Learning Lab

Example 1

Evaluate \(\frac{a+5}{b}\) if \(a=-9\) and \(b=2.\)
Solution: \[\begin{align*} \frac{a+5}{b} & =\frac{-9+5}{2}\\ & =\frac{-4}{2}\\ & =-2. \end{align*}\]

Example 2

Evaluate \(w^{2}-2z\) if \(w=-1\) and \(z=5.\)
Solution: \[\begin{align*} w^{2}-2z & =\left(-1\right)\times\left(-1\right)-2\times5\\ & =1-10\\ & =-9. \end{align*}\]

Example 3

Use the formula \[ C=\frac{5\left(F-32\right)}{9} \] to convert a temperature of \(212^{\circ}\)Fahrenheit (F) to Centigrade (C).
Solution:
Substituting \(F=212\) into the formula gives: \[\begin{align*} C & =\frac{5\left(F-32\right)}{9}\\ & =\frac{5\left(212-32\right)}{9}\\ & =\frac{5\times180}{9}\\ & =100^{\circ}\ \textrm{centigrade.} \end{align*}\]

Example 4

The current in an electrical circuit is given by \(V=IR\) where \(V\)is the voltage (in volts), \(I\) is the current (in amps) and \(R\) is the resistance (in ohms).

If the resistance is \(5\) ohms and the current is \(2\) amps, what is the voltage, \(V?\)
Solution:
We have \(r=5\) and \(I=2.\)Therefore, the voltage \[\begin{align*} V & =IR\\ & =2\times5\\ & =10\,\textrm{volts.} \end{align*}\]

Example 5

The volume \(V,\) of a right circular cone with base radius \(r\) and height \(h\) is given by \(V=\frac{1}{3}\pi r^{2}h.\)

If the radius of a cone is \(5\,cm\) and its height is \(15\:cm.\) what is the volume of the cone?
Solution:
In this case \(r=5\) and \(h=15\), so the volume of the cone is \[\begin{align*} V & =\frac{1}{3}\pi r^{2}h\\ & =\frac{1}{3}\times\pi\times5^{2}\times15\\ & \approx392.7\:cm^{3}. \end{align*}\]

Example 6

The formula relating distance traveled \(s,\) to initial speed \(u,\) acceleration \(a\) and time \(t\) is \[\begin{align*} s & =ut+\frac{1}{2}at^{2}. \end{align*}\] A car traveling at a speed of \(4\,m/s\) accelerates at a rate of \(2\:m/s^{2}\) for \(5\,s.\) How far does it travel during this time?
Solution:
In this case, \(u=4,\) \(a=2\) and \(t=5\). So the distance traveled is \[\begin{align*} s & =ut+\frac{1}{2}at^{2}\\ & =4\times5+\frac{1}{2}\times2\times5^{2}\\ & =20+\frac{1}{2}\times2\times25\\ & =45\:m. \end{align*}\]

Example 7

The area of a circle \(A\) is given by \[ A=\pi r^{2} \] where \(r\) is the radius. If a circle has an area of \(30\,cm^{2}\) what is it’s radius.
Solution:
Substituting \(A=30\) we have \[\begin{align*} 30 & =\pi r^{2}. \end{align*}\] We want \(r\). First we divide both sides by \(\pi\) to get \[\begin{align*} \frac{30}{\pi} & =\frac{\pi r^{2}}{\pi}.\\ & =r^{2}\\ r^{2} & =\frac{30}{\pi}\\ r & =\sqrt{\frac{30}{\pi}}\\ & \approx3.1\:cm. \end{align*}\]

2. Evaluate the following if \(a=-1,\,b=6,\,c=3,\,m=-2,\,n=2.\)

\[\begin{array}{ll} a)\,3a^{2}-7\quad & b)\,3a-b^{2}\\ c)\,\left(2m+1\right)^{2} & \quad d)\,3\left(a-b^{2}\right)\\ e)\,\frac{2m}{n} & \quad f)\,\left(m-n\right)^{2} \end{array}\]