Midterm 1: Version A Answer Key

$\begin{array}{l} \\ \\ \\ \\ \\ \\ \\ \\ -(-3)-\sqrt{(-3)^2-4(4)(-1)} \\ \\ 3-\sqrt{9+16} \\ \\ 3-\sqrt{25} \\ \\ 3-5 \\ \\ -2 \end{array}$
$\begin{array}{rrrrrrrrrrrr} \\ \\ \\ \\ \\ \\ \\ &2x&-&10&-&85&=&3&-&9x&-&54 \\ +&9x&+&10&+&85&&&+&9x&+&3 \\ &&&&&&&&&&+&85 \\ &&&&&&&&&&+&10 \\ \midrule &&&&&\dfrac{11x}{11}&=&\dfrac{44}{11}&&&& \\ \\ &&&&&x&=&4&&&& \end{array}$
$\begin{array}{rrl} \\ \\ \\ \\ \\ A(B-b)&=&h \\ \\ B-b&=&\dfrac{h}{A} \\ \\ -b&=&\dfrac{h}{A}-B \end{array}$
$\phantom{1}$
$\left(\dfrac{x+1}{4}-\dfrac{5}{8}=\dfrac{x-1}{8}\right)(8) \\$
$\begin{array}{rrrrrrrrr} 2(x&+&1)&-&5&=&x&-&1 \\ 2x&+&2&-&5&=&x&-&1 \\ -x&-&2&+&5&&-x&+&5 \\ &&&&&&&-&2 \\ \midrule &&&&x&=&2&& \end{array}$
$x=-2$
$\begin{array}{ll} \\ \\ \\ \begin{array}{rrl} m&=&\dfrac{\Delta y}{\Delta x} \\ \\ \dfrac{2}{5}&=&\dfrac{y--2}{x--1} \end{array} & \hspace{0.25in} \begin{array}{rrrrrrrrl} \\ \\ \\ &&2(x&+&1)&=&5(y&+&2) \\ &&2x&+&2&=&5y&+&10 \\ &&-5y&-&10&&-5y&-&10 \\ \midrule 2x&-&5y&-&8&=&0&& \\ &&&&0&=&2x&-&5y-8 \\ \\ &&&&y&=&\dfrac{2}{5}x&-&\dfrac{8}{5} \end{array} \end{array}$
$\begin{array}{ll} \\ \\ \\ \\ \\ \\ \\ \begin{array}{rrl} \text{1st slope} && \\ m&=&\dfrac{\Delta y}{\Delta x} \\ \\ m&=&\dfrac{4-0}{6--2} \\ \\ m&=&\dfrac{4}{8}\text{ or }\dfrac{1}{2} \end{array} & \hspace{0.25in} \begin{array}{rrl} \\ \\ \\ \\ \text{Now:} && \\ m&=&\dfrac{\Delta y}{\Delta x} \\ \\ \dfrac{1}{2}&=&\dfrac{y-0}{x--2} \\ \\ 1(x+2) &=&2(y-0) \\ x+2&=&2y \\ \\ \therefore y&=&\dfrac{1}{2}+1 \\ \\ \text{or }x-2y+2&=&0 \end{array} \end{array}$
$\begin{array}{rrrrrrr} \\ \\ \\ \\ \\ 6x&-&5&-&30x&>&67 \\ &+&5&&&&+5 \\ \midrule &&&&\dfrac{-24x}{-24}&>&\dfrac{72}{-24} \\ \\ &&&&x&<&-3 \end{array}$
$\begin{array}{rrrcrrr} \\ \\ \\ \\ \\ -10&\le &4x&-&2&\le &14 \\ +2&&&+&2&&+2 \\ \midrule \dfrac{-8}{4}&\le &&\dfrac{4x}{4}&&\le &\dfrac{16}{4} \\ \\ -2&\le &&x&&\le &4 \end{array}$
$\begin{array}{ll} \begin{array}{rrl} \dfrac{3x+2}{5}&=& 2 \\ \\ 3x+2&=&10 \\ -2&&-2 \\ \midrule \dfrac{3x}{3}&=&\dfrac{8}{3} \\ \\ x&=&\dfrac{8}{3} \end{array} & \hspace{0.25in} \begin{array}{rrl} \dfrac{3x+2}{5}&=& -2 \\ \\ 3x+2 &=&-10 \\ -2&&-2 \\ \midrule \dfrac{3x}{3}&=&\dfrac{-12}{3} \\ \\ x&=&-4 \end{array} \end{array}$
$5x+2y<15$

$x$ $y$

0 7.5

3 0

1 5

5 −5
$\begin{array}{rrrrrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ &(5L&+&3S&=&47)(2) \\ &(4L&-&2S&=&20)(3) \\ \\ &10L&+&6S&=&94 \\ +&12L&-&6S&=&60 \\ \midrule &&&\dfrac{22L}{22}&=&\dfrac{154}{22} \\ \\ &&&L&=&7 \\ \\ \therefore &4L&-&2S&=&\phantom{-}20 \\ &4(7)&-&2S&=&\phantom{-}20 \\ &28&-&2S&=&\phantom{-}20 \\ -&28&&&&-28 \\ \midrule &&&\dfrac{-2S}{-2}&=&\dfrac{-8}{-2} \\ \\ &&&S&=&4 \end{array}$
$\begin{array}{rrl} \\ \\ \\ \\ \\ \\ 36\text{ cm}&=&5x+x \\ 36\text{ cm}&=&6x \\ \\ x&=&\dfrac{36\text{ cm}}{6} \\ \\ x&=&6\text{ cm} \\ 5x&=&5(6)=30 \text{ cm} \end{array}$
$\begin{array}{ll} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \begin{array}{rrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ \text{\underline{1st}}&& \\ y&=&\dfrac{kmn}{d^2} \\ \\ \text{\underline{2nd}}&& \\ y&=&3 \\ k&=&\text{find} \\ m&=&2 \\ n&=&8 \\ d&=&4 \\ \\ y&=&\dfrac{kmn}{d^2} \\ \\ 3&=&\dfrac{k(2)(8)}{(4)^2} \\ \\ k&=&\dfrac{3\cdot (4)^2}{2\cdot 8} \\ \\ k&=& 3 \end{array} & \hspace{0.25in} \begin{array}{rrl} \\ \\ \\ \text{\underline{3rd}}&& \\ y&=&\text{find} \\ k&=&3 \\ m&=&15 \\ n&=&10 \\ d&=& 5 \\ \\ y&=&\dfrac{kmn}{d^2} \\ \\ y&=&\dfrac{(3)(15)(10)}{(5)^2} \\ \\ y&=&18 \end{array} \end{array}$

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