Midterm 1: Version D Answer Key
- \(\begin{array}{l}
\\ \\ \\ \\
3(4)-\sqrt{4^2-4(4)(1)} \\ \\
12-\sqrt{16-16} \\ \\
12
\end{array}\) - \(\begin{array}{rrrrrrrrrrr}
\\ \\ \\ \\ \\
2x&-&8&+&8&=&-6&+&3x&+&9 \\
&&&&2x&=&3x&+&3&& \\
&&&&-3x&&-3x&&&& \\
\hline
&&&&-x&=&3&&&& \\
&&&&x&=&-3&&&&
\end{array}\) - \(\phantom{1}\)
\(\left(\dfrac{1}{R}=\dfrac{1}{r_1}+\dfrac{1}{r_2}\right)(Rr_1r_2) \\ \)
\(\begin{array}{rrrrlrr}
&&r_1r_2&=&\phantom{-}Rr_2&+&Rr_1 \\
&&-Rr_2&&-Rr_2&& \\
\hline
r_1r_2&-&Rr_2&=&Rr_1&& \\
r_2(r_1&-&R)&=&Rr_1&& \\ \\
&&r_2&=&\dfrac{Rr_1}{r_1-R}&&
\end{array}\) - \(\phantom{1}\)
\(\left(\dfrac{x}{15}-\dfrac{x-3}{3}=\dfrac{1}{3}\right)(15) \\ \)
\(\begin{array}{rrcrrrr}
x&-&5(x&-&3)&=&5 \\
x&-&5x&+&15&=&5 \\
&&&-&15&&-15 \\
\hline
&&&&\dfrac{-4x}{-4}&=&\dfrac{-10}{-4} \\ \\
&&&&x&=&\dfrac{5}{2}
\end{array}\) - \(y=5\)
- \(\begin{array}{ll}
\\ \\ \\
\begin{array}{rrl}
m&=&\dfrac{\Delta y}{\Delta x} \\ \\
\dfrac{2}{3}&=&\dfrac{y-4}{x--2}
\end{array}
& \hspace{0.25in}
\begin{array}{rrrrrrlrr}
\\ \\ \\
&&2(x&+&2)&=&\phantom{-}3(y&-&4) \\
&&2x&+&4&=&\phantom{-}3y&-&12 \\
&-&3y&+&12&&-3y&+&12 \\
\hline
2x&-&3y&+&16&=&0&& \\ \\
&&&&y&=&\dfrac{2}{3}x&+&\dfrac{16}{3}
\end{array}
\end{array}\) - \(\begin{array}{ll}
\\ \\ \\ \\ \\ \\
\begin{array}{rrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
&&\textbf{1st slope:} \\
m&=&\dfrac{\Delta y}{\Delta x} \\ \\
m&=&\dfrac{-9--7}{8-12} \\ \\
m&=&\dfrac{-2}{-4} \\ \\
m&=&\dfrac{1}{2} \\ \\
&&\textbf{Now:} \\
m&=&\dfrac{\Delta y}{\Delta x} \\ \\
\dfrac{1}{2}&=&\dfrac{y--9}{x-8}
\end{array}
& \hspace{0.25in}
\begin{array}{rrrrrrrrr}
&&x&-&8&=&2(y&+&9) \\
&&x&-&8&=&2y&+&18 \\
&-&2y&-&18&&-2y&-&18 \\
\hline
x&-&2y&-&26&=&0&&
\end{array}
\end{array}\) 
- \(\begin{array}{rrrrrrr}
\\ \\ \\ \\ \\
-27&\le &6x&-&9&\le &3 \\
+9&&&+&9&&+9 \\
\hline
\dfrac{-18}{6}&\le &&\dfrac{6x}{6}&&\le &\dfrac{12}{6} \\ \\
-3&\le &&x&&\le &2
\end{array}\)
- \(\begin{array}{ll}
\\ \\ \\ \\ \\ \\ \\ \\ \\
\dfrac{2x+2}{6}=2& \hspace{0.75in} \dfrac{2x+2}{6}=-2 \\ \\
\begin{array}{rrrrl}
2x&+&2&=&2\cdot 6 \\
2x&+&2&=&12 \\
&-&2&&-2 \\
\hline
&&\dfrac{2x}{2}&=&\dfrac{10}{2} \\ \\
&&x&=&5
\end{array}
& \hspace{0.25in}
\begin{array}{rrrrl}
2x&+&2&=&-2\cdot 6 \\
2x&+&2&=&-12 \\
&-&2&&-2 \\
\hline
&&\dfrac{2x}{2}&=&\dfrac{-14}{2} \\ \\
&&x&=&-7
\end{array}
\end{array}\)
- \(\begin{array}{rrrrrrrrrrr}
\\ \\ \\ \\ \\ \\
2x&-&1&<&-6&\hspace{0.25in}&6&<&2x&-&1 \\
&+&1&&+1&&+1&&&+&1 \\
\hline
&&\dfrac{2x}{2}&<&\dfrac{-5}{2}&&\dfrac{7}{2}&<&\dfrac{2x}{2}&& \\ \\
&&x&<&-\dfrac{5}{2}&&x&>&\dfrac{7}{2}&&
\end{array}\)
INSERT IMAGE -
\(y=|2x|-1\) \(x\) \(y\) 3 5 2 3 1 1 0 −1 −1 1 −2 3 −3 5 
- \(\begin{array}{ll}
\\ \\ \\ \\ \\ \\ \\
\begin{array}{rrl}
A_1&=&A_2 \\
A_3&=&2A_1-10^{\circ} \\ \\
A_1+A_2+A_3&=&180^{\circ} \\
A_1+A_1+2A_1-10^{\circ}&=&180^{\circ} \\
+10^{\circ}&&+10^{\circ} \\
\hline
\dfrac{4A_1}{4}&=&\dfrac{190^{\circ}}{4}
\end{array}
& \hspace{0.25in}
\begin{array}{rrl}
A_1&=&47.5^{\circ} \\ \\
A_2&=&47.5^{\circ} \\ \\
A_3&=&2A_1-10^{\circ} \\
A_3&=&2(47.5^{\circ})-10^{\circ} \\
A_3&=&95^{\circ}-10^{\circ} \\
A_3&=&85^{\circ}
\end{array}
\end{array}\) - \(\phantom{1}\)
\(x, x+2 \\ \)
\(\begin{array}{rrrrrrrrr}
x&+&x&+&2&=&x&-&20 \\
&&2x&+&2&=&x&-&20 \\
&&-x&-&2&&-x&-&2 \\
\hline
&&&&x&=&-22&&
\end{array}\)
\(\phantom{1}\)
numbers are −22, −20 - \(\begin{array}{ll}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
\begin{array}{rrl}
y&=&\dfrac{kmn^2}{d} \\ \\
&&\textbf{1st data} \\
y&=&16 \\
k&=&\text{find 1st} \\
m&=&3 \\
n&=&4 \\
d&=&6 \\ \\
y&=&\dfrac{kmn^2}{d} \\ \\
16&=&\dfrac{k(3)(4)^2}{6} \\ \\
k&=&\dfrac{16\cdot 6}{3\cdot (4)^2} \\ \\
k&=&2
\end{array}
& \hspace{0.25in}
\begin{array}{rrl}
&&\textbf{2nd data} \\
y&=&\text{find 2nd} \\
k&=&2 \\
m&=&-2 \\
n&=&4 \\
d&=&8 \\ \\
y&=&\dfrac{kmn^2}{d} \\ \\
y&=&\dfrac{(2)(-2)(4)^2}{8} \\ \\
y&=&-8
\end{array}
\end{array}\)
