Midterm 1: Version C Answer Key
- \(\begin{array}{l}
\\ \\ \\ \\
-(4)-\sqrt{4^2-4(4)1} \\ \\
-4-\sqrt{16-16} \\ \\
-4
\end{array}\) - \(\begin{array}{rrrrrrrcrrrr}
\\ \\ \\ \\ \\
&2x&-&8&+&8&=&3&-&7x&-&21 \\
+&7x&&&&&&&+&7x&& \\
\hline
&&&&&\dfrac{9x}{9}&=&\dfrac{-18}{9}&&&& \\ \\
&&&&&x&=&-2&&&&
\end{array}\) - \(\phantom{1}\)
\(\left(A=\dfrac{h}{B+b}\right)(B+b) \\ \)
\(\begin{array}{rrrrrrr}
\\ \\ \\ \\ \\
\dfrac{A}{A}(B&+&b)&=&\dfrac{h}{A}&& \\ \\
B&+&b&=&\dfrac{h}{A}&& \\
&-&b&&&-&b \\
\hline
&&B&=&\dfrac{h}{A}&-&b
\end{array}\) - \(\phantom{1}\)
\(\left(\dfrac{x}{15}-\dfrac{x-3}{3}=\dfrac{1}{5}\right)(15) \\ \)
\(\begin{array}{rrcrrrl}
x&-&5(x&-&3)&=&3(1) \\
x&-&5x&+&15&=&\phantom{-1}3 \\
&&&-&15&&-15 \\
\hline
&&&&\dfrac{-4x}{-4}&=&\dfrac{-12}{-4} \\ \\
&&&&x&=&3
\end{array}\) - \(x=-2\)
- \(\begin{array}{rrl}
\\ \\ \\ \\
y&=&mx+6 \\
\therefore y&=&\dfrac{2}{3}x-3 \\ \\
\text{or } 3y&=&2x-9 \\
0&=&2x-3y-9 \\
\end{array}\) - \(\begin{array}{ll}
\begin{array}{rrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
&&\textbf{1st slope} \\
m&=&\dfrac{\Delta y}{\Delta x} \\ \\
m&=&\dfrac{4--6}{-1-14} \\ \\
m&=&\dfrac{10}{-15} \\ \\
m&=&-\dfrac{2}{3} \\ \\
&&\textbf{Now:} \\
m&=&\dfrac{\Delta y}{\Delta x} \\ \\
-\dfrac{2}{3}&=&\dfrac{y-4}{x--1}
\end{array}
& \hspace{0.25in}
\begin{array}{rrrrrrrlrr}
\\ \\ \\ \\ \\
&&&-2(x&+&1)&=&\phantom{-}3(y&-&4) \\
&&&-2x&-&2&=&\phantom{-}3y&-&12 \\
+&&&-3y&+&12&&-3y&+&12 \\
\hline
&-2x&-&3y&+&10&=&0&& \\
\text{or}&2x&+&3y&-&10&=&0&& \\ \\
&&&&&y&=&-\dfrac{2}{3}x&+&\dfrac{10}{3}
\end{array}
\end{array}\) -
\(2x-y=-2\) \(x\) \(y\) 0 2 −1 0 −2 −2 
- \(\begin{array}{rrrrrrr}
\\ \\ \\ \\ \\
0&\le &2x&+&4&<&8 \\
-4&&&-4&&&-4 \\
\hline
\dfrac{-4}{2}&\le &&\dfrac{2x}{2}&&<&\dfrac{4}{2} \\ \\
-2&\le &&x&&<&2
\end{array}\)
- \(\begin{array}{rrrrrrrrrrr}
\\ \\
y&-&1&>&3&\hspace{0.25in} \text{or}\hspace{0.25in}&y&-&1&<&-3 \\
&+&1&&+1&&&+&1&&+1 \\
\hline
&&y&>&4&\hspace{0.25in} \text{or}\hspace{0.25in}&&&y&<&-2
\end{array}\)
- \(\begin{array}{rrrrrrrrrrr}
\\ \\ \\
2x&-&3&<&-5&\hspace{0.5in}&2x&-&3&>&5 \\
&+&3&&+3&\hspace{0.5in}&&+&3&&+3 \\
\hline
&&2x&<&-2&\hspace{0.5in}&&&2x&>&8 \\
&&x&<&-1&\hspace{0.5in}&&&x&>&4
\end{array}\)
-
\(y=|x|-3\) \(x\) \(y\) 3 0 2 −1 1 −2 0 −3 −1 −2 −2 −1 −3 0 
- \(\begin{array}{ll}
\\ \\ \\ \\ \\ \\ \\
\begin{array}{rrrrl}
5L&+&3S&=&49 \\
4L&-&2S&=&26 \\ \\
\dfrac{4L}{2}&-&\dfrac{2S}{2}&=&\dfrac{26}{2} \\ \\
2L&-&S&=&13 \\ \\
&\text{or}&S&=&2L-13
\end{array}
& \hspace{0.25in}
\begin{array}{rrcrrrl}
\\ \\ \\
5L&+&3(2L&-&13)&=&49 \\
5L&+&6L&-&39&=&49 \\
&&&+&39&&+39 \\
\hline
&&&&\dfrac{11L}{11}&=&\dfrac{88}{11} \\ \\
&&&&L&=&8 \\ \\
&&&&\therefore S&=&2L-13 \\
&&&&S&=&2(8)-13 \\
&&&&S&=&16-13 \\
&&&&S&=&3
\end{array}
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\ \\
5x+x&=&42 \\
6x&=&42 \\
x&=&\dfrac{42}{6}\text{ or }7 \\ \\
\therefore 5x&=&5(7)\text{ or }35
\end{array}\) - \(\begin{array}{ll}
\\ \\ \\ \\ \\ \\ \\ \\ \\
\begin{array}{rrl}
\\ \\ \\ \\ \\ \\ \\
y&=&\dfrac{km}{d^2} \\ \\
&&\textbf{1st} \\
y&=&3 \\
k&=&\text{find 1st} \\
m&=&2 \\
d&=&4 \\ \\
y&=&\dfrac{km}{d^2} \\ \\
3&=&\dfrac{k(2)}{(4)^2} \\ \\
3&=&\dfrac{3(4)^2}{2} \\ \\
k&=&24
\end{array}
& \hspace{0.25in}
\begin{array}{rrl}
\\ \\ \\ \\ \\ \\ \\
&&\textbf{2nd} \\
y&=&\text{find} \\
k&=&24 \\
m&=&25 \\
d&=&5 \\ \\
y&=&\dfrac{km}{d^2} \\ \\
y&=&\dfrac{(24)(25)}{(5)^2} \\ \\
y&=&24
\end{array}
\end{array}\)
