Midterm 1: Version E Answer Key
- \(\phantom{1}\)
- −9
- 9
- −9
- 10
- −3
- \(\begin{array}{rrrrrrlrrrr}
\\ \\ \\ \\ \\
2x&-&8&+&18&=&-12&+&4x&+&12 \\
-2x&&&&&&&-&2x&& \\
\hline
&&&&\dfrac{10}{2}&=&\dfrac{2x}{2}&&&& \\ \\
&&&&x&=&5&&&&
\end{array}\) - \(\phantom{1}\)
\(\left(\dfrac{1}{R}-\dfrac{1}{r_1}=\dfrac{1}{r_2}\right)(Rr_1r_2) \\ \)
\(\begin{array}{rrrrlrr}
r_1r_2&-&Rr_2&=&\phantom{-}Rr_1&& \\
-Rr_1&+&Rr_2&&-Rr_1&+&Rr_2 \\
\hline
r_1r_2&-&Rr_1&=&Rr_2&& \\
r_1(r_2&-&R)&=&Rr_2&& \\ \\
&&r_1&=&\dfrac{Rr_2}{r_2-R}&&
\end{array}\) - \(\phantom{1}\)
\(\left(\dfrac{x}{12}-\dfrac{x-4}{3}=\dfrac{2}{3} \right)(12) \\ \)
\(\begin{array}{rrcrrrl}
x&-&4(x&-&4)&=&4(2) \\
x&-&4x&+&16&=&8 \\
&&&-&16&&-16 \\
\hline
&&&&\dfrac{-3x}{-3}&=&\dfrac{-8}{-3} \\ \\
&&&&x&=&\dfrac{8}{3}
\end{array}\) - \(y=-6\)
- \(\begin{array}{ll}
\\ \\ \\
\begin{array}{rrl}
m&=&\dfrac{\Delta y}{\Delta x} \\ \\
\dfrac{2}{5}&=&\dfrac{y-1}{x--1}
\end{array}
& \hspace{0.25in}
\begin{array}{rrrrrrlrr}
\\ \\ \\
&&2(x&+&1)&=&\phantom{-}5(y&-&1) \\
&&2x&+&2&=&\phantom{-}5y&-&5 \\
&-&5y&+&5&&-5y&+&5 \\
\hline
2x&-&5y&+&7&=&0&& \\ \\
&&&&y&=&\dfrac{2}{5}x&+&\dfrac{7}{5}
\end{array}
\end{array}\) - \(\begin{array}{ll}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
\begin{array}{rrl}
&&\textbf{1st slope:} \\ \\
m&=&\dfrac{\Delta y}{\Delta x} \\ \\
m&=&\dfrac{5--1}{2-0} \\ \\
m&=&\dfrac{6}{2} \\ \\
m&=&3
\end{array}
& \hspace{0.25in}
\begin{array}{rrl}
\\ \\ \\
&&\textbf{Now:} \\ \\
m&=&\dfrac{\Delta y}{\Delta x} \\ \\
3&=&\dfrac{y--1}{x-0} \\ \\
3x&=&y+1 \\ \\
3x-y-1&=&0 \\ \\
\therefore 0&=&3x-y-1 \\
y&=&3x-1
\end{array}
\end{array}\) 
- \(\begin{array}{rrrrrrr}
\\ \\ \\ \\ \\
-20&\le &8x&-&4&\le &28 \\
+4&&&+&4&&+4 \\
\hline
\dfrac{-16}{8}&\le &&\dfrac{8x}{8}&&\le &\dfrac{32}{8} \\ \\
-2&\le &&x&&\le &4
\end{array}\)
- \(\phantom{1}\)
\(\left(-2\le \dfrac{2x+2}{6} \le 2 \right)(6) \\ \)
\(\begin{array}{rrrrrrr}
-12&\le &2x&+&2&\le &12 \\
-2&&&-&2&&-2 \\
\hline
\dfrac{-14}{2}&\le &&\dfrac{2x}{2}&&\le &\dfrac{10}{2} \\ \\
-7&\le &&x&&\le &5
\end{array}\)
- \(\begin{array}{ll}
\\ \\ \\ \\ \\ \\ \\ \\
\left(\dfrac{3x-4}{5} < -1 \right)(5) \hspace{0.25in}& \text{or} \hspace{0.25in} \left(1 < \dfrac{3x-4}{5}\right)(5) \\ \\
\begin{array}{rrrrr}
3x&-&4&<&-5 \\
&+&4&&+4 \\
\hline
&&3x&<&-1 \\ \\
&&x&<&-\dfrac{1}{3}
\end{array}
& \hspace{0.25in}
\begin{array}{rrrrr}
5&<&3x&-&4 \\
+4&&&+&4 \\
\hline
\dfrac{9}{3}&<&\dfrac{3x}{3}&& \\ \\
x&>&3&&
\end{array}
\end{array}\)
-
\(3x-2y<12\) \(x\) \(y\) 0 −6 4 0 
- \(\phantom{1}\)
\(x, x+2, x+4 \\ \)
\(\begin{array}{rrcrrrcrrrl}
x&+&2(x&+&2)&+&3(x&+&4)&=&\phantom{-}94 \\
x&+&2x&+&4&+&3x&+&12&=&\phantom{-}94 \\
&&&&&&6x&+&16&=&\phantom{-}94 \\
&&&&&&&-&16&&-16 \\
\hline
&&&&&&&&\dfrac{6x}{6}&=&\dfrac{78}{6} \\ \\
&&&&&&&&x&=&13
\end{array}\)
\(\phantom{1}\)
numbers are 13, 15, 17 
\(\begin{array}{rrl}
\\ \\ \\ \\ \\
3x+x&=&800\text{ cm} \\
4x&=&800\text{ cm} \\
x&=&\dfrac{800\text{ cm}}{4}\text{ or }200\text{ cm} \\ \\
3x&=&600\text{ cm}
\end{array}\)- \(\begin{array}{ll}
\begin{array}{rrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
y&=&\dfrac{km}{n^2} \\ \\
&&\underline{\text{1st}} \\
y&=&12 \\
k&=&\text{find} \\
m&=&3 \\
n&=&4 \\ \\
12&=&\dfrac{k(3)}{(4)^2} \\ \\
k&=&\dfrac{12\cdot (4)^2}{3} \\ \\
k&=&64
\end{array}
& \hspace{0.25in}
\begin{array}{rrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
&&\underline{\text{2nd}} \\
y&=&\text{find} \\
k&=&64 \\
m&=&3 \\
n&=&-3 \\ \\
y&=&\dfrac{(64)(3)}{(-3)^2} \\ \\
y&=&\dfrac{64}{3}
\end{array}
\end{array}\)
