Mid Term 1: Review Questions Answer Key

True
Undefined
15
16
12
19
True
−18
18
−16
16
−16
$\begin{array}{l} \\ \\ \\ -(-6) - \sqrt{(-6)^2-4(8)(-2)} \\ 6 - \sqrt{36+64} \\ 6-10 \\ -4 \end{array}$
$\begin{array}{rrrrrrrrrrrr} \\ \\ \\ \\ \\ &3x&-&12&-&27&=&7&-&5x&-&30 \\ +&5x&+&12&+&27&&&+&5x&+&12 \\ &&&&&&&&&&+&7 \\ &&&&&&&&&&+&27 \\ \midrule &&&&&8x&=&16&&&& \\ &&&&&x&=&2&&&& \end{array}$
$\begin{array}{l} \\ \\ \\ \\ \\ \\ \\ \left(\dfrac{1}{R} = \dfrac{1}{r_1}+\dfrac{1}{r_2}\right)(Rr_1r_2) \\ \\ r_1r_2 = Rr_2 + Rr_1 \\ \\ r_1r_2 = R(r_2 + r_1) \\ \\ R=\dfrac{r_1r_2}{r_2 + r_1} \end{array}$
$\phantom{1}$
$\left(\dfrac{x+3}{8} - \dfrac{3}{4} = \dfrac {x+6}{10}\right)(40) \\$
$\begin{array}{rrrrrcrrrr} &5(x&+&3)&-&3(10)&=&4(x&+&6) \\ &5x&+&15&-&30&=&4x&+&24 \\ -&4x&-&15&+&30&&-4x&-&15 \\ &&&&&&&&+&30 \\ \midrule &&&&&x&=&39&& \end{array}$
Need graph drawn. $y=5$
$\begin{array}{ll} \begin{array}{rrl} \\ \\ m&=&\dfrac {\Delta y}{\Delta x}\\ \\ \dfrac{2}{3}& =& \dfrac{y-2}{x- -1} \end{array} & \hspace{0.25in} \begin{array}{rrrrrrrcl} \\ \\ \\ \\ \\ &&2(x&+&1)& =& 3(y&-&2) \\ &&2x &+& 2& = &3y& -& 6 \\ &&-3y& + &6 &&-3y &+ &6 \\ \midrule 2x&-&3y&+&8&=&0&& \\ \\ &&&&y&=&\dfrac{2}{3}x&+&\dfrac{8}{3} \end{array} \end{array}$
$\begin{array}{ll} \begin{array}{rrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ && \text{1st slope} \\ \\ m&=&\dfrac{\Delta y}{\Delta x} \\ \\ m&=&\dfrac{11--1}{2--2} \\ \\ m&=&\dfrac{12}{4} \\ \\ m&=& 3 \\ \\ && \text{2nd slope} \\ \\ m&=&\dfrac{\Delta y}{\Delta x} \\ \\ 3&=&\dfrac{y--1}{x--2} \end{array} & \hspace{0.25in} \begin{array}{rrrrrrrrr} \\ \\ \\ \\ \\ \\ \\ \\ &&3(x&+&2)&=&y&+&1 \\ &&3x&+&6&=&y&+&1 \\ &&-y&-&1&&-y&-&1 \\ \midrule 3x&-&y&+&5&=&0&& \\ &&&&y&=&3x&+&5 \end{array} \end{array}$
Use slop intercept method.
$\begin{array}{rrl} \\ \\ \\ d^2& =& \Delta x^2 + \Delta y^2\\ d^2 &=& (15-7)^2 + (3 - -3)^2\\ d^2& =& 8^2 + 6^2 \\ d&=&10 \end{array}$
$\phantom{1}$
$x=\dfrac{x_1+x_2}{2}=\dfrac{7+15}{2}=11 \\$
$y=\dfrac{y_1+y_2}{2}=\dfrac{-3+3}{2}=0 \\$
$(11,0)$
True
True
$\{r, s, t\}$
True
$\begin{array}{rrrrrrr} \\ \\ \\ 3x&-&6&-&36x&>&60 \\ &+&6&&&&+6 \\ \midrule &&&&-33x&>&66 \\ &&&&x&<&-2 \end{array}$

(−∞, −2)
$\begin{array}{rrrcrcr} \\ \\ \\ \\ \\ -18& \le& 4x& - &6& \le& 2\\ +6&&&+&6&&+6\\ \midrule \dfrac{-12}{4}&\le&&\dfrac{4x}{4}&&\le&\dfrac{8}{4}\\ \\ -3&\le&&x&&\le&2 \end{array}$

[−3, 2]
$\begin{array}{ll} \\ \\ \\ \\ \\ \begin{array}{rrrrr} 2x&-&1&<&-7 \\ &+&1&&+1 \\ \midrule &&\dfrac{2x}{2}&<&\dfrac{-6}{2} \\ \\ &&x&<&-3 \end{array} & \hspace{0.25in} \begin{array}{rrrrr} 7&<&2x&-&1 \\ +1&&&+&1 \\ \midrule \dfrac{8}{2}&<&\dfrac{2x}{2}&& \\ \\ 4&<&x&& \end{array} \end{array}$

(−∞, −3) or (4, ∞) WRONG IMAGE
$\phantom{1}$
$\left| \dfrac{3x - 4}{5} \right| < 1 \\$
$\left(-1 < \dfrac{3x-4}{5}< 1 \right)(5) \\$
$\begin{array}{rrrrrrr} -5&<&3x&-&4&<&5 \\ +4&&&+&4&&+4 \\ \midrule \dfrac{-1}{3}&<&&\dfrac{3x}{3}&&<&\dfrac{9}{3} \\ \\ -\dfrac{1}{3}&<&&x&&<&3 \end{array}$

$(-\dfrac{1}{3}, 3)$
$\begin{array}{rrrrrrrr} \\ \\ \\ \\ \\ &3x&-&2y&=&10&& \\ +&-10&+&2y&&-10&+&2y \\ \midrule &\dfrac{3x}{2}&-&\dfrac{10}{2}&=&\dfrac{2y}{2}&& \\ \\ &&&y&=&\dfrac{3}{2}x&-&5 \end{array}$

Slope intercept method. Check (0, 0): 3(0) − 2(0) < 10. Shade the (0, 0) side.
$y=|x-1|-2$

$x$ $y$

4 1

3 0

2 −1

1 −2

0 −1

−1 0

−2 1
$\begin{array}{rrrrrr} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ &6L&+&2S&=&38 \\ +&4L&-&2S&=&12 \\ \midrule &&&10L&=&50 \\ &&&L&=&5 \\ \\ \therefore &6(5)&+&2S&=&38 \\ &30&+&2S&=&38 \\ -&30&&&&-30 \\ \midrule &&&2S&=&8 \\ &&&S&=&4 \end{array}$
Insert diagram.
$\begin{array}{rrl} 5x+x&=&36 \\ 6x&=&36 \\ x&=&6 \\ \therefore 5x&=&30 \end{array}$
$\begin{array}{rrrrrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ &(d&+&q&=&14)(-10) \\ &10d&+&25q&=&260 \\ \\ &-10d&-&10q&=&-140 \\ +&10d&+&25q&=&\phantom{-}260 \\ \midrule &&&\dfrac{15q}{15}&=&\dfrac{120}{15} \\ \\ &&&q&=&8 \\ \\ \therefore &d&+&8&=&14 \\ &&-&8&&-8 \\ \midrule &&&d&=&6 \end{array}$
$\phantom{1}$
$x, x+2 \\$
$\begin{array}{rrrrrrrrr} x&+&x&+&2&=&x&-&10 \\ &&2x&+&2&=&x&-&10 \\ &&-x&-&2&&-x&-&2 \\ \midrule &&&&x&=&-12&& \\ \end{array}$
$\phantom{1}$
numbers are −12, −10
$\begin{array}{ll} \\ \\ \\ \\ \\ \\ \\ \\ \begin{array}{rrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ \text{\underline{1st}}&& \\ y&=&\dfrac{kmn^2}{d} \\ \\ \text{\underline{2nd}}&& \\ y&=&12 \\ k&=&\text{find} \\ m&=&3 \\ n&=&4 \\ d&=&8 \\ \\ 12&=&\dfrac{k(3)(4)^2}{8} \\ \\ k&=&\dfrac{12(8)}{3\cdot 16} \\ \\ k&=& 2 \end{array} & \hspace{0.25in} \begin{array}{rrl} \\ \\ \\ \text{\underline{3rd}}&& \\ y&=&\text{find} \\ k&=& 2 \\ m&=&-3 \\ n&=&3 \\ d&=& 6 \\ \\ y&=&\dfrac{(2)(-3)(3)^2}{6} \\ \\ y&=&-9 \end{array} \end{array}$

License

Icon for the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License

Intermediate Algebra Copyright © 2020 by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

Share This Book