13.3: Free damped vibrations
In practice a vibration is never completely undamped, since there are always small practical imperfections that cause some kind of dissipation due to nonconservative forces. The most common type of damping, that is also most easily analysed, is by a velocity proportional force \(\overrightarrow{\boldsymbol{F}}_{c}=-c \overrightarrow{\boldsymbol{v}}\), that is generated by a piston or other linear damper element. The force points in the opposite direction from the velocity vector of the mass, and thus the power done by such a force on the mass is given by \(P=\overrightarrow{\boldsymbol{F}}_{c} \cdot \overrightarrow{\boldsymbol{v}}=-c|\overrightarrow{\boldsymbol{v}}|^{2}\), which is always negative. This causes the kinetic and potential energy of the mass-spring system to reduce to zero.
When connecting a linear damper to the mass, in parallel to the spring, as shown in Figure 13.5, we obtain a mass-spring-damper system, we have from Newton’s second law that \(\overrightarrow{\boldsymbol{F}}_{k}+\overrightarrow{\boldsymbol{F}}_{c}=m_{A} \overrightarrow{\boldsymbol{a}}_{A}\), which when projected along the \(y\)-axis, results in the following equation of motion:
\[\begin{align} \sum F_{y}=-k y_{A}-c \dot{y}_{A} & =m_{A} \ddot{y}_{A} \tag{13.29} \label{13.29}\\[4pt] m_{A} \ddot{y}_{A}+c \dot{y}_{A}+k y_{A} & =0 \tag{13.30} \label{13.30}\end{align}\]
Here we note that normally \(m_{A}, c\) and \(k\) are all positive real constants and in static equilibrium \(\dot{y}_{A}=0\), such that the damper does generate force and does not affect \(y_{A, \mathrm{st}}\).
To solve this equation of motion we propose the following complex trial function:
\[y_{A}(t)=\Re A_{c} e^{\lambda t} \tag{13.31} \label{13.31}\]
In this equation both \(A_{c}\) and \(\lambda\) are complex numbers with a nonzero real and imaginary part. Note that we only use a complex trial function and don’t use a trial function with trigonometric equations since working with complex numbers is a lot easier when working with damped vibrations. Substitution in Equation 13.30 results in:
\[\Re\left(\left[m_{A} \lambda^{2}+c \lambda+k\right] A_{c} e^{\lambda t}\right)=0 \tag{13.32} \label{13.32}\]
Since this equation has to hold for all times \(t\), the quadratic equation in the square brackets has to be zero, and can be solved, to determine \(\lambda\), using the quadratic formula or ’ \(A B C\) formula’, with \(A=m_{A}, B=c\) and \(C=k\) :
\[\begin{align} & m_{A} \lambda^{2}+c \lambda+k=0 \tag{13.33} \label{13.33}\\[4pt] \lambda= & \frac{1}{2 A}\left(-B \pm \sqrt{B^{2}-4 A C}\right) \tag{13.34} \label{13.34}\\[4pt] \lambda= & \frac{1}{2 m_{A}}\left(-c \pm \sqrt{c^{2}-4 m_{A} k}\right)=\frac{-c \pm \sqrt{\Delta}}{2 m_{A}} \tag{13.35} \label{13.35}\end{align}\]
The term \(\Delta=c^{2}-4 m_{A} k\) is called the discriminant and depending on the sign of \(\Delta\), we obtain three different values of \(\lambda\) that correspond to three quite different types of solutions:
- \(\Delta>0\) : Overdamped vibration, the values of \(\lambda\) are both real and negative numbers with two solutions \(\lambda_{o, \pm}=\frac{-c \pm \sqrt{\Delta}}{2 m_{A}}\).
- \(\Delta<0\) : Underdamped vibration, the values of \(\lambda\) are complex numbers with a negative real part \(-c /\left(2 m_{A}\right)\). When taking the square-root of a negative number we get \(\sqrt{-1}=i\), such that:
\[\frac{\sqrt{\Delta}}{2 m_{A}}=i \frac{\sqrt{4 m_{A} k-c^{2}}}{2 m_{A}}=i \sqrt{\omega_{n}^{2}-\frac{c^{2}}{4 m_{A}^{2}}} \equiv i \omega_{d} \tag{13.36} \label{13.36}\]
On the right side of this equation we have defined the damped angular resonance frequency \(\omega_{d} \equiv \sqrt{\omega_{n}^{2}-c^{2} /\left(4 m_{A}^{2}\right)}\), and it can be seen that for small damping \(c\), this damped resonance frequency is almost equal to the natural angular resonance frequency \(\left(\omega_{d} \approx \omega_{n}\right)\). We then obtain two values for \(\lambda\) that solve the EoM: \(\lambda_{u, \pm}=\frac{-c}{2 m_{A}} \pm i \omega_{d}\).
3. \(\Delta=0\) : Critically damped vibration. In this case there is only one value \(\lambda=-c /\left(2 m_{A}\right)\). The value of the damping coefficient that belongs to this situation \(\Delta=c_{c}^{2}-4 m_{A} k=0\) is called the critical damping coefficient \(c_{c}=2 \sqrt{m_{A} k}\). For \(c>c_{c}\) we have overdamped vibrations and for \(c<c_{c}\) we get underdamped vibrations.
Let us now determine the different solutions Equation 13.31 of the EoM with these values of \(\lambda\).
13.3.1 Overdamped motion
Since \(\lambda\) is a real number for \(\Delta>0\), we need to add the solutions of the EoM for both values \(\lambda_{o, \pm}=\frac{-c \pm \sqrt{\Delta}}{2 m_{A}}\) to get a function that satisfies the equation of motion, while introducing two real constants \(A_{+}\)and \(A_{-}\)that can be adjusted to satisfy the initial position and velocity conditions. We find:
\[\begin{gathered} y_{A, o}(t)=A_{+} e^{\lambda_{o,+} t}+A_{-} e^{\lambda_{o,-} t} \tag{13.37} \label{13.37}\\[4pt] y_{A, o}(t)=A_{+} e^{\frac{-c+\sqrt{\Delta}}{2 m_{A}} t}+A_{-} e^{\frac{-c-\sqrt{\Delta}}{2 m_{A}} t} \tag{13.38} \label{13.38}\end{gathered}\]
13.3.2 Underdamped motion
For the case \(\Delta<0\), the complex number \(A_{c}=A e^{i \varphi_{0}}\) can be substituted in Equation 13.31. Using the positive value \({ }^{2} \lambda_{u,+}=\frac{-c}{2 m_{A}}+i \omega_{d}\) we obtain the solution:
\[y_{A, u}(t)=A e^{-\frac{c}{2 m_{A}} t} \Re e^{i\left(\omega_{d} t+\varphi_{0}\right)}=A e^{-\frac{c}{2 m_{A}} t} \cos \left(\omega_{d} t+\varphi_{0}\right) \tag{13.39} \label{13.39}\]
The motion consists of a fast oscillating cosine function with a slowly decaying amplitude given by the exponential function. Just like in Sec. 13.2.4 one might also choose to replace \(\cos \left(\omega_{d} t+\varphi_{0}\right)\) by the sum of a cosine and sine function, in which case the solution is written as:
\[y_{A, u}(t)=e^{-\frac{c}{2 m_{A}} t}\left[A_{1} \cos \left(\omega_{d} t\right)+A_{2} \sin \left(\omega_{d} t\right)\right] \tag{13.40} \label{13.40}\]
13.3.3 Critically damped motion*
For the general solution of critically damped motion we have a problem, since we have found only one value of \(\lambda\) that corresponds to the real solution \(y_{A}=A_{0} e^{-c /\left(2 m_{A}\right) t}\). For this specific case it can by shown, as can be checked by substitution in Equation 13.30, that the function \(y_{A}=A_{t} t e^{-c /\left(2 m_{A}\right) t}\) also obeys the EoM, such that the general solution for critically damped vibrations becomes:
\[y_{A, c}(t)=A_{0} e^{-\frac{c}{2 m_{A}} t}+A_{t} t e^{-\frac{c}{2 m_{A}} t} \tag{13.41} \label{13.41}\]
As can be seen in Figure 13.5 the critically damped vibrations approach zero faster than the overdamped vibrations with the same damping \(c / m_{A}\).
13.3.4 More complex free vibrations
Up to now we only dealt with mass-spring and mass-spring-damper systems. However, more complex systems can also result in the same equation of motion. For instance, applying Euler’s second law to rotating rigid bodies connected to springs leads to an EoM that is similar to that of the free vibration of a mass-spring-damper system. Even though the constants and variables in that EoM are different from \(y_{A}, m_{A}, k\) and \(c\), the EoM can be written similarly by introducing effective functions \(y_{e}, m_{e}, k_{e}\) and \(c_{e}\), such that the EoM becomes:
\[m_{e} \ddot{y}_{e}+c_{e} \dot{y}_{e}+k_{e} y_{e}=0 \tag{13.42} \label{13.42}\]
Here, the function \(y_{e}(t)\) can represent a position coordinate but can also be an angle, or any generalised coordinate that uniquely describes the position or orientation of the system. Mathematically, the motion of these more complex systems can be derived in exactly the same way as a mass-spring-damper system, like discussed in previous sections. Only the constants \(m_{e}, c_{e}\) and \(k_{e}\) and function name \(y_{e}(t)\) are different. So, everything we have derived for the mass-spring-damper system remains valid for these more complex systems.
13.3.5 Free vibrations: solution procedure
Finally, we discuss step-by-step the procedure for analysing free vibrations discussed in this section and the previous one.
- Sketch the system and coordinate system.
- Draw the free-body diagram and project the force and/or moment vectors on the coordinate axes. Determine the relevant constraint equations.
- Use Newton’s second law and Euler’s first law for translations and Euler’s second law for rotations to determine the scalar equation of motion along the relevant translation or rotation coordinate.
- Determine the static equilibrium position \(y_{A, \mathrm{st}}\) by setting all time derivatives in the EoM to zero \(\ddot{y}_{A, \text { tot }}=\dot{y}_{A, \text { tot }}=0\) and solving for \(y_{A}\).
- Define the time dependent part of the motion as \(y_{A}(t)=y_{A, \mathrm{tot}}-y_{A, \mathrm{st}}\).
- The system will only show free vibration if at least one of the EoMs can be written in the form \(m_{e} \ddot{y}_{e}+c_{e} \dot{y}_{e}+k_{e} y_{e}=0\).
- Determine the equations for the variable \(y_{e}\) and constants \(m_{e}, c_{e}\) and \(k_{e}\). Now one can either solve the ODE, like in the next step 8 , or directly go to step 9 and use the known solutions.
- To solve the ODE, substitute the trial function \(y_{e}=\Re A_{c} e^{\lambda t}\) and its time-derivatives into the EoM to obtain \(-m_{e} \lambda^{2}+c_{e} \lambda+k_{e}=0\). Solve for \(\lambda\) and substitute it back into the trial function to find the motion \(y_{e}(t)\) like explained in the previous sections.
- If \(c_{e}=0\), then we are dealing with free undamped vibration. In that case the solution of the EoM is \(y_{e}(t)=\Re A e^{i \omega_{n} t+\varphi_{0}}=A \cos \left(\omega_{n} t+\varphi_{0}\right)\), with natural resonance frequency \(\omega_{n}=\sqrt{k_{e} / m_{e}}\).
- If \(c_{e} \neq 0\) we are dealing with free damped vibration.
Determine the discriminant \(\Delta=c_{e}^{2}-4 m_{e} k_{e}\).
- If \(\Delta>0\) we are dealing with overdamped free vibration with \(y_{e}=A_{+} e^{\lambda_{+} t}+A_{-} e^{\lambda-t}\) and \(\lambda_{ \pm}=\frac{-c_{e} \pm \sqrt{\Delta}}{2 m_{e}}\).
- If \(\Delta<0\) we are dealing with underdamped free vibration with \(y_{e}=A e^{-\frac{c_{e}}{2 m_{e}} t} \cos \left(\omega_{d} t+\varphi_{0}\right)\) and \(\omega_{d}=\sqrt{\omega_{n}^{2}-c_{e}^{2} /\left(4 m_{e}^{2}\right)}\)
- If \(\Delta=0\) we are dealing with critically damped free vibration with \(y_{e}=A_{0} e^{-\frac{c_{e}}{2 m_{e}} t}+A_{t} t e^{-\frac{c_{e}}{2 m_{e}} t}\).
- Two initial conditions of the form \(y_{e}\left(t_{1}\right)=y_{1}\) or \(\dot{y}_{e}\left(t_{2}\right)=v_{2}\) need to be solved to determine the two unknown constants in the solution \(\left(A \& \varphi_{0}\right.\), \(A_{+} \& A_{-}\)or \(A_{0} \& A_{t}\) ). See Sec. 13.2.5.
- In some cases it is convenient to rewrite the cosine with phase-shift \(\varphi_{0}\) into a sum of a cosine and sine using the mathematical equation:
\[A \cos \left(\omega t+\varphi_{0}\right)=A_{1} \cos (\omega t)+A_{2} \sin (\omega t) \tag{13.43} \label{13.43}\]
where \(A^{2}=A_{1}^{2}+A_{2}^{2}\) and \(\tan \left(\varphi_{0}\right)=-A_{2} / A_{1}\).
Consider the car \(A\), in Figure 13.6 with a mass of \(m=1000 \mathrm{~kg}\) that can be considered as a point mass. It has a spring stiffness of \(k=50000 \mathrm{~N} / \mathrm{m}\) and a damping constant of \(c=1000 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}\). The car drops from a small height, such that it reaches its equilibrium position \(y_{A}(0)=0\) with a speed of \(v_{0}=-1 \mathrm{~m} / \mathrm{s}\) at \(t=0\). Determine the motion \(y_{A}(t)\) of the car.
Solution
We follow the steps from this section. The sketch and FBD are given in Figure 13.6.
So, we project the forces and write down the equation of motion along the \(y\)-axis:
\[\begin{align} \sum F_{y}=F_{k, y}+F_{c, y}+F_{g, y} & =m_{A} \ddot{y}_{A, \mathrm{tot}} \tag{13.44} \label{13.44}\\[4pt] -k\left(y_{A, \mathrm{tot}}-L_{0}\right)-c \dot{y}_{A, \mathrm{tot}}-m_{A} g & =m_{A} \ddot{y}_{A, \mathrm{tot}} \tag{13.45} \label{13.45}\\[4pt] -k y_{A}-c \dot{y}_{A} & =m_{A} \ddot{y}_{A} \tag{13.46} \label{13.46}\\[4pt] m_{A} \ddot{y}_{A}+k y_{A}+c \dot{y}_{A} & =0 \tag{13.47} \label{13.47}\end{align}\]
Here we used that \(y_{A, \mathrm{st}}=-\frac{m_{A} g}{k}+L_{0}\) like in Equation 13.2, and \(y_{A, \text { tot }}=y_{A}+y_{A, \mathrm{st}}\) to eliminate the static particular solution. Although, since we are only asked for the motion with respect to the equilibrium position, we could also have skipped this step, and directly eliminate the static terms. Now we see that the equation of motion Equation 13.47 can be written in the form of Equation 13.42, with \(y_{e}=y_{A}, m_{e}=m_{A}\), \(c_{e}=c\) and \(k_{e}=k\) (we will encounter more complex cases later). Now we can solve this EoM, like in Sec. 13.3. We will not repeat the derivation and directly substitute the values of \(m_{e}, c_{e}\) and \(k_{e}\) in the solution. We first determine that the discriminant \(\Delta=c^{2}-4 m_{A} k=-199 \times 10^{6} \mathrm{~kg}^{2} / \mathrm{s}^{2}\) is negative, such that we deal with an underdamped vibration. Then we determine the damped resonance frequency as \(\omega_{d}=\sqrt{k_{e} / m_{e}-c_{e} /\left(4 m_{e}^{2}\right)}\). The solution can then be written as:
\[y_{A}(t)=A e^{-\left(c_{e} / 2 m_{e}\right) t} \cos \left(\omega_{d} t+\varphi_{0}\right) \tag{13.48} \label{13.48}\]
Finally we need to determine the constants \(A\) and \(\varphi_{0}\) from the initial conditions. Since \(y_{A}(t=0)=0\), we find that \(\varphi_{0}=\pi / 2\). By taking the time derivative of \(y_{A}\), with the product rule and using that \(\dot{y}_{A}(0)=v_{0}\) we find:
\[\begin{align} \dot{y}_{A}(0) & =\left[-\left(c_{e} / 2 m_{e}\right) \cos \left(\omega_{d} 0+\pi / 2\right)-\omega_{d} \sin \left(\omega_{d} 0+\pi / 2\right)\right] A e^{-\left(c_{e} / 2 m_{e}\right) 0}=v_{0} \\[4pt] & =-\omega_{d} A=v_{0}=-1 \mathrm{~m} / \mathrm{s} \tag{13.49} \label{13.49}\end{align}\]
from which we find \(A\). Thus we have fully determined all constants in Equation 13.48, and fully determined the motion \(y_{A}(t)\).
A Example 13.3 For the car in the previous example we now design the optimal damping, that reduces the time for the amplitude to go to zero. The initial speed \(v_{0}\) and kinetic energy of the car is kept constant and the damping coefficient \(c\) is varied from underdamped \(c=\frac{1}{2} c_{c}\), to critically damped \(c=c_{c}\) to overdamped \(c=2 c_{c}\).
Plot the motion of the car for these three situations, and determine the optimal value of \(c\).
First we determine the constants \(\omega_{d}=\sqrt{\frac{k}{m}-\frac{c^{2}}{4 m^{2}}}, c_{c}=2 \sqrt{k m}\) and \(\Delta=c^{2}-4 m k\), and define \(\omega_{d o}=\sqrt{\Delta} /(2 m)\). Then we use the initial conditions \(y_{A}(0)=0\) and \(\dot{y}_{A}(0)=v_{0}=-1 \mathrm{~m} / \mathrm{s}\), to determine the constants \(A_{*}\) and \(\varphi_{0}\) for the underdamped, critically damped and overdamped motion. Check with some math that \(\phi_{0}=\Pi / 2\) this results in:
\[\begin{align} & y_{A, u}(t)=v_{0} t e^{-\frac{c}{2 m} t \frac{\sin \left(\omega_{d} t\right)}{\omega_{d} t}} \tag{13.50} \label{13.50}\\[4pt] & y_{A, c}(t)=v_{0} t e^{-\frac{c}{2 m} t} \tag{13.51} \label{13.51}\\[4pt] & y_{A, o}(t)=v_{0} t e^{-\frac{c}{2 m} t} \frac{\sinh \left(\omega_{d o} t\right)}{\omega_{d o} t} \tag{13.52} \label{13.52}\end{align}\]
We see that for a damping coefficient \(c=c_{c}\) the frequencies \(\omega_{d}\) and \(\omega_{d o}\) approach zero and because for small \(x\) we have \(\sin x \approx x\) and \(\sinh x \approx x\), the three expressions become equal. In Figure 13.7 it can be seen the that the solution approaches zero fastest when the damper is chosen to be critically damped with \(c=c_{c}\). The underdamped motion damps slower because of the smaller value of \(c\) in the exponent, and the overdamped motion is slower due to the slow decay of the \(A_{+}\)term in Equation 13.38.


5.5: Liquidity and Liquidity Ratios
Note:
Toward the end of chapter 5, the student will find a “Ratio Analysis Exercise.” As we go through and explain each of the twenty financial ratios below, the student will, one-by-one, calculate the ratios for the example given in the exercise. You will find that the example is constituted by a fictitious company’s Balance Sheet and Income Statement.
We are going to list 20 ratios – covering six categories – in this section. After this discussion, an exercise in calculating all the ratios is presented so that you may apply what you learned .
The first category is liquidity ratios. The idea of liquidity has to do with the ease and speed with which an asset may be converted into cash – without compromising its “true” worth or “intrinsic value.” It must satisfy both conditions in order to be considered liquid.
For instance, it is possible to sell virtually anything at a “ fire – sale price, ” if one needs the cash. The question is whether an item worth, say $100, will fetch $100 – or less. If less, we may say that the item is not liquid, as the seller did not obtain its true worth, even though s/he was able to obtain some cash for it.
An example of a liquid asset would be IBM stock. Let’s say it is trading for $ 1 90. Should one wish to sell 1 0 0 shares of the stock, s/he would obtain approximately $ 1 9,000 for it. However, if I wanted to sell my tie, for which I paid $50, it may be difficult for me to get that amount. Whether you wish to consider the asset’s “ value ” a matter of its original cost or its “current value,” is a matter of analytic choice and context .
The notion of liquidity is closely tied to another notion: “marketability.” This has to do with the availability of a market in which purchases and sales may be readily transacted. For instance, the stock market provides IBM common shares with a great deal of marketability. Should I however wish to sell my tie, I would not know where to go to find a “ used- tie market.” Real estate markets are dominated by brokers who advertise frequently, and I may easily sell my car to a used car dealer.
Now that we know what is meant by liquidity, let’s quantify the notion by detailing some relevant ratios. The greater the liquidity the less the risk that the firm will not have sufficient funds to defray its short-term requirements.
Current Ratio = Current Asset s Divided by Current Liabilities = CA ÷ CL
The current ratio gauges the extent to which a corporation has the liquidity with which it may honor its short-term obligations or liabilities. Does it have the cash to pay its suppliers when accounts payable are due? Are inventories and accounts receivable sufficient (and liquid) to cover its current liabilities? We will assume, for now that the inventory and accounts receivable are completely liquid.
It stands to reason that a watermark level for this ratio should be 1:1 or 1x (“one times”) ; one would not want to see this ratio go below that figure as it would imply that the company does not have sufficient current assets and , hence, the liquidity to meet its current obligations. Many analysts agree that 1.5x or 2x are reasonable figures, but we should be leery of such arbitrary minima. Companies that are very efficient in the manner in which they manage their current assets may have current ratios close to 1. In general, ratios will vary from industry to industry. Note that this ratio is affected by the choice of inventory costing method (e.g., FIFO vs. LIFO) because inventory is a constituent of current assets .
Note: Working Capital = Current Assets less Current Liabilities. If the current ratio is less than one, the working capital will be negative. (See Ratio Exercise.)
Quick Ratio (“Acid Test”) = (CA – Inv en t or y) ÷ CL
Some analysts take a more conservative approach in the assessment of liquidity coverage by assuming that inventory is worthless (even though it may, in fact, not be – or not be entirely) and hence that current liabilities will be covered by all current assets except inventory. Indeed, in some industries inventory quickly becomes obsolete due to technological advances ; in other cases, inventory may be easily damaged, or lose its fashion and sales appeal.
The current and quick ratios may be thought of as “coverage” ratios in that they indicate the extent to which liquid assets are adequate to “cover,” or pay off, current liabilities, especially accounts payable and notes payable (i.e., short-term loans).
Average Collection Period ( ACP ) = Days sales outstanding (DSO) = (A/R ÷ Credit Sales) (360)
“A/R” stands for Accounts Receivable. This ratio tells the analyst approximately how long it takes the firm to collect its receivables; the ratio is expressed in days. The shorter the collection period, the more liquid the firm’s accounts receivable.
Note that we use credit sales because only credit (and not cash) sales become accounts receivable. This ACP figure may then be compared to the firm’s typical credit terms. If, for argument’s sake, the ACP exceeds the firm’s credit terms, which may be 30 days, the analyst may first assess whether the firm is having collection problems. Alternatively, the analyst may investigate whether the firm is acting aggressively in its marketing, and choosing to live with the consequences of some late payments as a result. Perhaps the firm expects that the incremental profits will exceed any losses in its collections.
This ratio is our first case, so far, of “apples and oranges” in that, here, we note that both income statement and balance sheet number s appear within the same ratio. This inconsistency needs to be fixed ; we ought not to combine one datum from the Balance Sheet , which is a static figure , with an other from the Income Statement , which is a flow kind of number , within one ratio . The two are inconsistent with each other.
In such cases as this, what the analyst may choose to do is “average” the balance sheet number. One way of doing this would be by taking this year’s and last year’s year-end numbers and averaging them(i.e., add the two numbers up and divide by two). We could also use quarterlies or, if available, monthlies. In businesses where receivables fluctuate over the course of the year and may thus be either relatively high or low at certain times of year or during certain seasons, it may be better to average the quarterly numbers, if possible.
This technique would then capture a better sense of movement over the course of the seasonal business cycle and would thus provide a better comparison with the income statement number, which , again, flow over the course of the entire year. If the balance sheet number does not fluctuate – much, or at all – over time, there may be no need for averaging. This choice is at the discretion of the analyst. Compare Companies “A” and “B” in terms of their accounts receivable or inventory levels , as it were, over the course of four quarters. Is anything accomplished by averaging t h e Balance Sheet numbers for Company A?
You will also note that (usually) A ccounts R eceivable will display a smaller number than Sales . Normally, we would expect that accounts receivable are collected more or less monthly ; indeed, if sales terms are 30 days, we should get about 12 collections per year. This is illustrated on the next page.
We shall also assume a 360-day year, consisting of twelve 30-day rolling months , because most often collections are demanded within 30 days . S ome analysts will use a 365-day year on the rationale that C redit S ales accumulat e over a 365-day calendar year . It’s your choice, Ms/r. Analyst!
Final note: If you are analyzing data for just one quarter, you should use 90, rather than 360, as the multiplier for the ACP . For two or three quarters, you will use 180 and 270 respectively .
Inventory Turnover = COGS ÷ Inventory
This ratio tells us how often the inventory “ turns over , ” that is to say, how often the inventory – in its entirety – is replaced. Would you buy fresh tomatoes from a company whose inventory turnover is 10x a year (or every 36 days)? Note that this ratio is affected by the choice of inventory costing method. This ratio has some issues with mixing Balance Sheet and Income Statement numbers, which was discussed the prior section concerning ACP .


15.11B: Types of Cells in the Pancreas
The islets of Langerhans are the regions of the pancreas that contain many hormone-producing endocrine cells.
- Distinguish between the cell types of the pancreas
Key Points
- The pancreas reveals two different types of parenchymal tissue: exocrine acini ducts and the endocrine islets of Langerhans.
- The hormones produced in the islets of Langerhans are insulin, glucagon, somatostatin, pancreatic polypeptide, and ghrelin.
- The pancreatic hormones are secreted by alpha, beta, delta, gamma, and epsilon cells.
Key Terms
- somatostatin : A polypeptide hormone, secreted by the pancreas, that inhibits the production of certain other hormones.
- insulin : A polypeptide hormone that regulates carbohydrate metabolism.
- glucagon : A hormone, produced by the pancreas, that opposes the action of insulin by stimulating the production of sugar.
Pancreatic Cells
The pancreas is a glandular organ that belongs to both the digestive and the endocrine systems of vertebrates. It is an endocrine gland that produces several important hormones, including insulin, glucagon, somatostatin, and pancreatic polypeptide.
It is also a digestive, exocrine organ, that secretes pancreatic juice that contains digestive enzymes to assist with digestion and the absorption of nutrients in the small intestine. These enzymes help to further break down the carbohydrates, proteins, and lipids in the chyme.
Under a microscope, stained sections of the pancreas reveal two different types of parenchymal tissue. The light-stained clusters of cells are called islets of Langerhans, which produce hormones that underlie the endocrine functions of the pancreas.
The dark-stained cells form acini , connected to ducts. Acinar cells belong to the exocrine pancreas and secrete digestive enzymes into the gut via a system of ducts.
Islets of Langerhans
Islets of Langerhans : A porcine islet of Langerhans. On the left is a brightfield image created using hematoxylin stain; nuclei are dark circles and the acinar pancreatic tissue is darker than the islet tissue. The right image is the same section stained by immunofluorescence against insulin, indicating beta cells.
The pancreatic islets are small islands of cells that produce hormones that regulate blood glucose levels. Hormones produced in the pancreatic islets are secreted directly into the blood flow by five different types of cells. The endocrine cell subsets are:
- Alpha cells that produce glucagon, and make up 15–20% of total islet cells. Glucagon is a hormone that raises blood glucose levels by stimulating the liver to convert its glycogen into glucose.
- Beta cells that produce insulin and amylin, and make up 65–80% of the total islet cells. Insulin lowers blood glucose levels by stimulating cells to take up glucose out of the blood stream. Amylin slows gastric emptying, preventing spikes in blood glucose levels.
- Delta cells that produce somatostatin, and make up 3–10% of the total islet cells. Somatostatin is a hormone that suppresses the release of the other hormones made in the pancreas.
- Gamma cells that produce pancreatic polypeptide, and make up 3–5% of the total islet cells. Pancreatic polypeptide regulates both the endocrine and exocrine pancreatic secretions.
- Epsilon cells that produce ghrelin, and make up less than 1% of the total islet cells. Ghrelin is a protein that stimulates hunger.
The islets of Langerhans can influence each other through paracrine and autocrine communication. The paracrine feedback system is based on the following correlations:
- The insulin hormone activates beta cells and inhibits alpha cells.
- The hormone glucagon activates alpha cells which then activate beta cells and delta cells.
- Somatostatin hormone inhibits alpha cells and beta cells.


14.4: Nutritional Adaptations of Plants
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- Tom Buckley
- University of California, Davis
Skills to Develop
- Understand the nutritional adaptations of plants
- Describe mycorrhizae
- Explain nitrogen fixation
Plants obtain food in two different ways. Autotrophic plants can make their own food from inorganic raw materials, such as carbon dioxide and water, through photosynthesis in the presence of sunlight. Green plants are included in this group. Some plants, however, are heterotrophic: they are totally parasitic and lacking in chlorophyll. These plants, referred to as holo-parasitic plants, are unable to synthesize organic carbon and draw all of their nutrients from the host plant.
Plants may also enlist the help of microbial partners in nutrient acquisition. Particular species of bacteria and fungi have evolved along with certain plants to create a mutualistic symbiotic relationship with roots. This improves the nutrition of both the plant and the microbe. The formation of nodules in legume plants and mycorrhization can be considered among the nutritional adaptations of plants. However, these are not the only type of adaptations that we may find; many plants have other adaptations that allow them to thrive under specific conditions.
Nitrogen Fixation: Root and Bacteria Interactions
Nitrogen is an important macronutrient because it is part of nucleic acids and proteins. Atmospheric nitrogen, which is the diatomic molecule \(\ce{N2}\), or dinitrogen, is the largest pool of nitrogen in terrestrial ecosystems. However, plants cannot take advantage of this nitrogen because they do not have the necessary enzymes to convert it into biologically useful forms. However, nitrogen can be “fixed,” which means that it can be converted to ammonia (\(\ce{NH3}\)) through biological, physical, or chemical processes. As you have learned, biological nitrogen fixation (BNF) is the conversion of atmospheric nitrogen (\(\ce{N2}\)) into ammonia (\(\ce{NH3}\)), exclusively carried out by prokaryotes such as soil bacteria or cyanobacteria. Biological processes contribute 65 percent of the nitrogen used in agriculture. The following equation represents the process:
\[\ce { N2 + 16 ATP + 8 e^{-} + 8 H^{+} \rightarrow 2 NH3 + 16 ADP + 16 P_i + H_2} \nonumber\]
The most important source of BNF is the symbiotic interaction between soil bacteria and legume plants, including many crops important to humans (Figure \(\PageIndex{1}\)). The NH 3 resulting from fixation can be transported into plant tissue and incorporated into amino acids, which are then made into plant proteins. Some legume seeds, such as soybeans and peanuts, contain high levels of protein, and serve among the most important agricultural sources of protein in the world.
Exercise \(\PageIndex{1}\)
Farmers often rotate corn (a cereal crop) and soy beans (a legume) planting a field with each crop in alternate seasons. What advantage might this crop rotation confer?
- Answer
-
Soybeans are able to fix nitrogen in their roots, which are not harvested at the end of the growing season. The belowground nitrogen can be used in the next season by the corn.
Farmers often rotate corn (a cereal crop) and soy beans (a legume), planting a field with each crop in alternate seasons. What advantage might this crop rotation confer?
Soil bacteria, collectively called rhizobia , symbiotically interact with legume roots to form specialized structures called nodules , in which nitrogen fixation takes place. This process entails the reduction of atmospheric nitrogen to ammonia, by means of the enzyme nitrogenase . Therefore, using rhizobia is a natural and environmentally friendly way to fertilize plants, as opposed to chemical fertilization that uses a nonrenewable resource, such as natural gas. Through symbiotic nitrogen fixation, the plant benefits from using an endless source of nitrogen from the atmosphere. The process simultaneously contributes to soil fertility because the plant root system leaves behind some of the biologically available nitrogen. As in any symbiosis, both organisms benefit from the interaction: the plant obtains ammonia, and bacteria obtain carbon compounds generated through photosynthesis, as well as a protected niche in which to grow (Figure \(\PageIndex{2}\)).
Mycorrhizae: The Symbiotic Relationship between Fungi and Roots
A nutrient depletion zone can develop when there is rapid soil solution uptake, low nutrient concentration, low diffusion rate, or low soil moisture. These conditions are very common; therefore, most plants rely on fungi to facilitate the uptake of minerals from the soil. Fungi form symbiotic associations called mycorrhizae with plant roots, in which the fungi actually are integrated into the physical structure of the root. The fungi colonize the living root tissue during active plant growth.
Through mycorrhization, the plant obtains mainly phosphate and other minerals, such as zinc and copper, from the soil. The fungus obtains nutrients, such as sugars, from the plant root (Figure \(\PageIndex{3}\)). Mycorrhizae help increase the surface area of the plant root system because hyphae, which are narrow, can spread beyond the nutrient depletion zone. Hyphae can grow into small soil pores that allow access to phosphorus that would otherwise be unavailable to the plant. The beneficial effect on the plant is best observed in poor soils. The benefit to fungi is that they can obtain up to 20 percent of the total carbon accessed by plants. Mycorrhizae functions as a physical barrier to pathogens. It also provides an induction of generalized host defense mechanisms, and sometimes involves production of antibiotic compounds by the fungi.
There are two types of mycorrhizae: ectomycorrhizae and endomycorrhizae. Ectomycorrhizae form an extensive dense sheath around the roots, called a mantle. Hyphae from the fungi extend from the mantle into the soil, which increases the surface area for water and mineral absorption. This type of mycorrhizae is found in forest trees, especially conifers, birches, and oaks. Endomycorrhizae, also called arbuscular mycorrhizae, do not form a dense sheath over the root. Instead, the fungal mycelium is embedded within the root tissue. Endomycorrhizae are found in the roots of more than 80 percent of terrestrial plants.
Nutrients from Other Sources
Some plants cannot produce their own food and must obtain their nutrition from outside sources. This may occur with plants that are parasitic or saprophytic. Some plants are mutualistic symbionts, epiphytes, or insectivorous.
Plant Parasites
A parasitic plant depends on its host for survival. Some parasitic plants have no leaves. An example of this is the dodder (Figure \(\PageIndex{4}\)), which has a weak, cylindrical stem that coils around the host and forms suckers. From these suckers, cells invade the host stem and grow to connect with the vascular bundles of the host. The parasitic plant obtains water and nutrients through these connections. The plant is a total parasite (a holoparasite) because it is completely dependent on its host. Other parasitic plants (hemiparasites) are fully photosynthetic and only use the host for water and minerals. There are about 4,100 species of parasitic plants.
Saprophytes
A saprophyte is a plant that does not have chlorophyll and gets its food from dead matter, similar to bacteria and fungi (note that fungi are often called saprophytes, which is incorrect, because fungi are not plants). Plants like these use enzymes to convert organic food materials into simpler forms from which they can absorb nutrients (Figure \(\PageIndex{5}\)). Most saprophytes do not directly digest dead matter: instead, they parasitize fungi that digest dead matter, or are mycorrhizal, ultimately obtaining photosynthate from a fungus that derived photosynthate from its host. Saprophytic plants are uncommon; only a few species are described.
Symbionts
A symbiont is a plant in a symbiotic relationship, with special adaptations such as mycorrhizae or nodule formation. Fungi also form symbiotic associations with cyanobacteria and green algae (called lichens). Lichens can sometimes be seen as colorful growths on the surface of rocks and trees (Figure \(\PageIndex{6}\)). The algal partner (phycobiont) makes food autotrophically, some of which it shares with the fungus; the fungal partner (mycobiont) absorbs water and minerals from the environment, which are made available to the green alga. If one partner was separated from the other, they would both die.
Epiphytes
An epiphyte is a plant that grows on other plants, but is not dependent upon the other plant for nutrition (Figure \(\PageIndex{7}\)). Epiphytes have two types of roots: clinging aerial roots, which absorb nutrients from humus that accumulates in the crevices of trees; and aerial roots, which absorb moisture from the atmosphere.
Insectivorous Plants
An insectivorous plant has specialized leaves to attract and digest insects. The Venus flytrap is popularly known for its insectivorous mode of nutrition, and has leaves that work as traps (Figure \(\PageIndex{8}\)). The minerals it obtains from prey compensate for those lacking in the boggy (low pH) soil of its native North Carolina coastal plains. There are three sensitive hairs in the center of each half of each leaf. The edges of each leaf are covered with long spines. Nectar secreted by the plant attracts flies to the leaf. When a fly touches the sensory hairs, the leaf immediately closes. Next, fluids and enzymes break down the prey and minerals are absorbed by the leaf. Since this plant is popular in the horticultural trade, it is threatened in its original habitat.
Summary
Atmospheric nitrogen is the largest pool of available nitrogen in terrestrial ecosystems. However, plants cannot use this nitrogen because they do not have the necessary enzymes. Biological nitrogen fixation (BNF) is the conversion of atmospheric nitrogen to ammonia. The most important source of BNF is the symbiotic interaction between soil bacteria and legumes. The bacteria form nodules on the legume’s roots in which nitrogen fixation takes place. Fungi form symbiotic associations (mycorrhizae) with plants, becoming integrated into the physical structure of the root. Through mycorrhization, the plant obtains minerals from the soil and the fungus obtains photosynthate from the plant root. Ectomycorrhizae form an extensive dense sheath around the root, while endomycorrhizae are embedded within the root tissue. Some plants—parasites, saprophytes, symbionts, epiphytes, and insectivores—have evolved adaptations to obtain their organic or mineral nutrition from various sources.
Glossary
- epiphyte
- plant that grows on other plants but is not dependent upon other plants for nutrition
- insectivorous plant
- plant that has specialized leaves to attract and digest insects
- nitrogenase
- enzyme that is responsible for the reduction of atmospheric nitrogen to ammonia
- nodules
- specialized structures that contain Rhizobia bacteria where nitrogen fixation takes place
- parasitic plant
- plant that is dependent on its host for survival
- rhizobia
- soil bacteria that symbiotically interact with legume roots to form nodules and fix nitrogen
- saprophyte
- plant that does not have chlorophyll and gets its food from dead matter
- symbiont
- plant in a symbiotic relationship with bacteria or fungi


5.8: Building Simple Resistor Circuits
How to Build a Simple Series Circuit
If all we wish to construct is a simple single-battery, single-resistor circuit, we may easily use alligator clip jumper wires like this:
Jumper wires with “alligator” style spring clips at each end provide a safe and convenient method of electrically joining components together.
If we wanted to build a simple series circuit with one battery and three resistors, the same “point-to-point” construction technique using jumper wires could be applied:
Using a Solderless Breadboard for More Complex Circuits
This technique, however, proves impractical for circuits much more complex than this, due to the awkwardness of the jumper wires and the physical fragility of their connections. A more common method of temporary construction for the hobbyist is the solderless breadboard , a device made of plastic with hundreds of spring-loaded connection sockets joining the inserted ends of components and/or 22-gauge solid wire pieces. A photograph of a real breadboard is shown here, followed by an illustration showing a simple series circuit constructed on one:
Underneath each hole in the breadboard face is a metal spring clip, designed to grasp any inserted wire or component lead. These metal spring clips are joined underneath the breadboard face, making connections between inserted leads. The connection pattern joins every five holes along a vertical column (as shown with the long axis of the breadboard situated horizontally):
Thus, when a wire or component lead is inserted into a hole on the breadboard, there are four more holes in that column providing potential connection points to other wires and/or component leads. The result is an extremely flexible platform for constructing temporary circuits. For example, the three-resistor circuit just shown could also be built on a breadboard like this:
A parallel circuit is also easy to construct on a solderless breadboard:
Breadboards have their limitations, though. First and foremost, they are intended for temporary construction only. If you pick up a breadboard, turn it upside-down, and shake it, any components plugged into it are sure to loosen, and may fall out of their respective holes. Also, breadboards are limited to fairly low-current (less than 1 amp) circuits. Those spring clips have a small contact area, and thus cannot support high currents without excessive heating.
Soldering or Wire-Wrapping
For greater permanence, one might wish to choose soldering or wire-wrapping. These techniques involve fastening the components and wires to some structure providing a secure mechanical location (such as a phenolic or fiberglass board with holes drilled in it, much like a breadboard without the intrinsic spring-clip connections), and then attaching wires to the secured component leads. Soldering is a form of low-temperature welding, using a tin/lead or tin/silver alloy that melts to and electrically bonds copper objects. Wire ends soldered to component leads or to small, copper ring “pads” bonded on the surface of the circuit board serve to connect the components together. In wire wrapping, a small-gauge wire is tightly wrapped around component leads rather than soldered to leads or copper pads, the tension of the wrapped wire providing a sound mechanical and electrical junction to connect components together.
An example of a printed circuit board , or PCB , intended for hobbyist use is shown in this photograph:
This board appears copper-side-up: the side where all the soldering is done. Each hole is ringed with a small layer of copper metal for bonding to the solder. All holes are independent of each other on this particular board, unlike the holes on a solderless breadboard which are connected together in groups of five. Printed circuit boards with the same 5-hole connection pattern as breadboards can be purchased and used for hobby circuit construction, though.
Printed Circuit Boards (PCBs)
Production printed circuit boards have traces of copper laid down on the phenolic or fiberglass substrate material to form pre-engineered connection pathways which function as wires in a circuit. An example of such a board is shown here, this unit actually a “power supply” circuit designed to take 120 volt alternating current (AC) power from a household wall socket and transform it into low-voltage direct current (DC). A resistor appears on this board, the fifth component counting up from the bottom, located in the middle-right area of the board.
A view of this board’s underside reveals the copper “traces” connecting components together, as well as the silver-colored deposits of solder bonding the component leads to those traces:
A soldered or wire-wrapped circuit is considered permanent: that is, it is unlikely to fall apart accidently. However, these construction techniques are sometimes considered too permanent. If anyone wishes to replace a component or change the circuit in any substantial way, they must invest a fair amount of time undoing the connections. Also, both soldering and wire-wrapping require specialized tools which may not be immediately available.
Terminal Strips
An alternative construction technique used throughout the industrial world is that of the terminal strip . Terminal strips, alternatively called barrier strips or terminal blocks , are comprised of a length of nonconducting material with several small bars of metal embedded within. Each metal bar has at least one machine screw or other fastener under which a wire or component lead may be secured. Multiple wires fastened by one screw are made electrically common to each other, as are wires fastened to multiple screws on the same bar. The following photograph shows one style of terminal strip, with a few wires attached.
Another, smaller terminal strip is shown in this next photograph. This type, sometimes referred to as a “European” style, has recessed screws to help prevent accidental shorting between terminals by a screwdriver or other metal object:
In the following illustration, a single-battery, three-resistor circuit is shown constructed on a terminal strip:
If the terminal strip uses machine screws to hold the component and wire ends, nothing but a screwdriver is needed to secure new connections or break old connections. Some terminal strips use spring-loaded clips—similar to a breadboard’s except for increased ruggedness—engaged and disengaged using a screwdriver as a push tool (no twisting involved). The electrical connections established by a terminal strip are quite robust and are considered suitable for both permanent and temporary construction.
Translating a Schematic Diagram to a Circuit Layout
One of the essential skills for anyone interested in electricity and electronics is to be able to “translate” a schematic diagram to a real circuit layout where the components may not be oriented the same way. Schematic diagrams are usually drawn for maximum readability (excepting those few noteworthy examples sketched to create maximum confusion!), but practical circuit construction often demands a different component orientation. Building simple circuits on terminal strips is one way to develop the spatial-reasoning skill of “stretching” wires to make the same connection paths. Consider the case of a single-battery, three-resistor parallel circuit constructed on a terminal strip:
Progressing from a nice, neat, schematic diagram to the real circuit—especially when the resistors to be connected are physically arranged in a linear fashion on the terminal strip—is not obvious to many, so I’ll outline the process step-by-step. First, start with the clean schematic diagram and all components secured to the terminal strip, with no connecting wires:
Next, trace the wire connection from one side of the battery to the first component in the schematic, securing a connecting wire between the same two points on the real circuit. I find it helpful to over-draw the schematic’s wire with another line to indicate what connections I’ve made in real life:
Continue this process, wire by wire, until all connections in the schematic diagram have been accounted for. It might be helpful to regard common wires in a SPICE-like fashion: make all connections to a common wire in the circuit as one step, making sure each and every component with a connection to that wire actually has a connection to that wire before proceeding to the next. For the next step, I’ll show how the top sides of the remaining two resistors are connected together, being common with the wire secured in the previous step:
With the top sides of all resistors (as shown in the schematic) connected together, and to the battery’s positive (+) terminal, all we have to do now is connect the bottom sides together and to the other side of the battery:
Typically in industry, all wires are labeled with number tags, and electrically common wires bear the same tag number, just as they do in a SPICE simulation. In this case, we could label the wires 1 and 2:
Another industrial convention is to modify the schematic diagram slightly so as to indicate actual wire connection points on the terminal strip. This demands a labeling system for the strip itself: a “TB” number (terminal block number) for the strip, followed by another number representing each metal bar on the strip.
This way, the schematic may be used as a “map” to locate points in a real circuit, regardless of how tangled and complex the connecting wiring may appear to the eyes. This may seem excessive for the simple, three-resistor circuit shown here, but such detail is absolutely necessary for construction and maintenance of large circuits, especially when those circuits may span a great physical distance, using more than one terminal strip located in more than one panel or box.
Review
- A solderless breadboard is a device used to quickly assemble temporary circuits by plugging wires and components into electrically common spring-clips arranged underneath rows of holes in a plastic board.
- Soldering is a low-temperature welding process utilizing a lead/tin or tin/silver alloy to bond wires and component leads together, usually with the components secured to a fiberglass board.
- Wire-wrapping is an alternative to soldering, involving small-gauge wire tightly wrapped around component leads rather than a welded joint to connect components together.
- A terminal strip , also known as a barrier strip or terminal block is another device used to mount components and wires to build circuits. Screw terminals or heavy spring clips attached to metal bars provide connection points for the wire ends and component leads, these metal bars mounted separately to a piece of nonconducting material such as plastic, bakelite, or ceramic.


10.2: Other Organs of the Lymphatic System
The spleen is an important part of the lymphatic system. It is a deep red organ situated in the abdomen caudal to the stomach (see diagram 10.3). It is composed of two different types of tissue. The first type makes and stores lymphocytes, the cells of the immune system. The second type of tissue destroys worn out red blood cells, breaking down the haemoglobin into iron, which is recycled, and waste products that are excreted. The spleen also stores red blood cells. When severe blood loss occurs, it contracts and releases them into the circulation.
The thymus is a large pink organ lying just under the sternum (breastbone) just cranial to the heart (see diagram 10.1). It has an important function processing lymphocytes so they are capable of recognising and attacking foreign invaders like bacteria.
Other lymph organs are the bone marrow of the long bones where lymphocytes are produced and lymph nodules , which are like tiny lymph nodes. Large clusters of these are found in the wall of the small intestine (called Peyer’s Patches) and in the tonsils.


10.6: Multiplication and Division of Signed Numbers
- be able to multiply and divide signed numbers
- be able to multiply and divide signed numbers using a calculator
Multiplication of Signed Numbers
Let us consider first, the product of two positive numbers. Multiply: \(3 \cdot 5\).
\(3 \cdot 5\) means \(5 + 5 + 5 = 15\)
This suggests that (In later mathematics courses, the word "suggests" turns into the word "proof." One example does not prove a claim. Mathematical proofs are constructed to validate a claim for all possible cases.)
\(\text{(positive number)} \cdot \text{(positive number)} = \text{(positive number)}\)
More briefly,
(+) (+) = (+)
Now consider the product of a positive number and a negative number. Multiply: (3)(-5)
(3)(-5) means (-5) + (-5) + (-5) = -15
This suggests that
\(\text{(positive number)} \cdot \text{(negative number)} = \text{(negative number)}\)
More briefly,
(+) (-) = (-)
By the commutative property of multiplication, we get
\(\text{(negative number)} \cdot \text{(positive number)} = \text{(negative number)}\)
More briefly,
(-) (+) = (-)
The sign of the product of two negative numbers can be suggested after observing the following illustration.
Multiply -2 by, respectively, 4, 3, 2, 1, 0, -1, -2, -3, -4.
We have the following rules for multiplying signed numbers.
Rules for Multiplying Signed
Numbers
Multiplying signed numbers:
-
To multiply two real numbers that have the
same sign
, multiply their absolute values. The product is positive.
(+) (+) = (+)
(-) (-) = (+) -
To multiply two real numbers that have
opposite signs
, multiply their absolute values. The product is negative.
(+) (-) = (-)
(-) (+) = (-)
Find the following products.
\(8 \cdot 6\)
Solution
\(\begin{array} {ccl} {|8|} & = & {8} \\ {|6|} & = & {6} \end{array} \big \}\) Multiply these absolute values.
\(8 \cdot 6 = 48\)
Since the numbers have the same sign, the product is positive.
Thus, \(8 \cdot 6 = +48\), or \(8 \cdot 6 = 48\).
(-8)(-6)
Solution
\(\begin{array} {ccl} {|-8|} & = & {8} \\ {|-6|} & = & {6} \end{array} \big \}\) Multiply these absolute values.
\(8 \cdot 6 = 48\)
Since the numbers have the same sign, the product is positive.
Thus, \((-8)(-6) = +48\), or \((-8)(-6) = 48\).
(-4)(7)
Solution
\(\begin{array} {ccl} {|-4|} & = & {4} \\ {|7|} & = & {7} \end{array} \big \}\) Multiply these absolute values.
\(4 \cdot 7 = 28\)
Since the numbers have opposite signs, the product is negative.
Thus, (-4)(7) = -28.
6(-3)
Solution
\(\begin{array} {ccl} {|6|} & = & {6} \\ {|-3|} & = & {3} \end{array} \big \}\) Multiply these absolute values.
\(6 \cdot 3 = 18\)
Since the numbers have opposite signs, the product is negative.
Thus, 6(-3) = -18.
Practice Set A
Find the following products.
3(-8)
- Answer
-
-24
Practice Set A
4(16)
- Answer
-
64
Practice Set A
(-6)(-5)
- Answer
-
30
Practice Set A
(-7)(-2)
- Answer
-
14
Practice Set A
(-1)(4)
- Answer
-
-4
Practice Set A
(-7)7
- Answer
-
-49
Division of Signed Numbers
To determine the signs in a division problem, recall that
\(\dfrac{12}{3} = 4\) since \(12 = 3 \cdot 4\)
This suggests that
\(\dfrac{(+)}{(+)} = (+)\)
\(\dfrac{(+)}{(+)} = (+)\) since (+) = (+) (+)
What is \(\dfrac{12}{-3}\)?
12 = (-3)(-4) suggets that \(\dfrac{12}{-3} = -4\). That is,
\(\dfrac{(+)}{(-)} = (-)\)
(+) = (-) (-) suggets that \(\dfrac{(+)}{(-)} = (-)\)
What is \(\dfrac{-12}{3}\)?
-12 = (3)(-4) suggests that \(\dfrac{-12}{3} = -4\). That is,
\(\dfrac{(-)}{(+)} = (-)\)
(-) = (+) (-) suggets that \(\dfrac{(-)}{(+)} = (-)\)
What is \(\dfrac{-12}{-3}\)?
-12 = (-3)(4) suggets that \(\dfrac{-12}{-3} = 4\). That is,
\(\dfrac{(-)}{(-)} = (+)\)
(-) = (-)(+) suggests that \(\dfrac{(-)}{(-)} = (+)\)
We have the following rules for dividing signed numbers.
Rules for Dividing Signed Numbers
Dividing signed numbers:
-
To divide two real numbers that have the
same sign
, divide their absolute values. The quotient is positive.
\(\dfrac{(+)}{(+)} = (+)\dfrac{(-)}{(-)} = (+)\) -
To divide two real numbers that have
opposite signs
, divide their absolute values. The quotient is negative.
\(\dfrac{(-)}{(+)} = (-)\dfrac{(+)}{(-)} = (-)\)
Find the following quotients.
\(\dfrac{-10}{2}\)
Solution
\(\begin{array} {ccc} {|-10|} & = & {10} \\ {|2|} & = & {2} \end{array} \big \}\) Divide these absolute values.
\(\dfrac{10}{2} = 5\)
Since the numbers have opposite signs, the quotient is negative.
Thus \(\dfrac{-10}{2} = -5\).
\(\dfrac{-35}{-7}\)
Solution
\(\begin{array} {ccc} {|-35|} & = & {35} \\ {|-7|} & = & {7} \end{array} \big \}\) Divide these absolute values.
\(\dfrac{35}{7} = 5\)
Since the numbers have the same signs, the quotient is positive.
Thus \(\dfrac{-35}{-7} = 5\).
\(\dfrac{18}{-9}\)
Solution
\(\begin{array} {ccc} {|18|} & = & {18} \\ {|-9|} & = & {9} \end{array} \big \}\) Divide these absolute values.
\(\dfrac{18}{9} = 2\)
Since the numbers have opposite signs, the quotient is negative.
Thus \(\dfrac{18}{-9} = -2\).
Practice Set B
Find the following quotients.
\(\dfrac{-24}{-6}\)
- Answer
-
4
Practice Set B
\(\dfrac{30}{-5}\)
- Answer
-
-6
Practice Set B
\(\dfrac{-54}{27}\)
- Answer
-
-2
Practice Set B
\(\dfrac{51}{17}\)
- Answer
-
3
Find the value of \(\dfrac{-6(4 - 7) - 2(8 - 9)}{-(4 + 1) + 1}\).
Solution
Using the order of operations and what we know about signed numbers, we get,
\(\begin{array} {rcl} {\dfrac{-6(4 - 7) - 2(8 - 9)}{-(4 + 1) + 1}} & = & {\dfrac{-6(-3) - 2(-1)}{-(5) + 1}} \\ {} & = & {\dfrac{18 + 2}{-5 + 1}} \\ {} & = & {\dfrac{20}{-4}} \\ {} & = & {-5} \end{array}\)
Practice Set C
Find the value of \(\dfrac{-5(2 - 6) - 4(-8 - 1)}{2(3 - 10) - 9(-2)}\).
- Answer
-
14
Calculators
Calculators with the
key can be used for multiplying and dividing signed numbers.
Use a calculator to find each quotient or product.
\((-186) \cdot (-43)\)
Solution
Since this product involves a \(\text{(negative)} \cdot \text{(negative)}\), we know the result should be a positive number. We'll illustrate this on the calculator.
| Display Reads | ||
| Type | 186 | 186 |
| Press | -186 | |
| Press | \(\times\) | -186 |
| Type | 43 | 43 |
| Press | -43 | |
| Press | = | 7998 |
Thus, \((-186) \cdot (-43) = 7,998\)
\(\dfrac{158.64}{-54.3}\). Round to one decimal place.
Solution
Since this product involves a \(\text{(negative)} \cdot \text{(negative)}\), we know the result should be a positive number. We'll illustrate this on the calculator.
| Display Reads | ||
| Type | 158.64 | 158.64 |
| Press | \(\div\) | 158.64 |
| Type | 54.3 | 54.3 |
| Press | -54.3 | |
| Press | = | -2.921546961 |
Rounding to one decimal place we get -2.9.
Practice Set D
Use a calculator to find each value.
\((-51.3) \cdot (-21.6)\)
- Answer
-
1,108.08
Practice Set D
\(-2.5746 \div -2.1\)
- Answer
-
1.226
Practice Set D
\((0.006) \cdot (-0.241)\). Round to three decimal places.
- Answer
-
-0.001
Exercises
Find the value of each of the following. Use a calculator to check each result.
Exercise \(\PageIndex{1}\)
(-2)(-8)
- Answer
-
16
Exercise \(\PageIndex{2}\)
(-3)(-9)
Exercise \(\PageIndex{3}\)
(-4)(-8)
- Answer
-
32
Exercise \(\PageIndex{4}\)
(-5)(-2)
Exercise \(\PageIndex{5}\)
(3)(-12)
- Answer
-
-36
Exercise \(\PageIndex{6}\)
(4)(-18)
Exercise \(\PageIndex{7}\)
(10)(-6)
- Answer
-
-60
Exercise \(\PageIndex{8}\)
(-6)(4)
Exercise \(\PageIndex{9}\)
(-2)(6)
- Answer
-
-12
Exercise \(\PageIndex{10}\)
(-8)(7)
Exercise \(\PageIndex{11}\)
\(\dfrac{21}{7}\)
- Answer
-
3
Exercise \(\PageIndex{12}\)
\(\dfrac{42}{6}\)
Exercise \(\PageIndex{13}\)
\(\dfrac{-39}{3}\)
- Answer
-
-13
Exercise \(\PageIndex{14}\)
\(\dfrac{-20}{10}\)
Exercise \(\PageIndex{15}\)
\(\dfrac{-45}{-5}\)
- Answer
-
9
Exercise \(\PageIndex{16}\)
\(\dfrac{-16}{-8}\)
Exercise \(\PageIndex{17}\)
\(\dfrac{25}{-5}\)
- Answer
-
-5
Exercise \(\PageIndex{18}\)
\(\dfrac{36}{-4}\)
Exercise \(\PageIndex{19}\)
8 - (-3)
- Answer
-
11
Exercise \(\PageIndex{20}\)
14 - (-20)
Exercise \(\PageIndex{21}\)
20 - (-8)
- Answer
-
28
Exercise \(\PageIndex{22}\)
(-4) - (-1)
Exercise \(\PageIndex{23}\)
0 - 4
- Answer
-
-4
Exercise \(\PageIndex{24}\)
0 - (-1)
Exercise \(\PageIndex{25}\)
-6 + 1 - 7
- Answer
-
-12
Exercise \(\PageIndex{26}\)
15 - 12 - 20
Exercise \(\PageIndex{27}\)
1 - 6 - 7 + 8
- Answer
-
-4
Exercise \(\PageIndex{28}\)
2 + 7 - 10 + 2
Exercise \(\PageIndex{29}\)
3(4 - 6)
- Answer
-
-6
Exercise \(\PageIndex{30}\)
8(5 - 12)
Exercise \(\PageIndex{31}\)
-3(1 - 6)
- Answer
-
15
Exercise \(\PageIndex{32}\)
-8(4 - 12) + 2
Exercise \(\PageIndex{33}\)
-4(1 - 8) + 3(10 - 3)
- Answer
-
49
Exercise \(\PageIndex{34}\)
-9(0 - 2) + 4(8 - 9) + 0(-3)
Exercise \(\PageIndex{35}\)
6(-2 - 9) - 6(2 + 9) + 4(-1 - 1)
- Answer
-
-140
Exercise \(\PageIndex{36}\)
\(\dfrac{3(4 + 1) - 2 (5)}{-2}\)
Exercise \(\PageIndex{37}\)
\(\dfrac{4(8 + 1) - 3 (-2)}{-4 - 2}\)
- Answer
-
-7
Exercise \(\PageIndex{38}\)
\(\dfrac{-1(3 + 2) + 5}{-1}\)
Exercise \(\PageIndex{39}\)
\(\dfrac{-3(4 - 2) + (-3)(-6)}{-4}\)
- Answer
-
-3
Exercise \(\PageIndex{40}\)
-1(4 + 2)
Exercise \(\PageIndex{41}\)
-1(6 - 1)
- Answer
-
-5
Exercise \(\PageIndex{42}\)
-(8 + 21)
Exercise \(\PageIndex{43}\)
-(8 - 21)
- Answer
-
13
Exercises for Review
Exercise \(\PageIndex{44}\)
Use the order of operations to simplify \((5^2 + 3^2 + 2) \div 2^2\).
Exercise \(\PageIndex{45}\)
Find \(\dfrac{3}{8}\) of \(\dfrac{32}{9}\).
- Answer
-
\(\dfrac{4}{3} = 1 \dfrac{1}{3}\)
Exercise \(\PageIndex{46}\)
Write this number in decimal form using digits: “fifty-two three-thousandths”
Exercise \(\PageIndex{47}\)
The ratio of chlorine to water in a solution is 2 to 7. How many mL of water are in a solution that contains 15 mL of chlorine?
- Answer
-
\(52 \dfrac{1}{2}\)
Exercise \(\PageIndex{48}\)
Perform the subtraction -8 - (-20)


1.3.7: Partial Functions
It is sometimes useful to relax the definition of function so that it is not required that the output of the function is defined for all possible inputs. Such mappings are called partial functions .
A partial function \(f \colon A \pto B\) is a mapping which assigns to every element of \(A\) at most one element of \(B\) . If \(f\) assigns an element of \(B\) to \(x \in A\) , we say \(f(x)\) is defined , and otherwise undefined . If \(f(x)\) is defined, we write \(f(x) \fdefined\) , otherwise \(f(x) \fundefined\) . The domain of a partial function \(f\) is the subset of \(A\) where it is defined, i.e., \(\dom{f} = \Setabs{x \in A}{f(x) \fdefined}\) .
Every function \(f\colon A \to B\) is also a partial function. Partial functions that are defined everywhere on \(A\) —i.e., what we so far have simply called a function—are also called total functions.
The partial function \(f \colon \Real \pto \Real\) given by \(f(x) = 1/x\) is undefined for \(x = 0\) , and defined everywhere else.
Given \(f\colon A \pto B\) , define the partial function \(g\colon B \pto A\) by: for any \(y \in B\) , if there is a unique \(x \in A\) such that \(f(x) = y\) , then \(g(y) = x\) ; otherwise \(g(y) \fundefined\) . Show that if \(f\) is injective, then \(g(f(x)) = x\) for all \(x \in \dom{f}\) , and \(f(g(y)) = y\) for all \(y \in \ran{f}\) .
 The graph of \(f\) is the relation \(R_f \subseteq A \times B\) defined by \[R_f = \Setabs{\tuple{x,y}}{f(x) = y}.\nonumber\]
Suppose \(R \subseteq A \times B\) has the property that whenever \(Rxy\) and \(Rxy'\) then \(y = y'\) . Then \(R\) is the graph of the partial function \(f\colon X \pto Y\) defined by: if there is a \(y\) such that \(Rxy\) , then \(f(x) = y\) , otherwise \(f(x) \fundefined\) . If \(R\) is also serial , i.e., for each \(x \in X\) there is a \(y \in Y\) such that \(Rxy\) , then \(f\) is total.
Proof. Suppose there is a \(y\) such that \(Rxy\) . If there were another \(y' \neq y\) such that \(Rxy'\) , the condition on \(R\) would be violated. Hence, if there is a \(y\) such that \(Rxy\) , that \(y\) is unique, and so \(f\) is well-defined. Obviously, \(R_f = R\) and \(f\) is total if \(R\) is serial. ◻


16.7: Putting It Together- Accounting and Finance
Synthesis
Whether or not this module convinced you to pursue a career in accounting, by now you have acquired a working knowledge of some of the basics of financial accounting, as well as the importance of accuracy in the presentation of financial information to stakeholders. Whether you are running a bake sale or a multinational corporation, understanding the relationship between revenues and expenses is critical for success. The misdeeds of corporate executives and their accountants have peppered the news for the last decade; but, the vast majority of accountants and their clients follow a strict code of ethics and observe the laws and guidance provided by Congress, FASB, and AICPA. One of the best ways to protect yourself and your business against becoming involved in a financial scandal is to have a solid working knowledge of basic accounting principles so that you can recognize and correct any irregularities.
Summary
Accounting in Business
In short, accounting is the language of business—all business. Accounting represents all of the financial transactions of a business in a format that can be interpreted and understood by both internal and external stakeholders.
Key Financial Statements
When businesses present their financial condition to external stakeholders, taxing authorities, investors, and the general public, the most common format for this information is one of four key financial statements. These four statements are the Balance Sheet, Income Statement, Statement of Owners Equity, and Statement of Cash Flows. These four statements, although representing different facets of the company’s finances, are all interconnected and create a birds-eye view of the company’s financial position.
The Break-Even Point
Businesses, both large and small, are concerned with determining the point at which their revenues exceed their expenses and they begin to make a profit. The point at which revenue equals expenses (and profit is therefore $0) is called the break-even point.
Financial Ratios
Financial ratios allow business to represent the relationships between components of their financial operations as ratios. Financial ratios are used to measure a firm’s financial health in four areas: liquidity, long-term solvency, profitability tests, and the market. These ratios can be used to compare the company’s performance across periods (months, quarters, years) or to similar companies within the same industry.
Ethical Practices in Accounting
Certified public accountants (CPAs) and certified management accountants (CMAs) are bound to the Code of Ethics established by their licensing bodies. Generally accepted accounting principles (GAAP) and the Financial Accounting Standards Board (FASB) have established practices designed to ensure that the financial status of a company is “fairly and accurately” presented. Legislation such as the Sarbanes-Oxley Act has been passed by Congress to strengthen the emphasis on ethical practices in accountancy. Although stories of unethical conduct by companies such as Enron, WorldCom, and HP have made headlines, the overwhelming majority of individuals working as internal or external accountants follow the code of ethics and work hard to ensure that the information provided to stakeholders is fair and accurate.


2.2: Opera
Perhaps the single greatest musical development of the Baroque period is the creation a new genre of vocal music: opera. As with most genres in this era, opera undergoes significant stylistic evolution from its origins in the early 1600s to the opera seria of Handel in the 1730s. Please pay special attention to the following terms (if you would like to learn more about them, follow the links): libretto , recitative (both forms: secco and accompagnato ), and aria .
Introduction
Opera (English plural: operas; Italian plural: opere ) is an art form in which singers and musicians perform a dramatic work combining text (called a libretto ) and musical score, usually in a theatrical setting. Opera incorporates many of the elements of spoken theatre, such as acting, scenery, and costumes and sometimes includes dance. The performance is typically given in an opera house, accompanied by an orchestra or smaller musical ensemble.
Opera is part of the Western classical music tradition. It started in Italy at the end of the 16th century (with Jacopo Peri’s lost Dafne , produced in Florence in 1598) and soon spread through the rest of Europe: Schütz in Germany, Lully in France, and Purcell in England all helped to establish their national traditions in the seventeenth century. In the eighteenth century, Italian opera continued to dominate most of Europe (except France), attracting foreign composers such as Handel. Opera seria was the most prestigious form of Italian opera, until Gluckreacted against its artificiality with his “reform” operas in the 1760s. Today the most renowned figure of late 18th century opera is Mozart, who began with opera seria but is most famous for his Italian comic operas, especially The Marriage of Figaro (Le Nozze Di Figaro), Don Giovanni , and Così fan tutte , as well as The Magic Flute (Die Zauberflöte) , a landmark in the German tradition.
Operatic Terminology
The words of an opera are known as the libretto (literally “little book”). Some composers, notably Richard Wagner, have written their own libretti; others have worked in close collaboration with their librettists, e.g. Mozart with Lorenzo Da Ponte. Traditional opera, often referred to as “number opera,” consists of two modes of singing: recitative, the plot-driving passages sung in a style designed to imitate and emphasize the inflections of speech, and aria (an “air” or formal song) in which the characters express their emotions in a more structured melodic style. Duets, trios and other ensembles often occur, and choruses are used to comment on the action. In some forms of opera, such as Singspiel, opéra comique,operetta, and semi-opera, the recitative is mostly replaced by spoken dialogue. Melodic or semi-melodic passages occurring in the midst of, or instead of, recitative, are also referred to as arioso. During the Baroque and Classical periods, recitative could appear in two basic forms: secco (dry) recitative, sung with a free rhythm dictated by the accent of the words, accompanied only by continuo , which was usually a harpsichord and a cello; or accompagnato (also known as strumentato ) in which the orchestra provided accompaniment. By the nineteenth century, accompagnato had gained the upper hand, the orchestra played a much bigger role, and Richard Wagner revolutionised opera by abolishing almost all distinction between aria and recitative in his quest for what he termed “endless melody.” Subsequent composers have tended to follow Wagner’s example, though some, such as Stravinsky in his The Rake’s Progress have bucked the trend. The terminology of the various kinds of operatic voices is described in detail below.
Origins
The Italian word opera means “work,” both in the sense of the labour done and the result produced. The Italian word derives from the Latin opera , a singular noun meaning “work” and also the plural of the noun opus . According to the Oxford English Dictionary, the Italian word was first used in the sense “composition in which poetry, dance, and music are combined” in 1639; the first recorded English usage in this sense dates to 1648.
Dafne by Jacopo Peri was the earliest composition considered opera, as understood today. It was written around 1597, largely under the inspiration of an elite circle of literate Florentine humanists who gathered as the “Camerata de’ Bardi.” Significantly, Dafne was an attempt to revive the classical Greek drama, part of the wider revival of antiquity characteristic of the Renaissance. The members of the Camerata considered that the “chorus” parts of Greek dramas were originally sung, and possibly even the entire text of all roles; opera was thus conceived as a way of “restoring” this situation. Dafne is unfortunately lost. A later work by Peri, Euridice , dating from 1600, is the first opera score to have survived to the present day. The honour of being the first opera still to be regularly performed, however, goes to Claudio Monteverdi’s L’Orfeo , composed for the court of Mantua in 1607. The Mantua court of the Gonzagas, employers of Monteverdi, played a significant role in the origin of opera employing not only court singers of the concerto delle donne (till 1598), but also one of the first actual “opera singers”: Madama Europa.
The Baroque Era
Opera did not remain confined to court audiences for long. In 1637, the idea of a “season” (Carnival) of publicly attended operas supported by ticket sales emerged in Venice. Monteverdi had moved to the city from Mantua and composed his last operas, Il ritorno d’Ulisse in patria and L’incoronazione di Poppea , for the Venetian theatre in the 1640s. His most important follower Francesco Cavalli helped spread opera throughout Italy.
In these early Baroque operas, broad comedy was blended with tragic elements in a mix that jarred some educated sensibilities, sparking the first of opera’s many reform movements, sponsored by the Arcadian Academy, which came to be associated with the poet Metastasio, whose libretti helped crystallize the genre of opera seria, which became the leading form of Italian opera until the end of the eighteenth century. Once the Metastasian ideal had been firmly established, comedy in Baroque-era opera was reserved for what came to be called opera buffa. Before such elements were forced out of opera seria, many libretti had featured a separately unfolding comic plot as sort of an “opera-within-an-opera.” One reason for this was an attempt to attract members of the growing merchant class, newly wealthy, but still not as cultured as the nobility, to the public opera houses. These separate plots were almost immediately resurrected in a separately developing tradition that partly derived from the commedia dell’arte, a long-flourishing improvisatory stage tradition of Italy.
Just as intermedi had once been performed in-between the acts of stage plays, operas in the new comic genre of “intermezzi,” which developed largely in Naples in the 1710s and ’20s, were initially staged during the intermissions of opera seria. They became so popular, however, that they were soon being offered as separate productions.
Opera seria was elevated in tone and highly stylised in form, usually consisting of secco recitative interspersed with long da capo arias. These afforded great opportunity for virtuosic singing and during the golden age of opera seria the singer really became the star. The role of the hero was usually written for the castrato voice; castrati such as Farinelli and Senesino, as well as female sopranos such as Faustina Bordoni, became in great demand throughout Europe as opera seria ruled the stage in every country except France. Indeed, Farinelli was one of the most famous singers of the eighteenth century. Italian opera set the Baroque standard. Italian libretti were the norm, even when a German composer like Handelfound himself composing the likes of Rinaldo and Giulio Cesare for London audiences. Italian libretti remained dominant in the classical period as well, for example in the operas of Mozart, who wrote in Vienna near the century’s close. Leading Italian-born composers of opera seria include Alessandro Scarlatti, Vivaldi and Porpora.
Contributors and Attributions
- Authored by : Elliott Jones. Provided by : Santa Ana College. Located at : http://www.sac.edu . License : CC BY: Attribution
- Opera. Provided by : Wikipedia. Located at : http://en.Wikipedia.org/wiki/Opera . License : CC BY-SA: Attribution-ShareAlike


2.17: Simplifying Expressions with Roots
Simplifying Expressions with Roots
Remember that when a real number \(n\) is multiplied by itself, we write \(n^{2}\) and read it '\(n^{2}\) squared’. This number is called the square of \(n\), and \(n\) is called the square root . For example,
\(13^{2}\) is read "\(13\) squared"
\(169\) is called the square of \(13\), since \(13^{2}=169\)
\(13\) is a square root of \(169\)
Square
If \(n^{2}=m\), then \(m\) is the square of \(n\).
Square Root
If \(n^{2}=m\), then \(n\) is a square root of \(m\).
Notice \((−13)^{2} = 169\) also, so \(−13\) is also a square root of \(169\). Therefore, both \(13\) and \(−13\) are square roots of \(169\).
So, every positive number has two square roots—one positive and one negative. What if we only wanted the positive square root of a positive number? We use a radical sign , and write, \(\sqrt{m}\), which denotes the positive square root of \(m\). The positive square root is also called the principal square root .
We also use the radical sign for the square root of zero. Because \(0^{2}=0, \sqrt{0}=0\). Notice that zero has only one square root.
\(\sqrt{m}\) is read "the square root of \(m\)."
If \(n^{2}=m\), then \(n=\sqrt{m}\), for \(n\geq 0\).
We know that every positive number has two square roots and the radical sign indicates the positive one. We write \(\sqrt{169}=13\). If we want to find the negative square root of a number, we place a negative in front of the radical sign. For example, \(-\sqrt{169}=-13\).
Simplify:
- \(\sqrt{144}\)
- \(-\sqrt{289}\)
Solution :
- Since \(12^{2}=144\),
\(\sqrt{144}=12\)
- Since \(17^{2}=289\) and the negative is in front of the radical sign,
\(-\sqrt{289}=-17\)
Simplify:
- \(-\sqrt{64}\)
- \(\sqrt{225}\)
- Answer
-
- \(-8\)
- \(15\)
Can we simplify \(-\sqrt{49}\)? Is there a number whose square is \(-49\)?
\((\)___\( )^{2}=-49\)
Any positive number squared is positive. Any negative number squared is positive. There is no real number equal to \(\sqrt{-49}\). The square root of a negative number is not a real number.
Simplify:
- \(\sqrt{-196}\)
- \(-\sqrt{64}\)
Solution :
- There is no real number whose square is \(-196\).
\(\sqrt{-196}\) is not a real number.
- The negative is in front of the radical and \(8^2=64).
\(-\sqrt{64}=-8\)
Simplify:
- \(\sqrt{-169}\)
- \(-\sqrt{81}\)
- Answer
-
- not a real number
- \(-9\)
So far we have only talked about squares and square roots. Let’s now extend our work to include higher powers and higher roots.
Let’s review some vocabulary first.
\(\begin{array}{ll}{\text { We write: }} & {\text { We say: }} \\ {n^{2}} & {n \text { squared }} \\ {n^{3}} & {n \text { cubed }} \\ {n^{4}} & {n \text { to the fourth power }} \\ {n^{5}} & {n \text { to the fifth power }}\end{array}\)
The terms ‘squared’ and ‘cubed’ come from the formulas for area of a square and volume of a cube.
It will be helpful to have a table of the powers of the integers from \(−5\) to \(5\). See Figure \(\PageIndex{2}\)
Notice the signs in the table. All powers of positive numbers are positive, of course. But when we have a negative number, the even powers are positive and the odd powers are negative. We’ll copy the row with the powers of \(−2\) to help you see this.
We will now extend the square root definition to higher roots.
If \(b^{n}=a\), then \(b\) is an \(n^{th}\) root of \(a\).
The principal \(n^{th}\) root of \(a\) is written \(\sqrt[n]{a}\).
The \(n\) is called the index of the radical.
Just like we use the word ‘cubed’ for \(b^{3}\), we use the term ‘cube root’ for \(\sqrt[3]{a}\).
We can refer to Figure \(\PageIndex{2}\) to help find higher roots.
\(\begin{aligned} 4^{3} &=64 & \sqrt[3]{64}&=4 \\ 3^{4} &=81 & \sqrt[4]{81}&=3 \\(-2)^{5} &=-32 & \sqrt[5]{-32}&=-2 \end{aligned}\)
Could we have an even root of a negative number? We know that the square root of a negative number is not a real number. The same is true for any even root. Even roots of negative numbers are not real numbers. Odd roots of negative numbers are real numbers.
When \(n\) is an even number and
- \(a \geq 0\), then \(\sqrt[n]{a}\) is a real number.
- \(a<0\), then \(\sqrt[n]{a}\) is not a real number.
When \(n\) is an odd number, \(\sqrt[n]{a}\) is a real number for all the values of \(a\).
We will apply these properties in the next two examples.
Simplify:
- \(\sqrt[3]{64}\)
- \(\sqrt[4]{81}\)
- \(\sqrt[5]{32}\)
Solution :
- Since \(4^{3}=64\),
\(\sqrt[3]{64}=4\)
- Since \((3)^{4}=81\),
\(\sqrt[4]{81}=3\)
- Since \((2)^{5}=32\),
\(\sqrt[5]{32}=2\)
Simplify:
- \(\sqrt[3]{27}\)
- \(\sqrt[4]{256}\)
- \(\sqrt[5]{243}\)
- Answer
-
- \(3\)
- \(4\)
- \(3\)
In this example be alert for the negative signs as well as even and odd powers.
Simplify:
- \(\sqrt[3]{-125}\)
- \(\sqrt[4]{16}\)
- \(\sqrt[5]{-243}\)
Solution :
- Since \((-5)^{3}=-125\),
\(\sqrt[3]{-125}=-5\)
- Think, \((?)^{4}=-16\). No real number raised to the fourth power is negative.
\(\sqrt[4]{16}\) is not a real number.
- Since \((-3)^{5}=-243\),
\(\sqrt[5]{-243}=-3\)
Simplify:
- \(\sqrt[3]{-27}\)
- \(\sqrt[4]{-256}\)
- \(\sqrt[5]{-32}\)
- Answer
-
- \(-3\)
- not real
- \(-2\)
Estimating and Approximating Roots
When we see a number with a radical sign, we often don’t think about its numerical value. While we probably know that the \(\sqrt{4}=2\), what is the value of \(\sqrt{21}\) or \(\sqrt[3]{50}\)? In some situations a quick estimate is meaningful and in others it is convenient to have a decimal approximation.
To get a numerical estimate of a square root, we look for perfect square numbers closest to the radicand. To find an estimate of \(\sqrt{11}\), we see \(11\) is between perfect square numbers \(9\) and \(16\), closer to \(9\). Its square root then will be between \(3\) and \(4\), but closer to \(3\).
Similarly, to estimate \(\sqrt[3]{91}\), we see \(91\) is between perfect cube numbers \(64\) and \(125\). The cube root then will be between \(4\) and \(5\).
Estimate each root between two consecutive whole numbers:
- \(\sqrt{105}\)
- \(\sqrt[3]{43}\)
Solution :
- Think of the perfect square numbers closest to \(105\). Make a small table of these perfect squares and their squares roots.
| \(\sqrt{105}\) | |
| Locate \(105\) between two consecutive perfect squares. | \(100<\color{red}105 \color{black} <121\) |
| \(\sqrt{105}\) is between their square roots. | \(10< \color{red}\sqrt{105}< \color{black}11\) |
- Similarly we locate \(43\) between two perfect cube numbers.
| \(\sqrt[3]{43}\) | |
| Locate \(43\) between two consecutive perfect cubes. | |
| \(\sqrt[3]{43}\) is between their cube roots. |
Estimate each root between two consecutive whole numbers:
- \(\sqrt{38}\)
- \(\sqrt[3]{93}\)
- Answer
-
- \(6<\sqrt{38}<7\)
- \(4<\sqrt[3]{93}<5\)
There are mathematical methods to approximate square roots, but nowadays most people use a calculator to approximate square roots. To find a square root you will use the \(\sqrt{x}\) key on your calculator. To find a cube root, or any root with higher index, you will use the \(\sqrt[y]{x}\) key.
When you use these keys, you get an approximate value. It is an approximation, accurate to the number of digits shown on your calculator’s display. The symbol for an approximation is \(≈\) and it is read ‘approximately’.
Suppose your calculator has a \(10\) digit display. You would see that
\(\sqrt{5} \approx 2.236067978\) rounded to two decimal places is \(\sqrt{5} \approx 2.24\)
\(\sqrt[4]{93} \approx 3.105422799\) rounded to two decimal places is \(\sqrt[4]{93} \approx 3.11\)
How do we know these values are approximations and not the exact values? Look at what happens when we square them:
\(\begin{aligned}(2.236067978)^{2} &=5.000000002 &(3.105422799)^{4}&=92.999999991 \\(2.24)^{2} &=5.0176 & (3.11)^{4}&=93.54951841 \end{aligned}\)
Their squares are close to \(5\), but are not exactly equal to \(5\). The fourth powers are close to \(93\), but not equal to \(93\).
Round to two decimal places:
- \(\sqrt{17}\)
- \(\sqrt[3]{49}\)
- \(\sqrt[4]{51}\)
Solution :
- Use the calculator square root key, \(\sqrt{17} \approx 4.123105626 \dots\)
Round to two decimal places, \(\sqrt{17} \approx 4.12\)
- Use the calculator \(\sqrt[y]{x}\) key, \(\sqrt[3]{49} \approx 3.659305710 \dots\)
Round to two decimal places, \(\sqrt[3]{49} \approx 3.66\)
- Use the calculator \(\sqrt[y]{x}\) key, \(\sqrt[4]{51} \approx 2.6723451177 \dots\)
Round to two decimal places, \(\sqrt[4]{51} \approx 2.67\)
Round to two decimal places:
- \(\sqrt{11}\)
- \(\sqrt[3]{71}\)
- \(\sqrt[4]{127}\)
- Answer
-
- \(\approx 3.32\)
- \(\approx 4.14\)
- \(\approx 3.36\)
Simplifying Variable Expressions with Roots
The odd root of a number can be either positive or negative. For example,
But what about an even root? We want the principal root, so \(\sqrt[4]{625}=5\).
But notice,
How can we make sure the fourth root of \(−5\) raised to the fourth power is \(5\)? We can use the absolute value. \(|−5|=5\). So we say that when \(n\) is even \(\sqrt[n]{a^{n}}=|a|\). This guarantees the principal root is positive.
For any integer \(n\geq 2\),
when the index \(n\) is odd \(\sqrt[n]{a^{n}}=a\)
when the index \(n\) is even \(\sqrt[n]{a^{n}}=|a|\)
We must use the absolute value signs when we take an even root of an expression with a variable in the radical.
Simplify:
- \(\sqrt{x^{2}}\)
- \(\sqrt[3]{n^{3}}\)
- \(\sqrt[4]{p^{4}}\)
- \(\sqrt[5]{y^{5}}\)
Solution :
- We use the absolute value to be sure to get the positive root. Since the index \(n\) is even, \(\sqrt[n]{a^{n}}=|a|\),
\(\sqrt{x^{2}}=|x|\)
- This is an odd indexed root so there is no need for an absolute value sign. Since the index is \(n\) is odd, \(\sqrt[n]{a^{n}}=a\),
\(\sqrt[3]{m^{3}}=m\)
- Since the index \(n\) is even \(\sqrt[n]{a^{n}}=|a|\),
\(\sqrt[4]{p^{4}}=|p|\)
- Since the index \(n\) is odd, \(\sqrt[n]{a^{n}}=a\),
\(\sqrt[5]{y^{5}}=y\)
Simplify:
- \(\sqrt{b^{2}}\)
- \(\sqrt[3]{w^{3}}\)
- \(\sqrt[4]{m^{4}}\)
- \(\sqrt[5]{q^{5}}\)
- Answer
-
- \(|b|\)
- \(w\)
- \(|m|\)
- \(q\)
What about square roots of higher powers of variables? The Power Property of Exponents says \(\left(a^{m}\right)^{n}=a^{m \cdot n}\). So if we square \(a^{m}\), the exponent will become \(2m\).
\(\left(a^{m}\right)^{2}=a^{2 m}\)
Looking now at the square root.
\(\sqrt{a^{2 m}}\)
Since \(\left(a^{m}\right)^{2}=a^{2 m}\).
\(\sqrt{\left(a^{m}\right)^{2}}\)
Since \(n\) is even \(\sqrt[n]{a^{n}}=|a|\).
\(\left|a^{m}\right|\)
So \(\sqrt{a^{2 m}}=\left|a^{m}\right|\).
We apply this concept in the next example.
Simplify:
- \(\sqrt{x^{6}}\)
- \(\sqrt{y^{16}}\)
Solution :
- Since \(\left(x^{3}\right)^{2}=x^{6}\) and the index \(n\) is even \(\sqrt{a^{n}}=|a|\), \[\begin{align*}\sqrt{x^{6}}&=\sqrt{\left(x^{3}\right)^{2}}\\&=\left|x^{3}\right|\end{align*}\]
Hence, \(\sqrt{x^{6}}=\left|x^{3}\right|\)
- Since \(\left(y^{8}\right)^{2}=y^{16}\) and the index \(n\) is even \(\sqrt[n]{a^{n}}=|a|\), \[\begin{align*}\sqrt{y^{16}}&=\sqrt{\left(y^{8}\right)^{2}}\\&=y^{8}\end{align*}\]
Hence, \(\sqrt{y^{16}}=y^{8}\)
In this case the absolute value sign is not needed as \(y^{8}\) is positive.
Simplify:
- \(\sqrt{y^{18}}\)
- \(\sqrt{z^{12}}\)
- Answer
-
- \(|y^{9}|\)
- \(z^{6}\)
The next example uses the same idea for higher roots.
Simplify:
- \(\sqrt[3]{y^{18}}\)
- \(\sqrt[4]{z^{8}}\)
Solution :
- Since \(\left(y^{6}\right)^{3}=y^{18}\) and \(n\) is odd, \(\sqrt[n]{a^{n}}=a\), \[\begin{align*}\sqrt[3]{y^{18}}=\sqrt[3]{\left(y^{6}\right)^{3}}=y^{6}\end{align*}\]
Hence, \(\sqrt[3]{y^{18}}=y^{6}\)
- Since \(\left(z^{2}\right)^{4}=z^{8}\) and \(z^{2}\) is positive, we do not need an absolute value sign, \[\begin{align*}\sqrt[4]{z^{8}}&=\sqrt[4]{\left(z^{2}\right)^{4}}\\&=z^{2}\end{align*}\]
Hence, \(\sqrt[4]{z^{8}}=z^{2}\)
Simplify:
- \(\sqrt[4]{u^{12}}\)
- \(\sqrt[3]{v^{15}}\)
- Answer
-
- \(|u^{3}|\)
- \(v^{5}\)
In the next example, we now have a coefficient in front of the variable. The concept \(\sqrt{a^{2 m}}=\left|a^{m}\right|\) works in much the same way.
\(\sqrt{16 r^{22}}=4\left|r^{11}\right|\) because \(\left(4 r^{11}\right)^{2}=16 r^{22}\).
But notice \(\sqrt{25 u^{8}}=5 u^{4}\) and no absolute value sign is needed as \(u^{4}\) is always positive.
Simplify:
- \(\sqrt{16 n^{2}}\)
- \(-\sqrt{81 c^{2}}\)
Solution :
- Since \((4 n)^{2}=16 n^{2}\) and the index \(n\) is even \(\sqrt[n]{a^{n}}=|a|\), \[\begin{align*}\sqrt{16 n^{2}}&=\sqrt{(4 n)^{2}}\\&=4|n|\end{align*}\]
Hence, \(\sqrt{16 n^{2}}=4|n|\)
- Since \((9 c)^{2}=81 c^{2}\) and the index \(n\) is even \(\sqrt[n]{a^{n}}=|a|\), \[\begin{align*}-\sqrt{81 c^{2}}&=-\sqrt{(9 c)^{2}}\\&=-9|c| \end{align*}\]
Hence, \(-\sqrt{81 c^{2}}=-9|c|\)
Simplify:
- \(\sqrt{64 x^{2}}\)
- \(-\sqrt{100 p^{2}}\)
- Answer
-
- \(8|x|\)
- \(-10|p|\)
This example just takes the idea farther as it has roots of higher index.
Simplify:
- \(\sqrt[3]{64 p^{6}}\)
- \(\sqrt[4]{16 q^{12}}\)
Solution :
- Rewrite \(64p^{6}\) as \(\left(4 p^{2}\right)^{3}\), \[\sqrt[3]{64 p^{6}}=\sqrt[3]{\left(4 p^{2}\right)^{3}}\nonumber\]
Take the cube root, \[\sqrt[3]{64 p^{6}}=4p^{2}\nonumber\]
- Rewrite the radicand as a fourth power, \[\sqrt[4]{16 q^{12}}=\sqrt[4]{\left(2 q^{3}\right)^{4}}\nonumber\]
Take the fourth root, \[\sqrt[3]{64 p^{6}}=2|q^{3}|\nonumber\]
Simplify:
- \(\sqrt[3]{27 x^{27}}\)
- \(\sqrt[4]{81 q^{28}}\)
- Answer
-
- \(3x^{9}\)
- \(3|q^{7}|\)
The next examples have two variables.
Simplify:
- \(\sqrt{36 x^{2} y^{2}}\)
- \(\sqrt{121 a^{6} b^{8}}\)
- \(\sqrt[3]{64 p^{63} q^{9}}\)
Solution :
- Since \((6 x y)^{2}=36 x^{2} y^{2}\), \[\sqrt{36 x^{2} y^{2}}=\sqrt{(6 x y)^{2}}\nonumber\]
Take the square root, \(\sqrt{36 x^{2} y^{2}}=6|xy|\)
- Since \(\left(11 a^{3} b^{4}\right)^{2}=121 a^{6} b^{8}\), \[\sqrt{121 a^{6} b^{8}}=\sqrt{\left(11 a^{3} b^{4}\right)^{2}}\nonumber\]
Take the square root, \[\sqrt{121 a^{6} b^{8}}=11\left|a^{3}\right| b^{4}\nonumber\]
- Since \(\left(4 p^{21} q^{3}\right)^{3}=64 p^{63} q^{9}\), \[\sqrt[3]{64 p^{63} q^{9}}=\sqrt[3]{\left(4 p^{21} q^{3}\right)^{3}}\nonumber\]
Take the cube root, \[\sqrt[3]{64 p^{63} q^{9}}=4p^{21}q^{3}\nonumber\]
Simplify:
- \(\sqrt{100 a^{2} b^{2}}\)
- \(\sqrt{144 p^{12} q^{20}}\)
- \(\sqrt[3]{8 x^{30} y^{12}}\)
- Answer
-
- \(10|ab|\)
- \(12p^{6}q^{10}\)
- \(2x^{10}y^{4}\)


20.9.1: Genetic Variation
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Learning Objectives
- Assess the ways in which genetic variance affects the evolution of populations
Genetic Variation
Genetic variation is a measure of the genetic differences that exist within a population. The genetic variation of an entire species is often called genetic diversity. Genetic variations are the differences in DNA segments or genes between individuals and each variation of a gene is called an allele.For example, a population with many different alleles at a single chromosome locus has a high amount of genetic variation. Genetic variation is essential for natural selection because natural selection can only increase or decrease frequency of alleles that already exist in the population.
Genetic variation is caused by:
- mutation
- random mating between organisms
- random fertilization
- crossing over (or recombination) between chromatids of homologous chromosomes during meiosis
The last three of these factors reshuffle alleles within a population, giving offspring combinations which differ from their parents and from others.
Evolution and Adaptation to the Environment
Variation allows some individuals within a population to adapt to the changing environment. Because natural selection acts directly only on phenotypes, more genetic variation within a population usually enables more phenotypic variation. Some new alleles increase an organism’s ability to survive and reproduce, which then ensures the survival of the allele in the population. Other new alleles may be immediately detrimental (such as a malformed oxygen-carrying protein) and organisms carrying these new mutations will die out. Neutral alleles are neither selected for nor against and usually remain in the population. Genetic variation is advantageous because it enables some individuals and, therefore, a population, to survive despite a changing environment.
Geographic Variation
Some species display geographic variation as well as variation within a population. Geographic variation, or the distinctions in the genetic makeup of different populations, often occurs when populations are geographically separated by environmental barriers or when they are under selection pressures from a different environment. One example of geographic variation are clines: graded changes in a character down a geographic axis.
Sources of Genetic Variation
Gene duplication, mutation, or other processes can produce new genes and alleles and increase genetic variation. New genetic variation can be created within generations in a population, so a population with rapid reproduction rates will probably have high genetic variation. However, existing genes can be arranged in new ways from chromosomal crossing over and recombination in sexual reproduction. Overall, the main sources of genetic variation are the formation of new alleles, the altering of gene number or position, rapid reproduction, and sexual reproduction.
Key Points
- Genetic variation is an important force in evolution as it allows natural selection to increase or decrease frequency of alleles already in the population.
- Genetic variation can be caused by mutation (which can create entirely new alleles in a population), random mating, random fertilization, and recombination between homologous chromosomes during meiosis (which reshuffles alleles within an organism’s offspring).
- Genetic variation is advantageous to a population because it enables some individuals to adapt to the environment while maintaining the survival of the population.
Key Terms
- genetic diversity : the level of biodiversity, refers to the total number of genetic characteristics in the genetic makeup of a species
- crossing over : the exchange of genetic material between homologous chromosomes that results in recombinant chromosomes
- phenotypic variation : variation (due to underlying heritable genetic variation); a fundamental prerequisite for evolution by natural selection
- genetic variation : variation in alleles of genes that occurs both within and among populations