Midterm 1: Version B Answer Key
- \(\begin{array}{l}
\\ \\ \\ \\ \\ \\ \\ \\
-6-\sqrt{6^2-4(5)(1)} \\ \\
-6-\sqrt{36-20} \\ \\
-6-\sqrt{16} \\ \\
-6-4 \\ \\
-10
\end{array}\) - \(\begin{array}{rrrrrrrr}
\\ \\ \\ \\ \\ \\
&15x&-&18&=&4[-6&+&3x] \\
&15x&-&18&=&-24&+&12x \\
-&12x&+&18&&+18&-&12x \\
\hline
&&&\dfrac{3x}{3}&=&\dfrac{-6}{3} && \\ \\
&&&x&=&-2 &&
\end{array}\) - \(\begin{array}{l}
\\ \\ \\ \\ \\ \\
\left(A=\dfrac{h}{B\cdot b}\right)(b) \\ \\
\left(Ab=\dfrac{h}{B}\right)\div A \\ \\
\phantom{A}b=\dfrac{h}{BA}
\end{array}\) - \(\phantom{1}\)
\(\left(\dfrac{x+3}{5}-\dfrac{x}{2}=\dfrac{5-3x}{10}\right)(10) \\ \)
\(\begin{array}{rrcrcrrrr}
2(x&+&3)&-&5(x)&=&5&-&3x \\
2x&+&6&-&5x&=&5&-&3x \\
&-&3x&+&6&=&5&-&3x \\
&+&3x&-&6&&-6&+&3x \\
\hline
&&&&0&=&-1&&
\end{array}\)
\(\phantom{1}\)
No solution - \(y=4\)
- \(\begin{array}{ll}
\\ \\ \\
\begin{array}{rrl}
\text{slope}&=&\dfrac{\Delta y}{\Delta x} \\ \\
\dfrac{1}{3}&=&\dfrac{y-4}{x--1}
\end{array}
& \hspace{0.25in}
\begin{array}{rrrrrrlrr}
\\ \\ \\ \\ \\
&&1(x&+&1)&=&3(y&-&4) \\
&&x&+&1&=&3y&-&12 \\
&&&-&1&&&-&1 \\
\hline
&&&&x&=&3y&-&13 \\ \\
x&-&3y&+&13&=&0&& \\ \\
&&&&y&=&\dfrac{1}{3}x&+&\dfrac{13}{3}
\end{array}
\end{array}\) - \(\phantom{1}\)
\(\text{1st slope }\dots\text{ } m=\dfrac{\Delta y}{\Delta x}\Rightarrow \dfrac{5-4}{-3-0}\Rightarrow -\dfrac{1}{3} \\ \)
\(\text{now }\dots\text{ } m=\dfrac{\Delta y}{\Delta x}\Rightarrow -\dfrac{1}{3}\Rightarrow \dfrac{y-4}{x-0} \\ \)
\(\begin{array}{rrl}
-1(x)&=&3(y-4) \\
-x&=&3y-12 \\ \\
x+3y-12&=&0 \\
y&=&-\dfrac{1}{3}x+4 \\
\end{array}\) 
- \(\begin{array}{rrrrrrrrrrr}
\\ \\ \\ \\
6x&-&12&+&8x&>&15&-&20x&+&7 \\
&+&12&+&20x&&&+&20x&+&15 \\
&&&+&6x&&&&&+&12 \\
\hline
&&&&34x&>&34&&&& \\
&&&&x&>&1&&&&
\end{array}\)
- \(\begin{array}{rrrcrrr}
\\ \\ \\ \\ \\
-3&\le &2x&+&3&<&9 \\
-3&&&-&3&&-3 \\
\hline
\dfrac{-6}{2}&\le &&\dfrac{2x}{2}&&<&\dfrac{6}{2} \\ \\
-3&\le &&x&&<&3
\end{array}\)
- \(\phantom{1}\)
\(\left(-2<\dfrac{3x+2}{5}<2\right)(5) \\ \)
\(\begin{array}{rrrrrrr}
-10&<&3x&+&2&<&10 \\
-2&&&-&2&&-2 \\
\hline
\dfrac{-12}{3}&<&&\dfrac{3x}{3}&&<&\dfrac{8}{3} \\ \\
-4&<&&x&&<&\dfrac{8}{3}
\end{array}\)
-
\(5x+2y=10\) \(x\) \(y\) 2 0 0 5 −2 10 4 −5 
- \(\phantom{1}\)
\(x, x+2 \\ \)
\(\begin{array}{rrrrrrrrr}
x&+&(x&+&2)&=&5x&-&16 \\
&&2x&+&2&=&5x&-&16 \\
&+&-5x&-&2&&-5x&-&2 \\
\hline
&&&&\dfrac{-3x}{-3}&=&\dfrac{-18}{-3}&& \\ \\
&&&&x&=&6&& \\
\end{array}\)
\(\phantom{1}\)
numbers are 6, 8 - \(\begin{array}{rrl}
\\ \\ \\ \\ \\
4x+x&=&40\text{ cm} \\
5x&=&40\text{ cm} \\
x&=&\dfrac{40\text{ cm}}{5}\text{ or }8\text{ cm} \\ \\
\therefore 4x&=&4(8)\text{ or }32\text{ cm}
\end{array}\) - \(\begin{array}{ll}
\\ \\ \\ \\ \\ \\ \\ \\ \\
\begin{array}{rrl}
\\ \\ \\ \\ \\
&&\underline{\text{1st}} \\ \\
P&=&\dfrac{kT}{V} \\ \\
&&\underline{\text{2nd}} \\ \\
&&\textbf{1st data} \\
P&=&100 \\
k&=&\text{find 1st} \\
T&=&200 \\
V&=&500 \\ \\
P&=&\dfrac{kT}{V} \\ \\
100&=&\dfrac{k(200)}{500} \\ \\
k&=&\dfrac{\cancel{100}(500)}{\cancel{200}2} \\ \\
k&=&250
\end{array}
& \hspace{0.25in}
\begin{array}{rrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\
&&\textbf{2nd data} \\
P&=&\text{find 2nd} \\
k&=&250 \\
T&=&100 \\
V&=&500 \\ \\
P&=&\dfrac{kT}{V} \\ \\
P&=&\dfrac{(250)(100)}{500} \\ \\
P&=&50
\end{array}
\end{array}\)
