Midterm 1: Version E Answer Key

Terrance Berg

Midterm 1: Version E Answer Key

1. −9
2. 9
3. −9
4. 10
5. −3
$\begin{array}{rrrrrrlrrrr} 2 x & - & 8 & + & 18 & = & - 12 & + & 4 x & + & 12 \\ - 2 x & - & 2 x \\ \frac{10}{2} & = & \frac{2 x}{2} \\ x & = & 5 \end{array}$
$(\frac{1}{R} - \frac{1}{r_{1}} = \frac{1}{r_{2}}) (R r_{1} r_{2})$
$\begin{array}{rrrrlrr} r_{1} r_{2} & - & R r_{2} & = & R r_{1} \\ - R r_{1} & + & R r_{2} & - R r_{1} & + & R r_{2} \\ r_{1} r_{2} & - & R r_{1} & = & R r_{2} \\ r_{1} (r_{2} & - & R) & = & R r_{2} \\ r_{1} & = & \frac{R r_{2}}{r_{2} - R} \end{array}$
$(\frac{x}{12} - \frac{x - 4}{3} = \frac{2}{3}) (12)$
$\begin{array}{rrcrrrl} x & - & 4 (x & - & 4) & = & 4 (2) \\ x & - & 4 x & + & 16 & = & 8 \\ - & 16 & - 16 \\ \frac{- 3 x}{- 3} & = & \frac{- 8}{- 3} \\ x & = & \frac{8}{3} \end{array}$
$y = - 6$
$\begin{array}{ll} \begin{array}{rrl} m & = & \frac{Δ y}{Δ x} \\ \frac{2}{5} & = & \frac{y - 1}{x - - 1} \end{array} & \begin{array}{rrrrrrlrr} 2 (x & + & 1) & = & 5 (y & - & 1) \\ 2 x & + & 2 & = & 5 y & - & 5 \\ - & 5 y & + & 5 & - 5 y & + & 5 \\ 2 x & - & 5 y & + & 7 & = & 0 \\ y & = & \frac{2}{5} x & + & \frac{7}{5} \end{array} \end{array}$
$\begin{array}{ll} \begin{array}{rrl} 1st slope: \\ m & = & \frac{Δ y}{Δ x} \\ m & = & \frac{5 - - 1}{2 - 0} \\ m & = & \frac{6}{2} \\ m & = & 3 \end{array} & \begin{array}{rrl} Now: \\ m & = & \frac{Δ y}{Δ x} \\ 3 & = & \frac{y - - 1}{x - 0} \\ 3 x & = & y + 1 \\ 3 x - y - 1 & = & 0 \\ ∴ 0 & = & 3 x - y - 1 \\ y & = & 3 x - 1 \end{array} \end{array}$
$\begin{array}{rrrrrrr} - 20 & \leq & 8 x & - & 4 & \leq & 28 \\ + 4 & + & 4 & + 4 \\ \frac{- 16}{8} & \leq & \frac{8 x}{8} & \leq & \frac{32}{8} \\ - 2 & \leq & x & \leq & 4 \end{array}$
$(- 2 \leq \frac{2 x + 2}{6} \leq 2) (6)$
$\begin{array}{rrrrrrr} - 12 & \leq & 2 x & + & 2 & \leq & 12 \\ - 2 & - & 2 & - 2 \\ \frac{- 14}{2} & \leq & \frac{2 x}{2} & \leq & \frac{10}{2} \\ - 7 & \leq & x & \leq & 5 \end{array}$
$\begin{array}{ll} (\frac{3 x - 4}{5} < - 1) (5) & or (1 < \frac{3 x - 4}{5}) (5) \\ \begin{array}{rrrrr} 3 x & - & 4 & < & - 5 \\ + & 4 & + 4 \\ 3 x & < & - 1 \\ x & < & - \frac{1}{3} \end{array} & \begin{array}{rrrrr} 5 & < & 3 x & - & 4 \\ + 4 & + & 4 \\ \frac{9}{3} & < & \frac{3 x}{3} \\ x & > & 3 \end{array} \end{array}$
-1 over 3, x < 3" width="300" height="71" class="alignnone wp-image-648 size-medium">
$3 x - 2 y < 12$

$x$ $y$

0 −6

4 0
$x, x + 2, x + 4$
$\begin{array}{rrcrrrcrrrl} x & + & 2 (x & + & 2) & + & 3 (x & + & 4) & = & 94 \\ x & + & 2 x & + & 4 & + & 3 x & + & 12 & = & 94 \\ 6 x & + & 16 & = & 94 \\ - & 16 & - 16 \\ \frac{6 x}{6} & = & \frac{78}{6} \\ x & = & 13 \end{array}$

numbers are 13, 15, 17
$\begin{array}{rrl} 3 x + x & = & 800 cm \\ 4 x & = & 800 cm \\ x & = & \frac{800 cm}{4} or 200 cm \\ 3 x & = & 600 cm \end{array}$
$\begin{array}{ll} \begin{array}{rrl} y & = & \frac{k m}{n^{2}} \\ \underset{―}{1st} \\ y & = & 12 \\ k & = & find \\ m & = & 3 \\ n & = & 4 \\ 12 & = & \frac{k (3)}{(4)^{2}} \\ k & = & \frac{12 \cdot (4)^{2}}{3} \\ k & = & 64 \end{array} & \begin{array}{rrl} \underset{―}{2nd} \\ y & = & find \\ k & = & 64 \\ m & = & 3 \\ n & = & - 3 \\ y & = & \frac{(64) (3)}{(- 3)^{2}} \\ y & = & \frac{64}{3} \end{array} \end{array}$