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