Final Exam: Version B Answer Key
[latexpage]
Questions from Chapters 1 to 3
- \(\begin{array}{l}
\\ \\ \\ \\ \\ \\
-2(-3)-\sqrt{(-3)^2-4(4)(-1)} \\ \\
6-\sqrt{9+16} \\ \\
6-5 \\ \\
1
\end{array}\) - \(\begin{array}{rrrrrcllll}
\\ \\ \\ \\ \\ \\ \\
&18x&-&30&=&3[4&-&4x&-&7] \\
&18x&-&30&=&3[-4x&-&3]&& \\
&18x&-&30&=&-12x&-&\phantom{1}9&& \\
+&12x&+&30&&+12x&+&30&& \\
\midrule
&&&30x&=&21&&&& \\ \\
&&&x&=&\dfrac{21}{30}&=&\dfrac{7}{10}&&
\end{array}\) - \(\phantom{1}\)
\(\left(\dfrac{x+4}{2}-\dfrac{1}{3}=\dfrac{x+2}{6}\right)(6) \\ \)
\(\begin{array}{rrcrcrrrr}
(x&+&4)(3)&-&1(2)&=&x&+&2 \\
3x&+&12&-&2&=&x&+&2 \\
-x&-&10&&&&-x&-&10 \\
\midrule
&&&&\dfrac{2x}{2}&=&\dfrac{-8}{2}&& \\ \\
&&&&x&=&-4&&
\end{array}\) - \(\begin{array}{rrrrrrrrr}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\
&&&&m&=&\dfrac{\Delta y}{\Delta x}&& \\ \\
&&&&\dfrac{2}{3}&=&\dfrac{y-4}{x-1}&& \\ \\
&&2(x&-&1)&=&3(y&-&4) \\
&&2x&-&2&=&3y&-&12 \\
&&-3y&+&12&&-3y&+&12 \\
\midrule
2x&-&3y&+&10&=&0&& \\ \\
&&&&y&=&\dfrac{2}{3}x&+&\dfrac{10}{3} \\
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\ \\ \\
d^2&=&\Delta x^2+\Delta y^2 \\
&=&(4--4)^2+(4--2)^2 \\
&=&8^2+6^2 \\
&=&64+36 \\
&=&100 \\ \\
d&=&10
\end{array}\) -
\(3x-2y=6\) \(x\) \(y\) 2 0 0 −3 −2 −6 
- \(\begin{array}{rrrcrrr}
\\ \\ \\ \\ \\
3&\le &6x&+&3&<&9 \\
-3&&&-&3&&-3 \\
\midrule
\dfrac{0}{6}&\le &&\dfrac{6x}{6}&&<&\dfrac{6}{6} \\ \\
0&\le &&x&&<&1
\end{array}\)
\(\phantom{1}\)
\((0,1)\)
- \(\begin{array}{ll}
\\ \\ \\ \\ \\ \\ \\ \\
\left(\dfrac{3x+1}{4}=2\right)^4 & \hspace{0.25in} \left(\dfrac{3x+1}{4}=-2\right)^4 \\
\begin{array}{rrrrr}
\\ \\
3x&+&1&=&8 \\
&-&1&&-1 \\
\midrule
&&3x&=&7 \\ \\
&&x&=&\dfrac{7}{3}
\end{array}
& \hspace{0.25in}
\begin{array}{rrrrr}
3x&+&1&=&-8 \\
&-&1&&-1 \\
\midrule
&&3x&=&-9 \\
&&x&=&-3
\end{array}
\end{array}\)
- \(\begin{array}{ll}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
\hspace{0.75in} w_{\text{m}}=kw_e \\
\begin{array}{rrl}
\\ \\ \\
&&\text{1st data} \\ \\
w_{\text{m}}&=&38\text{ lb} \\
k&=&\text{find 1st} \\
w_{\text{e}}&=&95\text{ lb} \\ \\
w_{\text{m}}&=&kw_{\text{e}} \\
38&=&k(95) \\ \\
k&=&\dfrac{38}{95} \\ \\
k&=&0.4
\end{array}
& \hspace{0.25in}
\begin{array}{rrl}
&&\text{2nd data} \\ \\
w_{\text{m}}&=&\text{find} \\
k&=&0.4 \\
w_{\text{e}}&=&240\text{ lb} \\ \\
w_{\text{m}}&=&kw_{\text{e}} \\
w_{\text{m}}&=&(0.4)(240) \\
w_{\text{m}}&=&96\text{ lb}
\end{array}
\end{array}\) - \(\phantom{1}\)
\(x, x+2 \\ \)
\(\begin{array}{rrrrrrrrrrr}
x&+&x&+&2&=&x&+&2&-&20 \\
&&2x&+&2&=&x&-&18&& \\
&&-x&-&2&&-x&-&2&& \\
\midrule
&&&&x&=&-20&&&&
\end{array}\)
\(\phantom{1}\)
\(\text{numbers are }-20, -18\)
Questions from Chapters 4 to 6
- \(\begin{array}{rrrrrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
&(4x&-&3y&=&13)(5) \\
&(6x&+&5y&=&-9)(3) \\ \\
&20x&-&15y&=&\phantom{-}65 \\
+&18x&+&15y&=&-27 \\
\midrule
&&&38x&=&38 \\
&&&x&=&1 \\ \\
&4(1)&-&3y&=&13 \\
&-4&&&&-4 \\
\midrule
&&&-3y&=&9 \\
&&&y&=&-3
\end{array}\)
\((1,-3)\) - \(\begin{array}{rrrrrrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
&&&&x&=&-1-y \\ \\
\therefore 3(-1&-&y)&-&4y&=&-5 \\
-3&-&3y&-&4y&=&-5 \\
+3&&&&&&+3 \\
\midrule
&&&&-7y&=&-2 \\
&&&&y&=&\dfrac{2}{7} \\ \\
&&x&+&y&=&-1 \\
&&x&+&\dfrac{2}{7}&=&-1 \\ \\
&&&-&\dfrac{2}{7}&&-\dfrac{2}{7} \\
\midrule
&&&&x&=&-\dfrac{9}{7}
\end{array}\) - \(\begin{array}{rrrrrrrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
&&&(x&-&4z&=&0)(-1) \\ \\
&x&+&2y&&&=&0 \\
+&-x&&&+&4z&=&0 \\
\midrule
&&&(2y&+&4z&=&0)(\div 2) \\
&&&y&+&2z&=&0 \\ \\
&&&y&+&2z&=&0 \\
&&+&y&-&2z&=&0 \\
\midrule
&&&&&2y&=&0 \\
&&&&&y&=&0 \\ \\
&&&\cancel{y}0&-&2z&=&0 \\
&&&&&z&=&0 \\ \\
&&&x&+&\cancel{2y}0&=&0 \\
&&&&&x&=&0
\end{array}\)
\((0,0,0)\) - \(\begin{array}{l}
\\ \\ \\
28-\{5\cancel{x^0}1-\cancel{\left[6x-3(5-2x)\right]^0}1\}+5\cancel{x^0}1 \\
28-\{5-1\}+5 \\
28-4+5 \\
29
\end{array}\) - \(\begin{array}{rrrrrrrr}
\\ \\ \\ \\ \\
&x^2&-&3x&+&8\phantom{x}&& \\
\times&&&x&-&4\phantom{x}&& \\
\midrule
&x^3&-&3x^2&+&8x&& \\
+&&-&4x^2&+&12x&-&32 \\
\midrule
&x^3&-&7x^2&+&20x&-&32 \\
\end{array}\) - \(\begin{array}{l}
\\ \\
(x^{3n-6-3n})^{-1} \\
(x^{-6})^{-1} \\
x^6
\end{array}\) - \(5y(5y^2-3y+1)\)
- \(\begin{array}{l}
\\
x^3+(2y)^3 \\
(x+2y)(x^2-2xy+4y^2)
\end{array}\) -
Solution Amount Strength Equation Soda \(x\) 0 0 Juice 2 35 2 (35) Diluted \(x+2\) 8 \((x+2)8\) \(\begin{array}{rrrrrl}
&2(35)&=&8(x&+&2) \\
&70&=&8x&+&16 \\
-&16&&&-&16 \\
\midrule
&54&=&8x&& \\ \\
&x&=&\dfrac{54}{8}&\text{ or }&6\dfrac{3}{4}\text{ litres} \\
\end{array}\) - \(\begin{array}{rrrrrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
&(d&+&q&=&14)(-10) \\
&10d&+&25q&=&185 \\ \\
&-10d&-&10q&=&-140 \\
+&10d&+&25q&=&\phantom{-}185 \\
\midrule
&&&\dfrac{15q}{15}&=&\dfrac{45}{15} \\ \\
&&&q&=&3 \\ \\
&\therefore d&+&3&=&14 \\
&&&d&=&11
\end{array}\)
Questions from Chapters 7 to 10
- \(\dfrac{\cancel{9}\cancel{s^2}}{7t^3}\cdot \dfrac{15\cancel{t}}{\cancel{13}\cancel{s^2}}\cdot \dfrac{\cancel{26}2s}{\cancel{9}\cancel{t}}\Rightarrow \dfrac{15\cdot 2\cdot 5}{7t^3}\Rightarrow \dfrac{30s}{7t^3}\)
- \(\begin{array}{l}
\\ \\ \\ \\ \\ \\
\dfrac{(a-1)2a}{(a-1)(a-6)(a+6)}-\dfrac{5(a+6)}{(a-6)(a-1)(a+6)} \Rightarrow \dfrac{2a^2-2a-5a-30}{(a-1)(a-6)(a+6)} \\ \\
\Rightarrow \dfrac{2a^2-7a-30}{(a-1)(a-6)(a+6)}\Rightarrow \dfrac{2a^2-12a+5a-30}{(a-1)(a-6)(a+6)} \\ \\
\Rightarrow \dfrac{2a(a-6)+5(a-6)}{(a-1)(a-6)(a+6)}\Rightarrow \dfrac{\cancel{(a-6)}(2a+5)}{(a-1)\cancel{(a-6)}(a+6)}\Rightarrow \dfrac{2a+5}{(a-1)(a+6)}
\end{array}\) - \(\dfrac{\left(1-\dfrac{8}{x}\right)x^2}{\left(\dfrac{3}{x}-\dfrac{24}{x^2}\right)x^2}\Rightarrow \dfrac{x^2-8x}{3x-24}\Rightarrow \dfrac{x\cancel{(x-8)}}{3\cancel{(x-8)}}\Rightarrow \dfrac{x}{3}\)
- \(\begin{array}{l}
\\ \\ \\ \\
\sqrt{x^4\cdot x\cdot y^6\cdot y}+2xy\sqrt{16\cdot x\cdot y^2\cdot y}-\sqrt{x\cdot y^2\cdot y} \\ \\
x^2y^3\sqrt{xy}+2xy\cdot 4y\sqrt{xy}-y\sqrt{xy} \\ \\
(x^2y^3+8xy^2-y)\sqrt{xy}
\end{array}\) - \(\dfrac{2+x}{1-\sqrt{7}}\cdot \dfrac{1+\sqrt{7}}{1+\sqrt{7}}\Rightarrow \dfrac{2+2\sqrt{7}+x+x\sqrt{7}}{1-7}\Rightarrow \dfrac{2+x+2\sqrt{7}+x\sqrt{7}}{-6}\)
- \(\left(\dfrac{a^6b^3}{\cancel{c^0}1d^{-9}}\right)^{\frac{2}{3}}\Rightarrow \dfrac{a^{6\cdot \frac{2}{3}}b^{3\cdot \frac{2}{3}}}{d^{-9\cdot \frac{2}{3}}}\Rightarrow \dfrac{a^4b^2}{d^{-6}}\Rightarrow a^4b^2d^6\)
- \(\begin{array}{rrl}
\\
(x-5)(x+3)&=&0 \\
x&=&5, -3
\end{array}\) - \(\phantom{1}\)
\(\left(\dfrac{2x-1}{3x}=\dfrac{x-3}{x}\right)(3x) \\ \)
\(\begin{array}{rrrrrrrr}
&2x&-&1&=&(x&-&3)(3) \\ \\
&2x&-&1&=&3x&-&9\phantom{)(3)} \\
+&-3x&+&1&&-3x&+&1\phantom{)(3)} \\
\midrule
&&&-x&=&-8&& \\
&&&x&=&8&&
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
A&=&L\cdot W \\
L&=&5+2W \\ \\
75&=&W(5+2W) \\
75&=&5W+2W^2 \\ \\
0&=&2W^2+5W-75 \\
0&=&2W^2-10W+15W-75 \\
0&=&2W(W-5)+15(W-5) \\
0&=&(W-5)(2W+15) \\
W&=&5, \cancel{-\dfrac{15}{2}} \\ \\
L&=&5+2(5) \\
L&=&15
\end{array}\) - \(\phantom{1}\)
\(x, x+2, x+4 \\ \)
\(\begin{array}{rrcrrrrrrrr}
&&x(x&+&2)&=&8(x&+&4)&-&25 \\
x^2&+&2x&&&=&8x&+&32&-&25 \\
&-&8x&-&32&&-8x&-&32&+&25 \\
&&&+&25&&&&&& \\
\midrule
x^2&-&6x&-&7&=&0&&&& \\
(x&-&7)(x&+&1)&=&0&&&& \\
&&&&x&=&7,&-1&&& \\
\end{array}\)
\(\phantom{1}\)
\(\text{numbers are }7,9,11\text{ or }-1,1,3\)
