Midterm 3: Version A Answer Key

Terrance Berg

Midterm 3: Version A Answer Key

$\frac{15 m^{3}}{4 n^{2}} \cdot \frac{171 m^{3}}{124 n} \cdot \frac{31 m^{4}}{342 n^{2}} \Rightarrow \frac{15 m^{10}}{32 n^{5}}$
$\begin{array}{l} \frac{8 x - 8 y}{x^{3} + y^{3}} \cdot \frac{x^{2} - x y + y^{2}}{x^{2} - y^{2}} \\ \Rightarrow \frac{8 (x - y)}{(x + y) (x^{2} - x y + y^{2})} \cdot \frac{x^{2} - x y + y^{2}}{(x + y) (x - y)} \Rightarrow \frac{8}{(x + y)^{2}} \end{array}$
$\begin{array}{l} LCD = 6 (n - 3) \\ \frac{5 (n - 3) - 2 \cdot 6 (n - 3) - 5 \cdot 6}{6 (n - 3)} \\ \frac{5 n - 15 - 12 n + 36 - 30}{6 (n - 3)} \\ \frac{- 7 n - 9}{6 (n - 3)} \end{array}$
$\frac{(\frac{x^{2}}{y^{2}} - 4) y^{3}}{(\frac{x + 2 y}{y^{3}}) y^{3}} \Rightarrow \frac{x^{2} y - 4 y^{3}}{x + 2 y} \Rightarrow \frac{y (x^{2} - 4 y^{2})}{x + 2 y} \Rightarrow \frac{y (x - 2 y) (x + 2 y)}{(x + 2 y)}$
$\Rightarrow y (x - 2 y)$
$3 \cdot 5 + 2 \sqrt{36 \cdot 2} - 4$
$15 + 2 \cdot 6 \sqrt{2} - 4$
$11 + 12 \sqrt{2}$
$\frac{\sqrt{m^{7} n^{32}}}{\sqrt{2 n}} \cdot \frac{\sqrt{2}}{\sqrt{2}} \Rightarrow \frac{\sqrt{m^{6} \cdot m \cdot n^{2} \cdot 2}}{\sqrt{4}} \Rightarrow \frac{m^{3} n \sqrt{2 m}}{2}$
$\begin{array}{l} \frac{2 - x}{1 - \sqrt{3}} \cdot \frac{1 + \sqrt{3}}{1 + \sqrt{3}} \Rightarrow \frac{2 + 2 \sqrt{3} - x - x \sqrt{3}}{1 - 3} \\ [l a t e x] \Rightarrow \frac{2 + 2 \sqrt{3} - x - x \sqrt{3}}{- 2} or \frac{x + x \sqrt{3} - 2 - 2 \sqrt{3}}{2} \end{array}$
$\begin{array}{rrl} (\sqrt{7 x + 8})^{2} & = & (x)^{2} \\ 7 x + 8 & = & x^{2} \\ 0 & = & x^{2} - 7 x - 8 \\ 0 & = & (x - 8) (x + 1) \\ x & = & 8, - 1 \end{array}$
1. $\begin{array}{rrl} \frac{4 x^{2}}{4} & = & \frac{64}{4} \\ x^{2} & = & 16 \\ x & = & \pm 4 \end{array}$
2. $\begin{array}{rrl} 3 x^{2} - 12 x & = & 0 \\ 3 x (x - 4) & = & 0 \\ x & = & 0, 4 \end{array}$
1. $\begin{array}{rrl} (x - 5) (x - 1) & = & 0 \\ x & = & 5, 1 \end{array}$
2. $\begin{array}{rrl} x^{2} + 10 x + 9 & = & 0 \\ (x + 9) (x + 1) & = & 0 \\ x & = & - 9, - 1 \end{array}$
$(\frac{x + 4}{- 4} = \frac{8}{x}) (- 4) (x)$
$\begin{array}{rrl} x (x + 4) & = & - 4 (8) \\ x^{2} + 4 x & = & - 32 \\ 0 & = & x^{2} + 4 x + 32 Does not factor \end{array}$
$\begin{array}{rrl} Let u & = & x^{2} \\ u^{2} - 13 u + 36 & = & 0 \\ u^{2} - 4 u - 9 u + 36 & = & 0 \\ u (u - 4) - 9 (u - 4) & = & 0 \\ (u - 4) (u - 9) & = & 0 \\ (x^{2} - 4) (x^{2} - 9) & = & 0 \\ (x - 2) (x + 2) (x - 3) (x + 3) & = & 0 \\ x & = & \pm 2, \pm 3 \end{array}$
$\begin{array}{rrl} A & = & \frac{1}{2} b h \\ 300 & = & \frac{1}{2} (h + 10) h \\ 600 & = & h^{2} + 10 h \\ 0 & = & h^{2} + 10 h - 600 \\ 0 & = & (h - 20) (h + 30) \\ h & = & 20, - 30 \\ ∴ b & = & h + 10 = 30 \end{array}$
$x, x + 2, x + 4$
$\begin{array}{rrrrrrrrrrr} x (x & + & 4) & = & 38 & + & x & + & 2 \\ x^{2} & + & 4 x & = & x & + & 40 \\ - & x & - & 40 & - x & - & 40 \\ x^{2} & + & 3 x & - & 40 & = & 0 \\ 0 & = & (x & + & 8) (x & - & 5) \\ x & = & - 8, & 5 \end{array}$
∴ 5, 7, 9
$\begin{array}{rrl} r_{s} t_{s} & = & r_{f} t_{f} \\ r (4.5 h) & = & (r + 150) (3.0 h) \\ 4.5 r & = & 3.0 r + 450 \\ - 3.0 r & - 3.0 r \\ 1.5 r & = & 450 \\ r & = & \frac{450}{1.5} or 300 km/h \\ r_{f} & = & 300 + 150 \\ r_{f} & = & 450 km/h \end{array}$

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

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