Midterm 3: Version A Answer Key

Terrance Berg

Midterm 3: Version A Answer Key

$\dfrac{15m^3}{4n^2}\cdot \dfrac{\cancel{17}1m^3}{\cancel{12}4n}\cdot \dfrac{\cancel{3 }1m^4}{\cancel{34 }2n^2}\Rightarrow \dfrac{15m^{10}}{32n^5}$
$\dfrac{8x-8y}{x^3+y^3}\cdot \dfrac{x^2-xy+y^2}{x^2-y^2}$
$\Rightarrow \dfrac{8\cancel{(x-y)}}{(x+y)\cancel{(x^2-xy+y^2)}}\cdot \dfrac{\cancel{x^2-xy+y^2}}{(x+y)\cancel{(x-y)}}\Rightarrow \dfrac{8}{(x+y)^2}$
$\text{LCD}=6(n-3)$
$\begin{array}[t]{l} \dfrac{5(n-3)-2\cdot 6(n-3)-5\cdot 6}{6(n-3)} \\ \\ \dfrac{5n-15-12n+36-30}{6(n-3)} \\ \\ \dfrac{-7n-9}{6(n-3)} \end{array}$
$\dfrac{\left(\dfrac{x^2}{y^2}-4\right)y^3}{\left(\dfrac{x+2y}{y^3}\right)y^3} \Rightarrow \dfrac{x^2y-4y^3}{x+2y}\Rightarrow \dfrac{y(x^2-4y^2)}{x+2y}\Rightarrow \dfrac{y(x-2y)\cancel{(x+2y)}}{\cancel{(x+2y)}}$
$\Rightarrow y(x-2y)$
$3\cdot 5+2\sqrt{36\cdot 2}-4$
$15+2\cdot 6\sqrt{2}-4$
$11+12\sqrt{2}$
$\dfrac{\sqrt{m^7n^{\cancel{3}2}}}{\sqrt{2\cancel{n}}}\cdot \dfrac{\sqrt{2}}{\sqrt{2}}\Rightarrow \dfrac{\sqrt{m^6\cdot m\cdot n^2\cdot 2}}{\sqrt{4}}\Rightarrow \dfrac{m^3n\sqrt{2m}}{2}$
$\dfrac{2-x}{1-\sqrt{3}}\cdot \dfrac{1+\sqrt{3}}{1+\sqrt{3}}\Rightarrow \dfrac{2+2\sqrt{3}-x-x\sqrt{3}}{1-3}$
$\Rightarrow \dfrac{2+2\sqrt{3}-x-x\sqrt{3}}{-2}\text{ or }\dfrac{x+x\sqrt{3}-2-2\sqrt{3}}{2}$
$\phantom{a}$
$\begin{array}[t]{rrl} (\sqrt{7x+8})^2&=&(x)^2 \\ 7x+8&=&x^2 \\ 0&=&x^2-7x-8 \\ 0&=&(x-8)(x+1) \\ \\ x&=&8, \cancel{-1} \end{array}$
a
1. $\begin{array}{rrl}\\ \dfrac{4x^2}{4}&=&\dfrac{64}{4} x^2&=&16 \\ x&=&\pm 4 \end{array}$
2. $\begin{array}{rrl}\\ \\ 3x^2-12x&=&0 \\ 3x(x-4)&=&0 \\ \\ x&=&0,4 \end{array}$
a
1. $\begin{array}{rrl}\\ (x-5)(x-1)&=&0 \\ x&=&5, 1 \end{array}$
2. $\begin{array}{rrl}\\ \\ x^2+10x+9&=&0 \\ (x+9)(x+1)&=&0 \\ x&=&-9, -1 \end{array}$
$\left(\dfrac{x+4}{-4}=\dfrac{8}{x}\right)(-4)(x)$
$\begin{array}{rrl} x(x+4)&=&-4(8) \\ x^2+4x&=&-32 \\ 0&=&x^2+4x+32 \hspace{0.5in} \text{Does not factor} \end{array}$
$\text{Let }u=x^2$
$\begin{array}[t]{rrl} u^2-13u+36&=&0 \\ u^2-4u-9u+36&=&0 \\ u(u-4)-9(u-4)&=&0 \\ (u-4)(u-9)&=&0 \\ \\ (x^2-4)(x^2-9)&=&0 \\ (x-2)(x+2)(x-3)(x+3)&=&0 \\ x&=& \pm 2, \pm 3 \end{array}$
$\phantom{a}$
$\begin{array}[t]{rrl} A&=&\dfrac{1}{2}bh \\ \\ 300&=&\dfrac{1}{2}(h+10)h \\ \\ 600&=&h^2+10h \\ 0&=&h^2+10h-600 \\ 0&=&(h-20)(h+30) \\ \\ h&=& 20, \cancel{-30} \\ \\ \therefore b&=&h+10=30 \end{array}$
$x, x+2, x+4$
$\begin{array}[t]{rrrrrrrrrrr} &&x(x&+&4)&=&38&+&x&+&2 \\ x^2&+&4x&&&=&x&+&40&& \\ &-&x&-&40&&-x&-&40&& \\ \hline x^2&+&3x&-&40&=&0&&&& \\ \\ &&&&0&=&(x&+&8)(x&-&5) \\ &&&&x&=&\cancel{-8},&5&&& \\ \end{array}$
$\therefore$ 5, 7, 9
$\phantom{a}$
$\begin{array}[t]{rrl} r_st_s&=&r_ft_f \\ \\ r(4.5\text{ h})&=&(r+150)(3.0\text{ h}) \\ 4.5r&=&\phantom{-}3.0r+450 \\ -3.0r&&-3.0r \\ \hline 1.5r&=&450 \\ \\ r&=&\dfrac{450}{1.5}\text{ or }300\text{ km/h} \\ \\ r_f&=&300+150 \\ r_f&=&450\text{ km/h} \end{array}$

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