Practice 2 Solutions

5. y varies directly with the cube of x, when$y=54$ and $x=3.$ Find y when $x=–5.$

$\begin{array}{rcl}y& =& k{x}^3\\ 54& =& k{\left(3\right)}^3\\ 54& =& 27k\\ k& =& 2\\ y& =& 2{\left(–5\right)}^3\end{array}$

$y=–250$

6. y varies jointly with x and the square root of z and inversely as b, when$y=4,$ $x=7,$ $z=16,$ and $b=0.75.$ Find y when $x=–28,$ $z=25,$ and $b=2.$

$\begin{array}{c}\begin{array}{ccc}y=\frac{kx\sqrt{z}}{b}& & \\ 4=\frac{k\left(7\right)\sqrt{16}}{0.75}& & y=\frac{{\displaystyle \frac{3}{28}\left(–28\right)\sqrt{25}}}{2}\\ 4=\frac{k\left(7\right)\left(4\right)}{0.75}& & y=\frac{–3\left(5\right)}{2}\\ 3=28k& & \\ k=\frac{3}{28}& & \end{array}\end{array}$

$y=–\frac{15}{2}$

7. y varies inversely to x when$y=4$ and $x=0.125.$ Find x when$y=9.$

$$\begin{gathered} y=\frac{k}{x} \\ 4=\frac{k}{0.125} \\ k=0.5 \\ 9=\frac{0.5}{x} \\ 9x=0.5 \\ x=\frac{0.5}{9} \end{gathered}$$

$x=\frac{1}{18}=0.0\overline{5}$

8. x varies directly to the cube of y, when $x=16$ and$y=–2.$ Find x when$y=\frac{4}{5}.$

$\begin{array}{rcl}x& =& k{y}^3\\ 16& =& k{\left(–2\right)}^3\\ 16& =& –8k\\ k& =& –2\\ x& =& –2{\left(\frac{4}{5}\right)}^3\end{array}$

$x=–\frac{128}{125}$

9. y varies jointly with x and z, when$y=14,$ $x=–\frac{2}{3},$ and $z=28.$ Find y when $x=\frac{2}{3}$ and $z=\frac{1}{4}.$

$\begin{array}{rcl}y& =& kxz\\ 14& =& k\left(–\frac{2}{3}\right)\left(28\right)\\ 14& =& –\frac{56}{3}k\\ k& =& –\frac{3}{4}\\ y& =& –\frac{3}{4}\left(\frac{2}{3}\right)\left(\frac{1}{4}\right)\end{array}$

$y=–\frac{1}{8}$

10. z varies jointly with a and b and inversely with the square root of c, when $z=–1,$ $b=5,$ $a=4,$ and $c=9.$ Find c when $z=–2,$ $a=1,$ and $b=4.$

$\begin{array}{rcl}z& =& \frac{kab}{\sqrt{c}}\\ –1& =& \frac{k\left(4\right)\left(5\right)}{\sqrt{9}}\\ \left(\frac{3}{20}\right)\left(–1\right)& =& \frac{20}{3}k\left(\frac{3}{20}\right)\\ k& =& –\frac{3}{20}\\ –2& =& –\frac{{\displaystyle \frac{3}{20}}\left(1\right)\left(4\right)}{\sqrt{c}}\\ \left(\frac{1}{2}\right)2\sqrt{c}& =& \frac{12}{20}\left(\frac{1}{2}\right)\\ {\left(\sqrt{c}\right)}^2& =& {\left(\frac{3}{10}\right)}^2\end{array}$

$c=\frac{9}{100}$

11. a varies directly with the square of b and inversely as the cube of c, when $a=4,$ $b=2,$ and $c=–1.$ Find a when $b=–3$ and $c=2.$

$\begin{array}{rcl}a& =& \frac{k{b}^2}{c^3}\\ 4& =& \frac{k{\left(2\right)}^2}{{\left(–1\right)}^3}\\ 4& =& \frac{4k}{–1}\\ 4& =& –4k\\ k& =& –1\\ a& =& –\frac{1{\left(–3\right)}^2}{{\left(2\right)}^3}\\ a& =& –\frac{1\left(9\right)}{8}\end{array}$

$a=–\frac{9}{8}$

12. y varies directly as x, when$y=–6$ and $x=13.$ Find x when$y=\frac{5}{2}.$

$\begin{array}{rcl}y& =& kx\\ –6& =& 13k\\ k& =& –\frac{6}{13}\\ \left(–\frac{13}{6}\right)\frac{5}{2}& =& –\frac{6}{13}x\left(–\frac{13}{6}\right)\end{array}$

$x=–\frac{65}{12}$

13. The total charge to purchase, p, gasoline varies directly to the number of gallons of gasoline, g, purchased. Susan purchased 13 gallons of gasoline for $45.12. Determine the total number of gallons Susan could purchase for $25.

$\begin{array}{rcl}p& =& kg\\ 45.12& =& k\left(13\right)\\ k& =& 3.47\\ 25& =& 3.47g\\ g& =& 7.20 \end{array}$

Susan could purchase 7.20 gallons of gasoline for $25.

14. The volumetric flow rate of water, W, varies inversely as time, t. At 2 seconds, the volume flow rate is 15 cubic meters/second. What is the flow rate at 7 seconds?

$\begin{array}{rcl}W& =& \frac{k}{t}\\ 15& =& \frac{k}{2}\\ k& =& 30\\ W& =& \frac{30}{7}\\ W& =& 4.2857\end{array}$

At 7 seconds, the flow rate is approximately 4.29 cubic meters/second.

15. Kinetic energy, KE, varies jointly as the mass, m, and the square of velocity, v. When the mass of an object is 15 kilograms, the velocity is 4 meters per second and has a kinetic energy of 240 Newtons. Determine the velocity of an object when the kinetic energy is 315 Newtons with a mass of 12 kilograms.

$\begin{array}{rcl}KE& =& km{v}^2\\ 240& =& k\left(15\right){\left(4\right)}^2\\ 240& =& 240k\\ k& =& 1\\ 315& =& 12\left(v^2\right)\\ \sqrt{26.25}& =& \sqrt{v^2}\\ v& =& \pm 5.1234\\ v& =& 5.1234\end{array}$

The velocity of an object with a mass of 12 kilograms is 5.12 meters per second.

16. When an object slides along a surface, kinetic friction opposes the object’s motion. Kinetic friction, $F_f,$ varies directly to the normal force of the object, $F_N.$ If the force of kinetic friction is 0.35 Newtons when the normal force is 0.22 Newtons, determine the force of kinetic friction when the normal force is 0.4 Newtons.

$\begin{array}{rcl}{F}_f& =& k{F}_N\\ 0.35& =& k\left(0.22\right)\\ 1.6& =& k\\ F_f& =& 1.6\left(0.4\right)\\ F_f& =& 0.64\end{array}$

The force of kinetic friction is 0.64 Newtons

17. x varies directly to the sum of z and y, and inversely to the square of a.

$\begin{array}{rcl}x& =& \frac{k\left(y+z\right)}{a^2}\\ x{a}^2& =& k\left(y+z\right)\\ \frac{x{a}^2}{k}& =& y+z\end{array}$

$y=\frac{x{a}^2}{k}–z$

18. x varies inversely to the joint variation of z and y.

$\begin{array}{rcl}x& =& \frac{k}{yz}\\ \left(\frac{1}{xz}\right)xyz& =& k\left(\frac{1}{xz}\right)\end{array}$

$y=\frac{k}{xz}$

19. y varies directly as the sum of x and z, and inversely as the square root of b.

$\begin{array}{rcl}y& =& \frac{k\left(x+z\right)}{\sqrt{b}}\\ \left(\frac{1}{x+z}\right)y\sqrt{b}& =& k\left(x+z\right)\left(\frac{1}{x+z}\right)\end{array}$

$k=\frac{y\sqrt{b}}{x+z}$

20. y varies jointly with x and z, and inversely as the difference of a and b.

$\begin{array}{rcl}y& =& \frac{kxz}{a–b}\\ \left(\frac{1}{xz}\right)y\left(a–b\right)& =& kxz\left(\frac{1}{xz}\right)\end{array}$

$k=\frac{y\left(a–b\right)}{xz}$