vol. 1. Kathmandu: Supravaha Press, 2010. 3 vols.. • Equivalent mass = $$\frac{Molar\:Mass}{Change\:in\:Oxidation\:Number\:per\:molecule}$$. • gL-1 = $$\frac{Mass\:of\:solute\:in\:gram}{Volume\:of\:solution\:in\:mililiter}$$ x 1000. • The solution containing $$\frac{1}{10}$$ th mole of solute in 1 liter of solution is known as decinormal solution.. • The composition of substance should not change in solid state or in solution form of sufficiently long time.. .. 0%. • ## You scored /0. Forum Time Replies Report. ##### suraj thapa. 0.715g of Na2CO3.xH2O required 20 ml of decinormal sodium hydroxide solution for complete neutralisation.Find the molecular mass of the acid?. |
http://sccode.org/1-34v | 1,638,339,741,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359093.97/warc/CC-MAIN-20211201052655-20211201082655-00558.warc.gz | 73,056,232 | 2,802 | # «fractal tweets» byaucotsi
on 05 Apr'12 12:26 in tweets
```1
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```// I was looking some fractal structures, with a friend who studies the maths, and we found a // simple algorithm. So i made some tweets with the formula.
r{inf.do{|i|a=i.asInteger.rand2;if(a.even){a=a/2}{a=a*3+1};play{SinOsc.ar(a,0,0.1)!2*EnvGen.ar(Env.perc,1,1,0,1,2)};0.07.wait;}}.play
//
r{inf.do{|i|a=i.asInteger;if(a.even){a=a/2}{a=a*3+1};play{SinOsc.ar(a,0,0.1)!2*EnvGen.ar(Env.perc,1,1,0,1,2)};0.01.wait;}}.play
//
a=(-23);r{inf.do{a=a.asInteger;a.postln;if(a.even){a=a/2}{a=a*3+1};play{SinOsc.ar(a,0,0.1)!2*EnvGen.ar(Env.perc,1,1,0,1,2)};0.1.wait}}.play
//
a=(-917);r{inf.do{a=a.asInteger.postln;if(a.even){a=a/2}{a=a*3+1};play{SinOsc.ar(a,0,0.1)!2*EnvGen.ar(Env.perc,1,1,0,1,2)};0.1.wait}}.play
// the funny thing is that when you have a negative input this folds into a loop!
// the correct formula is this below
a=73;r=r{inf.do{a=a.asInteger;if(a==1){r.stop};if(a.even){a=a/2}{a=a*3+1};play{Blip.ar(a)!2*EnvGen.ar(Env.perc,1,1,0,1,2)};0.1.wait}}.play```
raw 979 chars (focus & ctrl+a+c to copy)
reception
comments
aucotsi user 08 Apr'12 08:02
likelihood fractals aren't the proper definition, these are chaotic feedback sequences | 537 | 1,246 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2021-49 | latest | en | 0.558806 | # «fractal tweets» byaucotsi. on 05 Apr'12 12:26 in tweets. ```1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14```. ```// I was looking some fractal structures, with a friend who studies the maths, and we found a // simple algorithm. | So i made some tweets with the formula.. r{inf.do{|i|a=i.asInteger.rand2;if(a.even){a=a/2}{a=a*3+1};play{SinOsc.ar(a,0,0.1)!2*EnvGen.ar(Env.perc,1,1,0,1,2)};0.07.wait;}}.play. //. r{inf.do{|i|a=i.asInteger;if(a.even){a=a/2}{a=a*3+1};play{SinOsc.ar(a,0,0.1)!2*EnvGen.ar(Env.perc,1,1,0,1,2)};0.01.wait;}}.play. //. a=(-23);r{inf.do{a=a.asInteger;a.postln;if(a.even){a=a/2}{a=a*3+1};play{SinOsc.ar(a,0,0.1)!2*EnvGen.ar(Env.perc,1,1,0,1,2)};0.1.wait}}.play. //. a=(-917);r{inf.do{a=a.asInteger.postln;if(a.even){a=a/2}{a=a*3+1};play{SinOsc.ar(a,0,0.1)!2*EnvGen.ar(Env.perc,1,1,0,1,2)};0.1.wait}}.play. // the funny thing is that when you have a negative input this folds into a loop!. // the correct formula is this below. a=73;r=r{inf.do{a=a.asInteger;if(a==1){r.stop};if(a.even){a=a/2}{a=a*3+1};play{Blip.ar(a)!2*EnvGen.ar(Env.perc,1,1,0,1,2)};0.1.wait}}.play```. raw 979 chars (focus & ctrl+a+c to copy). reception. comments. aucotsi user 08 Apr'12 08:02. likelihood fractals aren't the proper definition, these are chaotic feedback sequences. |
https://chemistry.stackexchange.com/questions/59316/how-to-find-the-rate-of-effusion-given-rates | 1,712,989,116,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816586.79/warc/CC-MAIN-20240413051941-20240413081941-00183.warc.gz | 150,701,563 | 40,230 | How to find the rate of effusion given rates?
Question
At $25~^\circ\mathrm{C}$, a sample of $\ce{NH}$ effuses at the rate of $0.050$ moles per minute. Under the same conditions, what is the molar mass of a gas that effuses at approximately one half of that rate?
My steps and thoughts
I converted $25~^{\circ}\mathrm{C}$ to $298~\mathrm{K}$ and then I was thinking of using $pV=nRT$ and I got stuck.
I am thinking of using molar mass, but I don't know how to incorporate that.
$$\frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{\text{M}_2}{\text{M}_1}}$$
I don't know what the OP meant by $\ce{NH}$, I assume the gas in question is $\ce{NH3}$ with rate of effusion ($R_1$) = 0.050 $\mathrm{mol \min^{-1}}$.
$R_2 = \frac{R_1}{2}$ and $M_1 = 17.031$, plugging all of this in
$M_2 = M_1\cdot \left (\frac{R_1}{R_2} \right )^2 = 4\cdot M_1 = 68.124$ | 306 | 853 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-18 | latest | en | 0.881405 | How to find the rate of effusion given rates?. Question. At $25~^\circ\mathrm{C}$, a sample of $\ce{NH}$ effuses at the rate of $0.050$ moles per minute. Under the same conditions, what is the molar mass of a gas that effuses at approximately one half of that rate?. My steps and thoughts. I converted $25~^{\circ}\mathrm{C}$ to $298~\mathrm{K}$ and then I was thinking of using $pV=nRT$ and I got stuck. | I am thinking of using molar mass, but I don't know how to incorporate that.. $$\frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{\text{M}_2}{\text{M}_1}}$$. I don't know what the OP meant by $\ce{NH}$, I assume the gas in question is $\ce{NH3}$ with rate of effusion ($R_1$) = 0.050 $\mathrm{mol \min^{-1}}$.. $R_2 = \frac{R_1}{2}$ and $M_1 = 17.031$, plugging all of this in. $M_2 = M_1\cdot \left (\frac{R_1}{R_2} \right )^2 = 4\cdot M_1 = 68.124$. |
http://mathhelpforum.com/advanced-math-topics/214608-sketching-regions-complex-plane.html | 1,496,115,919,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463613780.89/warc/CC-MAIN-20170530031818-20170530051818-00199.warc.gz | 283,491,625 | 10,727 | # Thread: Sketching regions in the complex plane
H
2. ## Re: Sketching regions in the complex plane
Originally Posted by redtdc
How would I sketch the region S={z in the complex plane: abs(z-3i)=abs(z-2)} ?
And is S open or closed?
Thanks!
Hi redtdc!
abs(z-3i) is the same as the distance in R2 of a point (x,y) to (0,3).
The region S corresponds to the points in R2 that have the same distance to (0,3) as to (2,0).
What kind of region is that?
3. ## Re: Sketching regions in the complex plane
Originally Posted by redtdc
How would I sketch the region S={z in the complex plane: abs(z-3i)=abs(z-2)} ?
And is S open or closed?
Do you understand that $|z-z_0|$ is the distance from $z\text{ to }z_0~?$
So what is the set $S=\{z:|z-3i|=|z-2|\}~?$
4. ## Re: Sketching regions in the complex plane
|z- 3i| is the distance from z to 3i and |x- 2| is the distance from z to 2 so S consists of points that are equally distant from 3i and 2. Geometrically, the set of all points equally distant from point P and Q is the perpendicular bisector of the segment PQ.
Algebraically, taking z= x+ iy, |z- 3i|= |z- 2| is the same as $\sqrt{(x^2+ (y- 3))^2}= \sqrt{((x- 2)^2+ y^2}$ which is the same as $x^2+ (y- 3)^2= (x- 2)^2+ y^2$.
As for whether it is open or close, what topology are you using? | 434 | 1,298 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2017-22 | longest | en | 0.930794 | # Thread: Sketching regions in the complex plane. H. 2. ## Re: Sketching regions in the complex plane. Originally Posted by redtdc. How would I sketch the region S={z in the complex plane: abs(z-3i)=abs(z-2)} ?. And is S open or closed?. Thanks!. Hi redtdc!. abs(z-3i) is the same as the distance in R2 of a point (x,y) to (0,3).. The region S corresponds to the points in R2 that have the same distance to (0,3) as to (2,0).. What kind of region is that?. 3. | ## Re: Sketching regions in the complex plane. Originally Posted by redtdc. How would I sketch the region S={z in the complex plane: abs(z-3i)=abs(z-2)} ?. And is S open or closed?. Do you understand that $|z-z_0|$ is the distance from $z\text{ to }z_0~?$. So what is the set $S=\{z:|z-3i|=|z-2|\}~?$. 4. ## Re: Sketching regions in the complex plane. |z- 3i| is the distance from z to 3i and |x- 2| is the distance from z to 2 so S consists of points that are equally distant from 3i and 2. Geometrically, the set of all points equally distant from point P and Q is the perpendicular bisector of the segment PQ.. Algebraically, taking z= x+ iy, |z- 3i|= |z- 2| is the same as $\sqrt{(x^2+ (y- 3))^2}= \sqrt{((x- 2)^2+ y^2}$ which is the same as $x^2+ (y- 3)^2= (x- 2)^2+ y^2$.. As for whether it is open or close, what topology are you using?. |
https://socratic.org/questions/how-do-you-write-the-combined-function-as-a-composition-of-several-functions-if-#219841 | 1,716,211,570,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058278.91/warc/CC-MAIN-20240520111411-20240520141411-00264.warc.gz | 471,629,041 | 6,154 | # How do you write the combined function as a composition of several functions if f(g(x)) = sqrt (1-x^2) +2?
##### 1 Answer
Feb 1, 2016
If I remember correctly there are many "right" ways to solve this equation. $f \left(g \left(x\right)\right)$ merely means you are plugging a function of $x$ into $g$ into $f$ with each letter being a different function.
For instance: your equation for $g$ could possibly be $\sqrt{1 - {x}^{2}}$.
After, we plug in $x$ into that equation we get the same function ($\sqrt{1 - {x}^{2}}$).
Next step when solving multiple step functions is to plug in your result of your first function (in our case $g$) into the new function ($f$). This is where we need another function to get $\sqrt{1 - {x}^{2}}$ into $\sqrt{1 - {x}^{2}}$$+ 2$.
For this step we may say $\left(f\right) = x + 2$.
When we plug in $\sqrt{1 - {x}^{2}}$ into our new $\left(f\right) = x + 2$ our result is
$\sqrt{1 - {x}^{2}}$$+ 2$, the equation we needed to get to.
Another solution would be to have $\left(g\right) = 1 - {x}^{2}$ and $\left(f\right) = \sqrt{x} + 2$
There are many more solution, just be creative, these are only two! | 359 | 1,141 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 20, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-22 | latest | en | 0.865837 | # How do you write the combined function as a composition of several functions if f(g(x)) = sqrt (1-x^2) +2?. ##### 1 Answer. Feb 1, 2016. If I remember correctly there are many "right" ways to solve this equation. $f \left(g \left(x\right)\right)$ merely means you are plugging a function of $x$ into $g$ into $f$ with each letter being a different function.. For instance: your equation for $g$ could possibly be $\sqrt{1 - {x}^{2}}$.. After, we plug in $x$ into that equation we get the same function ($\sqrt{1 - {x}^{2}}$). | Next step when solving multiple step functions is to plug in your result of your first function (in our case $g$) into the new function ($f$). This is where we need another function to get $\sqrt{1 - {x}^{2}}$ into $\sqrt{1 - {x}^{2}}$$+ 2$.. For this step we may say $\left(f\right) = x + 2$.. When we plug in $\sqrt{1 - {x}^{2}}$ into our new $\left(f\right) = x + 2$ our result is. $\sqrt{1 - {x}^{2}}$$+ 2$, the equation we needed to get to.. Another solution would be to have $\left(g\right) = 1 - {x}^{2}$ and $\left(f\right) = \sqrt{x} + 2$. There are many more solution, just be creative, these are only two!. |
https://wikieducator.org/AfroPhysics/Subject_Materials/Physics/Mechanics/Page3E | 1,618,713,878,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038464146.56/warc/CC-MAIN-20210418013444-20210418043444-00511.warc.gz | 696,959,250 | 8,139 | # Mechanics11/Page3
## interpretation of time local diagrams
Assignment
Desiree and Marc go by bike to the plan lake. They drive off at 8.00 o'clock and arrive around 19.00 again at home.
Along the road are kilometer stones. They note each full hour their position x.
• a) calculation it for each time interval of 1 hour the average speed! Use calculation formulas!
• b) describing it for each time interval the process of the movement in words! Possible interpretation?
• c) calculation it the average speed for the first 4 hours! How can this be plotted?
## 1. Motion equation
We deduce a general equation for straight-line-homogeneous movements: from
$v = \frac{\Delta x}{\Delta t} = \frac{x - x zero}{\Delta t}$
follows
x = x0 + v$\Delta t$
Thus the current position at the end of one time interval can be computed e.g. with well-known initial place x and well-known speed of v.
At present if we take for x0 the place x (0) to t = 0, then those results
1. Motion equation:
x (t) = x0 + v t
## sample calculation
Assignment
Problem 1.6: Flight competition Bussard flies with a falcon around the bet. The bussard gets 14 km projection/lead from the starting point measured, because it is clearly slower with 95 km/h than the falcon (153 km/h).
• a) when and where the falcon catches up the bussard?
• b) who wins the flight up to the summit cross of the Hörnle (40 km from the start )? | 365 | 1,406 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2021-17 | latest | en | 0.862832 | # Mechanics11/Page3. ## interpretation of time local diagrams. Assignment. Desiree and Marc go by bike to the plan lake. They drive off at 8.00 o'clock and arrive around 19.00 again at home.. Along the road are kilometer stones. They note each full hour their position x.. • a) calculation it for each time interval of 1 hour the average speed! Use calculation formulas!. • b) describing it for each time interval the process of the movement in words! Possible interpretation?. • c) calculation it the average speed for the first 4 hours! How can this be plotted?. ## 1. Motion equation. We deduce a general equation for straight-line-homogeneous movements: from. $v = \frac{\Delta x}{\Delta t} = \frac{x - x zero}{\Delta t}$. follows. | x = x0 + v$\Delta t$. Thus the current position at the end of one time interval can be computed e.g. with well-known initial place x and well-known speed of v.. At present if we take for x0 the place x (0) to t = 0, then those results. 1. Motion equation:. x (t) = x0 + v t. ## sample calculation. Assignment. Problem 1.6: Flight competition Bussard flies with a falcon around the bet. The bussard gets 14 km projection/lead from the starting point measured, because it is clearly slower with 95 km/h than the falcon (153 km/h).. • a) when and where the falcon catches up the bussard?. • b) who wins the flight up to the summit cross of the Hörnle (40 km from the start )?. |
http://www.sxlist.com/techref/scenix/lib/math/div/div16or32by16to16_sx.htm | 1,601,436,179,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402101163.62/warc/CC-MAIN-20200930013009-20200930043009-00506.warc.gz | 198,823,801 | 9,415 | # SXMicrocontrollerMathMethod
## 32 by 16 Divison from Nikolai Golovchenko, or How to make use of a 16 bit division routine for 32 bit division
Division routines sometimes can be used for more than they were designed for! For example, it is possible to divide 16 or 32 bit dividend by 16 bit divisor to 16 bit quotient in just one routine.
The idea is simple: you do 16 bit division as usually, but when the dividend is 32 bit, the lower two bytes are used as dividend and the higher two bytes are used as initial value for remainder (which is zero when doing 16 bit division)
To understand how initializing remainder relates to dividend, let's examine the algorithms for two cases:
``` x
q = ---
y
```
1) when x, y, and q are 16 bit unsigned integers, and
2) when x and q are 32 bit unsigned integers, but y is a 16 bit unsigned integer
Note that in some cases it is known that the result of 32 bit by 16 bit division is 16 bit maximum. The trick described on this page works only for 16 bit result.
For example, when speed of a motor is calculated by a Hall sensor pulses (one pulse per revolution) the following formula is used:
``` 60 x 1,000,000 (conversions for minutes and us,
rpm = ------------------ timer is clocked by 1 MHz clock)
16 bit timer value
```
Timer is used to measure the Hall sensor pulse period. For this motor and timer the rpm range is within 915-20,000 rpm, which fits perfectly into 16 bits.
#### 1) 16 bit division
x - 16 bit dividend;
y - 16 bit divisor;
q - 16 bit quotient;
rem - 16 bit remainder;
counter - loop counter.
``` rem q
------- -------
|00|00| |??|??|
------- -------
b1 b0 b1 b0
```
2. Shift left dividend x, and shift left remainder rem, by 1 bit, so that MSb of x was shifted to the remainder's LSb
``` rem x
------- -------
|r1|r0| <- |x1|x0| <- shift left
------- -------
b1 b0 b1 b0
```
3. Subtract divisor y from remainder. If subtraction successful (no borrow), next bit of q is 1. If borrow, next bit of q is 0, and remainder should be restored.
``` -------
|r1|r0| rem
-------
b1 b0
-
-------
|y1|y0| y
-------
b1 b0
```
4. Shift left q and set its LSb according to subtraction result from previous step.
``` q carry(inverted borrow) flag
------- ---
|q1|q0| <- |C| shift in next result bit
------- ---
b1 b0
```
5. Decrement counter and if it's not zero repeat from step 2 (16 iterations)
After that, q contains quotient, rem - remainder, x is garbage (it was shifted out completely), and y is untouched.
#### 2) 32 bit division with 16 bit divisor
x - 32 bit dividend;
y - 16 bit divisor;
q - 32 bit quotient;
rem - 17 bit remainder (remainder may be bigger than 16 bit, for example, when x=0x80000000 and y=0xFFFF);
counter - loop counter.
``` rem q
---------- -------------
|00|00|00| |??|??|??|??|
---------- -------------
b2 b1 b0 b3 b2 b1 b0
```
2. Shift left dividend x, and shift left remainder rem, by 1 bit, so that MSb of x was shifted to the remainder's LSb
``` rem x
---------- -------------
|r2|r1|r0| <- |x3|x2|x1|x0| <- shift left
---------- -------------
b2 b1 b0 b3 b2 b1 b0
```
3. Subtract divisor y from remainder. If subtraction successful (no borrow), next bit of q is 1. If borrow, next bit of q is 0, and remainder should be restored.
``` ----------
|r2|r1|r0| rem
----------
b2 b1 b0
-
-------
|y1|y0| y
-------
b1 b0
```
4. Shift left q and set its LSb according to subtraction result from previous step.
``` q carry(inverted borrow) flag
------------- ---
|q3|q2|q1|q0| <- |C| shift in next result bit
------------- ---
b3 b2 b1 b0
```
5. Decrement counter and if it's not zero repeat from step 2 (32 iterations)
After that, q contains quotient, rem - remainder (16 bit), x is garbage (it was shifted out completely), and y is untouched.
Now imagine that you know the result is always 16 bit (for example). That means that the first 16 iterations through steps 1-5 produce 16 zero bits in quotient (because we know that higher 2 bytes of quotient are zero in that case), x was shifted 16 times to remainder, remainder was never subtracted from. So we can say that with these 16 iterations we actually did the following:
From:
``` rem x
---------- -------------
| 0| 0| 0| <- |x3|x2|x1|x0| <- shift left
---------- -------------
b2 b1 b0 b3 b2 b1 b0
-
y q
------- -------------
|y1|y0| |??|??|??|??|
------- -------------
b1 b0 b3 b2 b1 b0
```
To:
``` rem x
---------- -------------
| 0|x3|x2| <- |x1|x0|??|??| <- shift left
---------- -------------
b2 b1 b0 b3 b2 b1 b0
- y q
------- -------------
|y1|y0| |??|??| 0| 0|
------- -------------
b1 b0 b3 b2 b1 b0
```
And after the next 16 iterations:
``` rem x
---------- -------------
| 0|r1|r0| <- |??|??|??|??| <- shift left
---------- -------------
b2 b1 b0 b3 b2 b1 b0
- y q
------- -------------
|y1|y0| | 0| 0|q1|q0|
------- -------------
b1 b0 b3 b2 b1 b0
```
This is very similar to 16 bit division:
From:
``` rem x
------- -------
| 0| 0| <- |x1|x0| <- shift left
------- -------
b1 b0 b1 b0
- y q
------- -------
|y1|y0| |??|??|
------- -------
b1 b0 b1 b0
```
To:
``` rem x
------- -------
|r1|r0| <- |??|??| <- shift left
------- -------
b1 b0 b1 b0
- y q
------- -------
|y1|y0| |q1|q0|
------- -------
b1 b0 b1 b0
```
``` rem x
------- -------
|x3|x2| <- |x1|x0| <- shift left
------- -------
b1 b0 b1 b0
- y q
------- -------
|y1|y0| | 0| 0|
------- -------
b1 b0 b1 b0
```
x3 should be less than 0x80 though, so that after first left shift the higher bit wouldn't disappear. So we actually can do 31-by-16-to-16 division with a 16-by-16-to-16 one!
### Example code
This example uses slightly different algorithm from the one described above. It does not restore the remainder immidiately after a subtraction causes a borrow. However, the trick still aplies, and because this variant of division routine has an extended remainder (necessary for the non-restoring method to hold the current remainder sign), it can do the full 32-by-16-to-16-bit division.
```; x = 60*1e6/y
;
; x, x+1 - rpm
; y, y+1 - pulse width in 1 us units
;
FindRPM
clr x
mov W, #\$87
mov x+1, W
mov W, #\$93
mov x+2, W
mov W, #\$03
mov x+3, W
jmp div32by16to16
; uint16 x = uint32 x / uint16 y
;
; Input:
; x, x+1, x+2, x+3 - 32 bit unsigned integer dividend (x - lsb, x+3 - msb)
; y, y+1 - 16 bit unsigned integer divisor
; Output:
; x, x+1 - 16 bit unsigned integer quotient
; Temporary:
; counter
; temp - remainder extension
;
; Note: result must fit in 16 bits for routine to work
; correctly
div32by16to16
jmp div16by16loopinit ;or just move the label
; uint16 x = uint16 x / uint16 y
;
; Input:
; x, x+1 - 16 bit unsigned integer dividend (x - lsb, x+1 - msb)
; y, y+1 - 16 bit unsigned integer divisor
; Output:
; x, x+1 - 16 bit unsigned integer quotient
; Temporary:
; counter
; x+2, x+3 - 16 bit remainder
; temp - remainder extension
; Size: 36 instructions
; Max timing: 6+16*(5+14+4)-2+2+3=377 cycles
div16by16
clr x+2 ;clear
clr x+3 ;remainder
div16by16loopinit
clr temp ;clear remainder extension
mov W, #16
mov counter, W
stc ;first iteration will be subtraction
div16by16loop
;shift in next result bit and shift out next
;dividend bit to remainder
rl x ;shift lsb
rl x+1 ;shift msb
rl x+2
rl x+3
rl temp
mov W, y
sb x.0
;subtract divisor from remainder
sub x+2, W
mov W, y+1
sc
movsz W, ++y+1
sub x+3, W
mov W, #1
sc
sub temp, W
jmp div16by16next
mov W, y+1
snc
movsz W, ++y+1
mov W, #1
snc
div16by16next
;carry is next result bit
decsz counter
jmp div16by16loop
;shift in last bit
rl x
rl x+1
ret
```
file: /Techref/scenix/lib/math/div/div16or32by16to16_sx.htm, 10KB, , updated: 2004/6/10 14:40, local time: 2020/9/29 20:22, owner: NG--944, TOP NEW HELP FIND: 3.237.71.23:LOG IN
©2020 These pages are served without commercial sponsorship. (No popup ads, etc...).Bandwidth abuse increases hosting cost forcing sponsorship or shutdown. This server aggressively defends against automated copying for any reason including offline viewing, duplication, etc... Please respect this requirement and DO NOT RIP THIS SITE. Questions?Please DO link to this page! Digg it! / MAKE! How to make use of a 16 bit division routine for 32 bit division
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### Welcome to sxlist.com!
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. | 2,902 | 9,629 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2020-40 | latest | en | 0.899351 | # SXMicrocontrollerMathMethod. ## 32 by 16 Divison from Nikolai Golovchenko, or How to make use of a 16 bit division routine for 32 bit division. Division routines sometimes can be used for more than they were designed for! For example, it is possible to divide 16 or 32 bit dividend by 16 bit divisor to 16 bit quotient in just one routine.. The idea is simple: you do 16 bit division as usually, but when the dividend is 32 bit, the lower two bytes are used as dividend and the higher two bytes are used as initial value for remainder (which is zero when doing 16 bit division). To understand how initializing remainder relates to dividend, let's examine the algorithms for two cases:. ``` x. q = ---. y. ```. 1) when x, y, and q are 16 bit unsigned integers, and. 2) when x and q are 32 bit unsigned integers, but y is a 16 bit unsigned integer. Note that in some cases it is known that the result of 32 bit by 16 bit division is 16 bit maximum. The trick described on this page works only for 16 bit result.. For example, when speed of a motor is calculated by a Hall sensor pulses (one pulse per revolution) the following formula is used:. ``` 60 x 1,000,000 (conversions for minutes and us,. rpm = ------------------ timer is clocked by 1 MHz clock). 16 bit timer value. ```. Timer is used to measure the Hall sensor pulse period. For this motor and timer the rpm range is within 915-20,000 rpm, which fits perfectly into 16 bits.. #### 1) 16 bit division. x - 16 bit dividend;. y - 16 bit divisor;. q - 16 bit quotient;. rem - 16 bit remainder;. counter - loop counter.. ``` rem q. ------- -------. |00|00| |??|??|. ------- -------. b1 b0 b1 b0. ```. 2. Shift left dividend x, and shift left remainder rem, by 1 bit, so that MSb of x was shifted to the remainder's LSb. ``` rem x. ------- -------. |r1|r0| <- |x1|x0| <- shift left. ------- -------. b1 b0 b1 b0. ```. 3. Subtract divisor y from remainder. If subtraction successful (no borrow), next bit of q is 1. If borrow, next bit of q is 0, and remainder should be restored.. ``` -------. |r1|r0| rem. -------. b1 b0. -. -------. |y1|y0| y. -------. b1 b0. ```. 4. Shift left q and set its LSb according to subtraction result from previous step.. ``` q carry(inverted borrow) flag. ------- ---. |q1|q0| <- |C| shift in next result bit. ------- ---. b1 b0. ```. 5. Decrement counter and if it's not zero repeat from step 2 (16 iterations). After that, q contains quotient, rem - remainder, x is garbage (it was shifted out completely), and y is untouched.. #### 2) 32 bit division with 16 bit divisor. x - 32 bit dividend;. y - 16 bit divisor;. q - 32 bit quotient;. rem - 17 bit remainder (remainder may be bigger than 16 bit, for example, when x=0x80000000 and y=0xFFFF);. counter - loop counter.. ``` rem q. ---------- -------------. |00|00|00| |??|??|??|??|. ---------- -------------. b2 b1 b0 b3 b2 b1 b0. ```. 2. Shift left dividend x, and shift left remainder rem, by 1 bit, so that MSb of x was shifted to the remainder's LSb. ``` rem x. ---------- -------------. |r2|r1|r0| <- |x3|x2|x1|x0| <- shift left. ---------- -------------. b2 b1 b0 b3 b2 b1 b0. ```. 3. Subtract divisor y from remainder. If subtraction successful (no borrow), next bit of q is 1. If borrow, next bit of q is 0, and remainder should be restored.. ``` ----------. |r2|r1|r0| rem. ----------. b2 b1 b0. -. -------. |y1|y0| y. -------. b1 b0. ```. 4. Shift left q and set its LSb according to subtraction result from previous step.. ``` q carry(inverted borrow) flag. ------------- ---. |q3|q2|q1|q0| <- |C| shift in next result bit. ------------- ---. b3 b2 b1 b0. ```. 5. Decrement counter and if it's not zero repeat from step 2 (32 iterations). After that, q contains quotient, rem - remainder (16 bit), x is garbage (it was shifted out completely), and y is untouched.. Now imagine that you know the result is always 16 bit (for example). That means that the first 16 iterations through steps 1-5 produce 16 zero bits in quotient (because we know that higher 2 bytes of quotient are zero in that case), x was shifted 16 times to remainder, remainder was never subtracted from. So we can say that with these 16 iterations we actually did the following:. From:. ``` rem x. ---------- -------------. | 0| 0| 0| <- |x3|x2|x1|x0| <- shift left. ---------- -------------. b2 b1 b0 b3 b2 b1 b0. -. y q. ------- -------------. |y1|y0| |??|??|??|??|. ------- -------------. b1 b0 b3 b2 b1 b0. ```. To:. ``` rem x. ---------- -------------. | 0|x3|x2| <- |x1|x0|??|??| <- shift left. ---------- -------------. b2 b1 b0 b3 b2 b1 b0. - y q. ------- -------------. |y1|y0| |??|??| 0| 0|. ------- -------------. b1 b0 b3 b2 b1 b0. ```. And after the next 16 iterations:. ``` rem x. ---------- -------------. | 0|r1|r0| <- |??|??|??|??| <- shift left. ---------- -------------. b2 b1 b0 b3 b2 b1 b0. - y q. ------- -------------. |y1|y0| | 0| 0|q1|q0|. ------- -------------. b1 b0 b3 b2 b1 b0. ```. This is very similar to 16 bit division:. | From:. ``` rem x. ------- -------. | 0| 0| <- |x1|x0| <- shift left. ------- -------. b1 b0 b1 b0. - y q. ------- -------. |y1|y0| |??|??|. ------- -------. b1 b0 b1 b0. ```. To:. ``` rem x. ------- -------. |r1|r0| <- |??|??| <- shift left. ------- -------. b1 b0 b1 b0. - y q. ------- -------. |y1|y0| |q1|q0|. ------- -------. b1 b0 b1 b0. ```. ``` rem x. ------- -------. |x3|x2| <- |x1|x0| <- shift left. ------- -------. b1 b0 b1 b0. - y q. ------- -------. |y1|y0| | 0| 0|. ------- -------. b1 b0 b1 b0. ```. x3 should be less than 0x80 though, so that after first left shift the higher bit wouldn't disappear. So we actually can do 31-by-16-to-16 division with a 16-by-16-to-16 one!. ### Example code. This example uses slightly different algorithm from the one described above. It does not restore the remainder immidiately after a subtraction causes a borrow. However, the trick still aplies, and because this variant of division routine has an extended remainder (necessary for the non-restoring method to hold the current remainder sign), it can do the full 32-by-16-to-16-bit division.. ```; x = 60*1e6/y. ;. ; x, x+1 - rpm. ; y, y+1 - pulse width in 1 us units. ;. FindRPM. clr x. mov W, #\$87. mov x+1, W. mov W, #\$93. mov x+2, W. mov W, #\$03. mov x+3, W. jmp div32by16to16. ; uint16 x = uint32 x / uint16 y. ;. ; Input:. ; x, x+1, x+2, x+3 - 32 bit unsigned integer dividend (x - lsb, x+3 - msb). ; y, y+1 - 16 bit unsigned integer divisor. ; Output:. ; x, x+1 - 16 bit unsigned integer quotient. ; Temporary:. ; counter. ; temp - remainder extension. ;. ; Note: result must fit in 16 bits for routine to work. ; correctly. div32by16to16. jmp div16by16loopinit ;or just move the label. ; uint16 x = uint16 x / uint16 y. ;. ; Input:. ; x, x+1 - 16 bit unsigned integer dividend (x - lsb, x+1 - msb). ; y, y+1 - 16 bit unsigned integer divisor. ; Output:. ; x, x+1 - 16 bit unsigned integer quotient. ; Temporary:. ; counter. ; x+2, x+3 - 16 bit remainder. ; temp - remainder extension. ; Size: 36 instructions. ; Max timing: 6+16*(5+14+4)-2+2+3=377 cycles. div16by16. clr x+2 ;clear. clr x+3 ;remainder. div16by16loopinit. clr temp ;clear remainder extension. mov W, #16. mov counter, W. stc ;first iteration will be subtraction. div16by16loop. ;shift in next result bit and shift out next. ;dividend bit to remainder. rl x ;shift lsb. rl x+1 ;shift msb. rl x+2. rl x+3. rl temp. mov W, y. sb x.0. ;subtract divisor from remainder. sub x+2, W. mov W, y+1. sc. movsz W, ++y+1. sub x+3, W. mov W, #1. sc. sub temp, W. jmp div16by16next. mov W, y+1. snc. movsz W, ++y+1. mov W, #1. snc. div16by16next. ;carry is next result bit. decsz counter. jmp div16by16loop. ;shift in last bit. rl x. rl x+1. ret. ```. file: /Techref/scenix/lib/math/div/div16or32by16to16_sx.htm, 10KB, , updated: 2004/6/10 14:40, local time: 2020/9/29 20:22, owner: NG--944, TOP NEW HELP FIND: 3.237.71.23:LOG IN. ©2020 These pages are served without commercial sponsorship. (No popup ads, etc...).Bandwidth abuse increases hosting cost forcing sponsorship or shutdown. This server aggressively defends against automated copying for any reason including offline viewing, duplication, etc... Please respect this requirement and DO NOT RIP THIS SITE. Questions?Please DO link to this page! Digg it! / MAKE! How to make use of a 16 bit division routine for 32 bit division. After you find an appropriate page, you are invited to your to this massmind site! (posts will be visible only to you before review) Just type in the box and press the Post button. (HTML welcomed, but not the <A tag: Instead, use the link box to link to another page. A tutorial is available Members can login to post directly, become page editors, and be credited for their posts.. Attn spammers: All posts are reviewed before being made visible to anyone other than the poster.. Did you find what you needed? "No. I'm looking for: " "No. Take me to the search page." "No. Take me to the top so I can drill down by catagory" "No. I'm willing to pay for help, please refer me to a qualified consultant" "No. But I'm interested. me at when this page is expanded.". ### Welcome to sxlist.com!. Site supported by. & kind contributors. just like you!. (here's why. Copies of the site on CD. are available at minimal cost.. . |
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# To try 5 management wants an estimate of the
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Unformatted text preview: rom all the newspapers printed during a single day. In this sample, 35 contain some type of nonconforming attribute. Construct and interpret a 90% confidence interval for the proportion of newspapers printed during the day that have a nonconforming attribute. To Try… 5. Management wants an estimate of the proportion of the corporation’s employees who favour a bonus plan. From a random sample of 344 employees, it was found that 261 were in favour of this particular plan. Find a 90% confidence interval estimate of the population proportion that favours this modified bonus plan. p α=.10; zα/2 = z0.05 =1.645; Confidence interval: p ± zα 2 p( − p) 1 n = 261/344 = 0.7587; n = 344 Sample: 344 proportion pbar: 0.75872093 significance α: 0.1 confidence: 0.7587⋅ 0.2413 344 0.7587 ± 0.0379 size n: 90% standard error of the proportion: 0.023068621 0.7587 ± 1.645 Interval Estimation Sampling Distribution: 0.7208 ≤ p ≤ 0.7966 Or using MS Excel: Interval Estimate: Lower: =.7587– NORM.S.INV(1–.10/2)*SQRT(.7587*(1–.7587)/344)=0.720754373 zα/2: 1.644853627 margin of error: 0.037944505 To Try… The operations manager at a large newspaper wants to estimate the proportion of newspapers printed that have a nonconforming attribute (e.g., excessive ruboff, improper page setup, missing or duplicate pages). A random sample of 200 newspapers is selected from all the newspapers printed during a single day. In this sample, 35 contain some type of nonconforming attribute. Construct and interpret a 90% confidence interval for the proportion of newspapers printed during the day that have a nonconforming attribute. p α=.10; zα/2 = z0.05 =1.645; = 35/200 = 0.175; n = 200 6. Confidence interval: p ± zα 2 p (1− p) n Sample: size n: 200 proportion pbar: 0.175 significance α: 0.1 confidence: 90% Lower: =.175– Interval Estimate: NORM.S.INV(1–.10/2)*SQRT(.175*(1–.175)/200)=0.130806514 standard error of the proportion: 0.026867732 0.175 ± 1.645 0.175⋅0.825 200 Interval Estimation 0.175 ± 0.0442 0.1308 ≤ p ≤ 0.2192 Sampling Distribution: Or using MS Excel: Upper: zα/2: 1.644853627...
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Ask a homework question - tutors are online | 709 | 2,383 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2016-50 | longest | en | 0.815327 | 8.3 Estimation Proportion. # To try 5 management wants an estimate of the. This preview shows page 1. Sign up to view the full content.. This is the end of the preview. Sign up to access the rest of the document.. Unformatted text preview: rom all the newspapers printed during a single day. In this sample, 35 contain some type of nonconforming attribute. Construct and interpret a 90% confidence interval for the proportion of newspapers printed during the day that have a nonconforming attribute. To Try… 5. Management wants an estimate of the proportion of the corporation’s employees who favour a bonus plan. | From a random sample of 344 employees, it was found that 261 were in favour of this particular plan. Find a 90% confidence interval estimate of the population proportion that favours this modified bonus plan. p α=.10; zα/2 = z0.05 =1.645; Confidence interval: p ± zα 2 p( − p) 1 n = 261/344 = 0.7587; n = 344 Sample: 344 proportion pbar: 0.75872093 significance α: 0.1 confidence: 0.7587⋅ 0.2413 344 0.7587 ± 0.0379 size n: 90% standard error of the proportion: 0.023068621 0.7587 ± 1.645 Interval Estimation Sampling Distribution: 0.7208 ≤ p ≤ 0.7966 Or using MS Excel: Interval Estimate: Lower: =.7587– NORM.S.INV(1–.10/2)*SQRT(.7587*(1–.7587)/344)=0.720754373 zα/2: 1.644853627 margin of error: 0.037944505 To Try… The operations manager at a large newspaper wants to estimate the proportion of newspapers printed that have a nonconforming attribute (e.g., excessive ruboff, improper page setup, missing or duplicate pages). A random sample of 200 newspapers is selected from all the newspapers printed during a single day. In this sample, 35 contain some type of nonconforming attribute. Construct and interpret a 90% confidence interval for the proportion of newspapers printed during the day that have a nonconforming attribute. p α=.10; zα/2 = z0.05 =1.645; = 35/200 = 0.175; n = 200 6. Confidence interval: p ± zα 2 p (1− p) n Sample: size n: 200 proportion pbar: 0.175 significance α: 0.1 confidence: 90% Lower: =.175– Interval Estimate: NORM.S.INV(1–.10/2)*SQRT(.175*(1–.175)/200)=0.130806514 standard error of the proportion: 0.026867732 0.175 ± 1.645 0.175⋅0.825 200 Interval Estimation 0.175 ± 0.0442 0.1308 ≤ p ≤ 0.2192 Sampling Distribution: Or using MS Excel: Upper: zα/2: 1.644853627.... View Full Document. Ask a homework question - tutors are online. |
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