tsarratori
A dot labeled with a letter; A precise location
Line Segment
Two points connected in a line that ends at both points.
Straight line that continues in both directions forever.
A line with 1 end point that extends in the other direction forever.
The place where 2 lines or rays meet.
An angle that is 90 degrees
An angle that is less than 90 degrees
An angle that is more that 90 degrees
Another name for the angle.
Straight Angle
An angle that is 180 degrees
Perpendicular Angle
Lines that meet to form right angles.
Two lines that would never touch.
A triangle where all 3 sides are the same length.
A triangle where 2 of the sides are the same length.
A triangle where none of the sides are the same length.
A triangle where all 3 angles are acute.
A triangle with 1 obtuse angle.
A triangle with 1 right angle.
A polygon with 4 sides
A quadrilateral with 1 set of parallel lines.
A quadrilateral with 2 sets of parallel sides.
A quadrilateral with 4 right angles. Not all sides are the same length.
A quadrilateral with 4 right angles where all of the sides are the same length.
A quadrilateral with 4 equal sides and NO right angles.
List 12 - "Er" Sounds
List 10 - Double Consonants
List 9 - Long E Sounds
List 8 - Long O Sounds
Either find a number k such that the given function is a probability density function or explain why no such number exists. $$ f(x)=\left\{\begin{array}{ll}{x-k x^{2}} & {\text { for } 0 \leq x \leq 1} \\ {0} & {\text { otherwise }}\end{array}\right. $$
Use the method of reduction of order to find the general solution of the following differential equations. $t^{2} \frac{d^{2} y}{d t^{2}}+t \frac{d y}{d t}+\left(t^{2}-\frac{1}{4}\right) y=0 \quad\left(y_{1}(t)=\frac{\sin t}{\sqrt{t}}\right)$
Suppose that an object can be at any one of n+1 equally spaced points $x_{0}, x_{1}, \ldots, x_{n}.$ When an object is at location $x_i,$ it is equally likely to move to either $x_{i-1} \text { or } x_{i+1}$ and cannot directly move to any other location. Consider the probabilities $\left\{P_{i}\right\}_{i=0}^{n}$ that an object starting at location $x_i$ will reach the left endpoint $x_0$ before reaching the right endpoint $x_n.$ Clearly, $P_{0}=1 \text { and } P_{n}=0.$ Since the object can move to $x_i$ only from $x_{i-1} \text { or } x_{i+1}$ and does so with probability $\frac{1}{2}$ for each of these locations, $P_{i}=\frac{1}{2} P_{i-1}+\frac{1}{2} P_{i+1}$ for each $i=1,2, \ldots, n-1$ a. Show that $$ \left[\begin{array}{rrrrrrr}{1} & -\frac{1}{2} & {0} \cdots & \cdots & \cdots & \cdots \\ -\frac{1}{2} & {1} & -\frac{1}{2} & \ddots & {} & {} & \vdots \\ {0} & -\frac{1}{2} & {1} & \ddots & \ddots & {} & \vdots \\ \vdots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots \\ \vdots & {} & \ddots & \ddots & \ddots & \ddots & \ddots & {0} \\ \vdots & {} & {} & \ddots & -\frac{1}{2} & {1} & -\frac{1}{2} \\ {0} & \cdots & \cdots & \cdots & \cdots & 0 -\frac{1}{2} & {1}\end{array}\right]\left[\begin{array}{c}P_{1} \\ {P_{2}} \\ \vdots \\ P_{n-1}\end{array}\right]=\left[\begin{array}{c}\frac{1}{2} \\ {0} \\ \vdots \\ {0}\end{array}\right] $$ b. Solve this system using n = 10, 50, and 100. c. Change the probabilities to $\alpha$ and $1−\alpha$ for movement to the left and right, respectively, and derive the linear system similar to the one in part (a). d. Repeat part (b) with $\alpha = \frac{1}{3}.$
These three points are on the same line: (1,2,3),(2,4,−6),(−3,−6,x). Find x. Explain your answer fully.
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