Practice 1 Solutions

5. M varies directly to R and inversely to T, when $M=–11,$ $R=22,$ and $T=4.$ Find M when $R=–3$ and $T=–2.$

$\begin{array}{rcl}M& =& \frac{kR}{T}\\ –11& =& \frac{k\left(22\right)}{4}\\ –44& =& 22k\\ k& =& –2\\ M& =& \frac{\left(–2\right)\left(–3\right)}{–2}\end{array}$

$M=–3$

6. y varies directly as the cube root of x, when $x=27$ and$y=4.$ Find y when $x=64.$

$$\begin{gathered} y=k \sqrt[3]{x} \\ 4=k \sqrt[3]{27} \\ 4=3k \\ k=\frac{4}{3} \\ y=\frac{4}{3}\sqrt[3]{64} \\ y=\frac{4}{3}\left(4\right) \end{gathered}$$

$y=\frac{16}{3}$

7. z varies jointly as the square of y and the cube of x, when $z=–8,$$y=2$, and $x=–3.$ Find z when$y=1$ and $x=2.$

$\begin{array}{rcl}z& =& k{y}^2{x}^3\\ –8& =& k{\left(2\right)}^2{\left(–3\right)}^3\\ –8& =& k\left(4\right)\left(–27\right)\\ k& =& \frac{2}{27}\\ z& =& \frac{2}{27}{\left(1\right)}^2{\left(2\right)}^3\end{array}$

$z=\frac{16}{27}$

8. a varies inversely as the square of b, when $a=16$ and $b=2.$ Find a when $b=7.$

$\begin{array}{rcl}a& =& \frac{k}{b^2}\\ 16& =& \frac{k}{{\left(2\right)}^2}\\ 16& =& \frac{k}{4}\\ 64& =& k\\ a& =& \frac{64}{7^2}\end{array}$

$a=\frac{64}{49}$

9. y varies directly as x, when$y=8$ and $x=0.5.$ Find x when$y=–5.$

$\begin{array}{rcl}y& =& kx\\ 8& =& k\left(0.5\right)\\ k& =& 16\\ –5& =& 16x\end{array}$

$x=–\frac{5}{16}$

10. g varies inversely as h, when $g=–9$ and $h=–12.$ Find g when $h=–33.$

$\begin{array}{rcl}g& =& \frac{k}{h}\\ –9& =& \frac{k}{–12}\\ k& =& 108\\ g& =& \frac{108}{–33}\end{array}$

$g=–\frac{36}{11}$

11. C varies jointly as V and T when $C=100,$$T=5,$ and $V=2000.$ Find C when $T=8$ and $V=3200.$

$\begin{array}{rcl}C& =& kVT\\ 100& =& k\left(5\right)\left(2000\right)\\ 100& =& 10,000k\\ k& =& \frac{1}{100}\\ C& =& \frac{1}{100}\left(8\right)\left(3200\right)\end{array}$

$C=256$

12.  f varies directly as g and inversely as the square root of h when$f=6,$ $g=–4,$ and $h=25.$ Find f when $g=10$ and $h=9.$

$\begin{array}{rcl}f& =& \frac{kg}{\sqrt{h}}\\ 6& =& \frac{k\left(–4\right)}{\sqrt{25}}\\ 6& =& –\frac{4}{5}k\\ k& =& –\frac{15}{2}\\ f& =& –\frac{{\displaystyle \frac{15}{2}}\left(10\right)}{\sqrt{9}}\\ f& =& –\frac{15\left(5\right)}{3}\end{array}$

$f=–25$

13. The amount of money earned (before taxes) in an hourly employee’s paycheck is directly related to the number of hours they work. If an employee earned $61.13 for 7.5 hours of work, how much will they earn for working 20 hours? What is their hourly rate?

$\begin{array}{rcl}& & \begin{array}{ccc}y=kx& & \\ \frac{61.13}{7.5}=\frac{y}{20}& \mathrm{or}& \frac{61.13}{7.5}=8.15\\ \left(61.13\right)\left(20\right)=7.5y& & y=8.15x\\ \frac{1222.6}{7.5}=\frac{7.5y}{7.5}& & y=8.15\left(20\right)\\ y=163.01& & y=163.01\end{array}\end{array}$

When an employee works 20 hours, they will earn $163.01 at $8.15 per hour.

14. The pressure, P, of a spray can varies inversely to its volume, V. When the volume is 10 cubic inches, the pressure is 14 pounds per square inch. Find the volume when the pressure is 24 pounds per square inch.

$\begin{array}{rcl}P& =& \frac{k}{V}\\ 14& =& \frac{k}{10}\\ k& =& 140\\ 24& =& \frac{140}{V}\\ 24V& =& 140\\ V& =& \frac{140}{24}\end{array}$

The volume is $\frac{35}{6} {\mathrm{in}}^3$ when the pressure is 24 pounds per square inch.

15. The owners of Steve’s Surfboards determined that the monthly quantity sold, S, varies directly with their advertising budget, A, and inversely with the price, P, of the surfboards. When $75,000 is spent on advertising and the price of each surfboard is $250, the monthly quantity sold is 6,000 surfboards. How much needs to be spent on advertising to increase the monthly quantity sold to 10,000 surfboards?

$\begin{array}{rcl}S& =& \frac{kA}{P}\\ 6000& =& \frac{k\left(75000\right)}{250}\\ 6000& =& 300k\\ k& =& 20\\ 10000& =& \frac{20A}{250}\\ 2500000& =& 20A\\ A& =& 125000\end{array}$

A total of $125,000 should be spent on advertising to increase the monthly sales to 10,000 surfboards.

16. Centrifugal force, F, varies directly with velocity squared, $v^2,$ and inversely as the radius, r. The body’s mass, m, is constant (in this problem, m replaces k). When an object is moving in a circular path with a radius of 5 meters and a velocity of $8 \raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.,$ it has a centrifugal force of 32 Newtons. Find the centrifugal force of the object moving in a circular path with a radius of 10 meters and velocity of $6 \raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right..$ (The mass unit will be kilograms.)

$\begin{array}{rcl}F& =& \frac{m{v}^2}{r}\\ 32& =& \frac{m{\left(8\right)}^2}{5}\\ 160& =& 64m\\ m& =& 2.5 \mathrm{kg}\\ F& =& \frac{2.5{\left(6\right)}^2}{10}\\ F& =& 9\end{array}$

The centrifugal force is 9 Newtons for an object moving in a circular path with a radius of 10 meters.

17. x varies directly to the cube root of z and inversely to y.

$\begin{array}{rcl}x& =& \frac{k \sqrt[3]{z}}{y}\\ xy& =& k \sqrt[3]{z} \end{array}$

$y=\frac{k \sqrt[3]{z}}{x}$

18. x varies jointly as z and the sum of y and w.

$\begin{array}{rcl}x& =& kz\left(y+w\right)\\ \frac{x}{kz}& =& y+w\end{array}$

$y=\frac{x}{kz}–w$

19. x varies directly as the square of y and inversely with the cube root of z.

$\begin{array}{rcl}x& =& \frac{k{y}^2}{\sqrt[3]{z}}\\ x \sqrt[3]{z}& =& k{y}^2\\ \frac{x \sqrt[3]{z}}{y^2}& =& k\\ & & \end{array}$

$k=\frac{x \sqrt[3]{z}}{y^2}$

20. y varies jointly with x and the difference of c and b, and inversely with the square of z.

$\begin{array}{rcl}y& =& \frac{kx\left(c–b\right)}{z^2}\\ y{z}^2& =& kx\left(c–b\right)\\ \frac{y{z}^2}{x\left(c–b\right)}& =& k\end{array}$

$k=\frac{y{z}^2}{x\left(c–b\right)}$