Midterm 3: Version C Answer Key

$\dfrac{\cancel{15}3\cancel{m^3}}{4\cancel{n^2}}\cdot \dfrac{\cancel{17}\cancel{n^3}}{\cancel{30}\cancel{10}2\cancel{m^3}}\cdot \dfrac{\cancel{3}1m^4}{\cancel{34}2n^{\cancel{2}}}\Rightarrow \dfrac{3m^4}{16n}$
$\dfrac{5v^2-25v}{5v+25}\cdot \dfrac{10v}{v^2-11v+30}\Rightarrow \dfrac{\cancel{5}v\cancel{(v-5)}}{\cancel{5}(v+5)}\cdot \dfrac{10v}{\cancel{(v-5)}(v-6)}\Rightarrow \dfrac{10v^2}{(v+5)(v-6)}$
$\phantom{1}$
$\left(\dfrac{8}{2x}=\dfrac{2}{x}+1\right)(2x) \\$
$\begin{array}{rrl} 8&=&2\cdot 2+1(2x) \\ 8&=&\phantom{-}4+2x \\ -4&&-4 \\ \midrule \dfrac{4}{2}&=&\dfrac{2x}{2} \\ \\ x&=&2 \end{array}$
$\begin{array}{l} \dfrac{\left(\dfrac{x^2}{y^2}-16\right)y^3}{\left(\dfrac{x+4y}{y^3}\right)y^3}\Rightarrow \dfrac{x^2y-16y^3}{x+4y}\Rightarrow \dfrac{y(x^2-16y^2)}{x+4y}\Rightarrow\dfrac{y(x-4y)\cancel{(x+4y)}}{\cancel{x+4y}} \\ \\ \Rightarrow y(x-4y) \end{array}$
$\begin{array}{l} \\ 5y+2\cdot 9y+6y\sqrt{y} \\ 23y+6y\sqrt{y} \end{array}$
$\dfrac{(28)(7+3\sqrt{5})}{(7-3\sqrt{5})(7+3\sqrt{5})}\Rightarrow \dfrac{196+84\sqrt{5}}{49-9\cdot 5}\Rightarrow \dfrac{196+84\sqrt{5}}{4}\Rightarrow 49+21\sqrt{5}$
$\begin{array}{l} \\ \\ \\ \\ \\ \\ \\ (27a^{-\frac{3}{8}})^{\frac{1}{3}} \\ \\ 27^{\frac{1}{3}}a^{-\frac{3}{8}\cdot \frac{1}{3}} \\ \\ 3a^{-\frac{1}{8}} \\ \\ \dfrac{3}{a^{\frac{1}{8}}}\Rightarrow \dfrac{3}{\sqrt[8]{a}} \end{array}$
$\phantom{1}$
$(\sqrt{3x-2})^2=(\sqrt{5x+4})^2 \\$
$\begin{array}{rrrrrrrr} &3x&-&2&=&5x&+&4 \\ -&3x&-&4&&-3x&-&4 \\ \midrule &&&-6&=&2x&& \\ \\ &&&x&=&\dfrac{-6}{2}&=&-3 \end{array}$
1. $\begin{array}{rrl} \\ \\ \\ \dfrac{2x^2}{2}&=&\dfrac{72}{2} \\ \\ x^2&=&36 \\ x&=&\pm 6 \end{array}$
2. $\begin{array}{rrl} \\ \\ 2x^2-8x&=&0 \\ 2x(x-4)&=&0 \\ x&=&0, 4 \end{array}$
1. $\begin{array}{rrl} \\ (x+5)(x+1)&=&0 \\ x&=&-1, -5 \end{array}$
2. $\begin{array}{l} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \text{Quadratic:} \\ \\ x^2-10x+4=0 \\ \\ \dfrac{-(-10)\pm \sqrt{(-10)^2-4(1)(4)}}{2} \\ \\ \dfrac{-10\pm \sqrt{100-16}}{2} \\ \\ \dfrac{10\pm \sqrt{84}}{2} \\ \\ \dfrac{10\pm 2\sqrt{21}}{2}\Rightarrow 5 \pm \sqrt{21} \end{array}$
$\phantom{1}$
$\left(\dfrac{8}{4x}=\dfrac{2}{x}+3\right)(4x) \\$
$\begin{array}{rrrrl} 8&=&8&+&3(4x) \\ -8&&-8&& \\ \midrule \dfrac{0}{12}&=&\dfrac{12x}{12}&& \\ \\ x&=&0&&\therefore \text{Undefined. No solution} \end{array}$
$\begin{array}{rrl} \\ \\ \\ \\ \\ \\ \\ \text{Let }u&=&x^2 \\ \\ u^2-17u+16&=&0 \\ (u-16)(u-1)&=&0 \\ \\ (x^2-16)(x^2-1)&=&0 \\ (x-4)(x+4)(x-1)(x+1)&=&0 \\ x&=& \pm 1, \pm 4 \end{array}$
$\begin{array}{ll} \begin{array}{rrl} \\ \\ \\ \\ L&=&W+6 \\ \text{Area}&=&12+\text{Perimeter} \\ \\ L\cdot W&=& 12+2L+2W \\ (W+6)W&=&12+2(W+6)+2W \\ W^2+6W&=&12+2W+12+2W \end{array} & \hspace{0.25in} \begin{array}{rrrrcrr} \\ \\ \\ \\ \\ \\ \\ 0&=&W^2&+&6W&& \\ &&&-&4W&-&24 \\ \midrule 0&=&W^2&+&2W&-&24 \\ 0&=&(W&+&6)(W&-&4) \\ W&=&\cancel{-6},&4&&& \\ \\ L&=&W&+&6&& \\ L&=&4&+&6&& \\ L&=&10&&&& \\ \end{array} \end{array}$
$\phantom{1}$
$x, x+2, x+4 \\$
$\begin{array}{ll} \begin{array}{rrrrrrrrrrr} &&x(x&+&4)&=&31&+&x&+&2 \\ x^2&+&4x&&&=&33&+&x&& \\ &-&x&-&33&&-33&-&x&& \\ \midrule x^2&+&3x&-&33&=&0&&&& \end{array} & \hspace{0.25in} \begin{array}{l} \dfrac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \\ \dfrac{-3\pm \sqrt{3^2-4(1)(-33)}}{2} \\ \\ \dfrac{-3\pm \sqrt{141}}{2} \end{array} \end{array}$
$\begin{array}{rrrrcl} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ &d&=&r&\cdot &t \\ \text{To outpost:}&60&=&(B&-&C)5 \\ \text{Back:}&60&=&(B&+&C)3 \\ \\ &60&=&5B&-&5C \\ &60&=&3B&+&3C \\ \\ &12&=&B&-&C \\ +&20&=&B&+&C \\ \midrule &32&=&2B&& \\ &\therefore B&=&16&\text{ km/h}& \\ \\ &\therefore B&+&C&=&\phantom{-}20 \\ &16&+&C&=&\phantom{-}20 \\ -&16&&&&-16 \\ \midrule &&&C&=&4\text{ km/h} \end{array}$

License

Icon for the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License

Intermediate Algebra Copyright © 2020 by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

Share This Book