Final Exam: Version A Answer Key
[latexpage]
Questions from Chapters 1 to 3
- \(\begin{array}{l}
\\ \\ \\ \\ \\ \\
-(6)-\sqrt{6^2-4(4)(2)} \\ \\
-6-\sqrt{36-32} \\ \\
-6-\sqrt{4} \\ \\
-6-2=-8
\end{array}\) - \(\begin{array}{rrrrrrrrrrr}
\\ \\ \\ \\
6x&+&24&=&35&-&5x&-&8&+&12x \\
6x&+&24&=&27&+&7x&&&& \\
-7x&-&24&&-24&-&7x&&&& \\
\midrule
&&-x&=&3&&&&&& \\
&&\therefore x&=&-3&&&&&& \\
\end{array}\) - \(\phantom{1}\)
\(\left(\dfrac{x+4}{2}-\dfrac{1}{2}=\dfrac{x+2}{4}\right)(4) \\ \)
\(\begin{array}{crrrcrrrl}
2(x&+&4)&-&1(2)&=&x&+&2 \\
2x&+&8&-&2&=&x&+&2 \\
-x&-&8&+&2&&-x&-&8+2 \\
\midrule
&&&&x&=&-4&&
\end{array}\) - \(x=-2\)
- \(\begin{array}{rrl}
\\ \\ \\ \\ \\ \\
d^2&=&\Delta x^2+\Delta y^2 \\
&=&(2--4)^2+(6--2)^2 \\
&=&6^2+8^2 \\
&=&36+64 \\
&=&100 \\ \\
\therefore d&=&\sqrt{100}=10
\end{array}\) -
\(2x-3y=6\) \(x\) \(y\) 0 −2 3 0 6 2 
- \(\begin{array}{rrrrrrrrr}
\\ \\ \\ \\ \\ \\
x&-&2x&+&10&\le &18&+&3x \\
&&-x&+&10&\le &18&+&3x \\
+&&-3x&-&10&&-10&-&3x \\
\midrule
&&&&\dfrac{-4x}{-4}&\le &\dfrac{8}{-4}&& \\ \\
&&&&x&\ge &-2&& \\
\end{array}\)
\(\left[-2, \infty)\)
or equal to -2" width="300" height="69" class="alignnone wp-image-1361 size-medium"> - \(\phantom{1}\)
\(\left(-1 < \dfrac{3x-2}{7}<1 \right)(7) \\ \)
\(\begin{array}{rrrcrrr}
\\ \\ \\ \\
-7&<&3x&-&2&<&7 \\
+2&&&+&2&&+2 \\
\midrule
\dfrac{-5}{3}&<&&\dfrac{3x}{3}&&<&\dfrac{9}{3} \\ \\
-\dfrac{5}{3}&<&&x&&<&3
\end{array}\)
\(\phantom{1}\)
\(\left(-\dfrac{5}{3}, 3\right)\)
- \(\begin{array}{ll}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
t=\dfrac{k}{r} \\
\begin{array}{rrl}
\\
&&\text{1st data} \\ \\
t&=&45\text{ min} \\
k&=&\text{find 1st} \\
r&=&600\text{ kL/min} \\ \\
t&=&\dfrac{k}{r} \\ \\
45&=&\dfrac{k}{600} \\ \\
k&=&45(600) \\
k&=&27000\text{ kL}
\end{array}
& \hspace{0.25in}
\begin{array}{rrl}
&&\text{2nd data} \\ \\
t&=&\text{find} \\
k&=&27000 \\
r&=&1000\text{ kL/min} \\ \\
t&=&\dfrac{k}{r} \\ \\
t&=&\dfrac{27000}{1000} \\ \\
t&=&27\text{ min}
\end{array}
\end{array}\) - \(\phantom{1}\)
\(x, x+2 \\ \)
\(\begin{array}{rrrrrrrrr}
x&+&x&+&2&=&4(x)&-&12 \\
&&2x&+&2&=&4x&-&12 \\
&-&2x&+&12&&-2x&+&12 \\
\midrule
&&&&\dfrac{14}{2}&=&\dfrac{2x}{2}&& \\ \\
&&&&x&=&7&&
\end{array}\)
\(\phantom{1}\)
\(\text{numbers are }7,9\)
Questions from Chapters 4 to 6
- \(\begin{array}{rrrrrr}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\
&2x&+&5y&=&-18 \\
+&-2x&+&y&=&6 \\
\midrule
&&&\dfrac{6y}{6}&=&\dfrac{-12}{6} \\ \\
&&&y&=&-2 \\ \\
&\therefore y&-&6&=&2x \\
&-2&-&6&=&2x \\
&&&2x&=&-8 \\
&&&x&=&-4
\end{array}\)
Answer: \((-4, -2)\) - \(\begin{array}{ll}
\begin{array}{rrrrrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\
&(8x&+&7y&=&51)(-2) \\
&(5x&+&2y&=&20)(7) \\ \\
&-16x&-&14y&=&-102 \\
+&35x&+&14y&=&\phantom{-}140 \\
\midrule
&&&\dfrac{19x}{19}&=&\dfrac{38}{19} \\ \\
&&&x&=&2 \\ \\
\end{array}
& \hspace{0.25in}
\begin{array}{rrrrr}
\\ \\ \\ \\ \\
\therefore 5x&+&2y&=&20 \\
5(2)&+&2y&=&20 \\
10&+&2y&=&20 \\
-10&&&&-10 \\
\midrule
&&2y&=&10 \\
&&y&=&5
\end{array}
\end{array}\)
Answer: \((2, 5)\) - \(\begin{array}{ll}
\\ \\ \\ \\ \\ \\
\begin{array}{rrrrrrrl}
\\ \\ \\ \\
&-2x&-&2y&-&12z&=&-10 \\
+&2x&&&-&3z&=&\phantom{-}4 \\
\midrule
&&&(-2y&-&15z&=&-6)(3) \\
&&&(3y&+&4z&=&\phantom{-}9)(2) \\ \\
&&&-6y&-&45z&=&-18 \\
&&+&6y&+&8z&=&\phantom{-}18 \\
\midrule
&&&&&-37z&=&0 \\
&&&&&z&=&0 \\ \\
\end{array}
&\hspace{0.25in}
\begin{array}{rrrrl}
2x&-&\cancel{3z}0&=&4 \\
&&x&=&\dfrac{4}{2}\text{ or }2 \\ \\
3y&+&\cancel{4z}0&=&9 \\
&&y&=&\dfrac{9}{3}\text{ or }3
\end{array}
\end{array}\)
Answer \((2, 3, 0)\) - \(\begin{array}{l}
\\ \\
24+\{-3x-\cancel{\left[6x-3(5-2x)\right]^0}1\}+3x \\
24-3x-1+3x \\
23
\end{array}\) - \(2ab^3(a^2-16)\Rightarrow 2a^3b^3-32ab^3\)
- \(\begin{array}{l}
\\ \\ \\ \\ \\ \\
(x^{1--2}y^{-3-4})^{-1} \\ \\
(x^3y^{-7})^{-1} \\ \\
x^{-3}y^7 \\ \\
\dfrac{y^7}{x^3}
\end{array}\) - \(\begin{array}{l}
\\ \\
3x^2+3x+8x+8 \\
3x(x+1)+8(x+1) \\
(x+1)(3x+8)
\end{array}\) - \((4x)^3-y^3\Rightarrow (4x-y)(16x^2+4xy+y^2)\)
- \(\begin{array}{rrrcrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
&(A&+&B&=&\phantom{191.}50)(-370) \\
&(3.95A&+&3.70B&=&191.25)(100) \\ \\
&-370A&-&370B&=&-18500 \\
+&395A&+&370B&=&\phantom{-}19125 \\
\midrule
&&&25A&=&625 \\ \\
&&&A&=&\dfrac{625}{25}\text{ or }25 \\ \\
&A&+&B&=&50 \\
&25&+&B&=&50 \\
&&&B&=&25 \\
\end{array}\) - \(\begin{array}{rrrrrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\
&(d&+&q&=&16)(-10) \\
&10d&+&25q&=&235 \\ \\
&-10d&-&10q&=&-160 \\
+&10d&+&25q&=&\phantom{-}235 \\
\midrule
&&&\dfrac{15q}{15}&=&\dfrac{75}{15} \\ \\
&&&q&=&5 \\
&&&\therefore d&=&16-5=11 \\
\end{array}\)
Questions from Chapters 7 to 10
- \(\dfrac{\cancel{15}3s^{\cancel{3}2}}{\cancel{3t^2}1}\cdot \dfrac{\cancel{17}1\cancel{s^3}}{\cancel{5}1\cancel{t}}\cdot \dfrac{\cancel{3t^3}}{\cancel{34}2\cancel{s^4}}\Rightarrow \dfrac{3s^2}{2}\)
- \(\begin{array}{l}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
\text{LCD}=(x+2)(x-2) \\ \\
\dfrac{2x(x-2)-4x(x+2)+20}{(x+2)(x-2)} \\ \\
\dfrac{2x^2-4x-4x^2-8x+20}{(x+2)(x-2)} \\ \\
\dfrac{-2x^2-12x+20}{(x+2)(x-2)} \\ \\
\dfrac{-2(x^2+6x-10)}{(x+2)(x-2)}
\end{array}\) - \(\begin{array}{l}
\dfrac{\left(\dfrac{x^2}{y^2}-9\right)y^3}{\left(\dfrac{x+3y}{y^3}\right)y^3}\Rightarrow \dfrac{x^2y-9y^3}{x+3y}\Rightarrow \dfrac{y(x^2-9y^2)}{x+3y}\Rightarrow \dfrac{y(x-3y)\cancel{(x+3y)}}{\cancel{(x+3y)}} \\ \\
\Rightarrow y(x-3y)
\end{array}\) - \(\begin{array}{l}
\\ \\ \\ \\
3\cdot 5\sqrt{x}-2\sqrt{36\cdot 2x}-\sqrt{16\cdot x^2\cdot x} \\ \\
15\sqrt{x}-2\cdot 6\sqrt{2x}-4x\sqrt{x} \\ \\
15\sqrt{x}-12\sqrt{2x}-4x\sqrt{x}
\end{array}\) - \(\dfrac{\sqrt{m^6\cancel{n}}}{\sqrt{3\cancel{n}}}\Rightarrow \dfrac{m^3}{\sqrt{3}}\cdot \dfrac{\sqrt{3}}{\sqrt{3}}\Rightarrow \dfrac{m^3\sqrt{3}}{3}\)
- \(\left(\dfrac{\cancel{a^0}1b^4}{c^8d^{-12}}\right)^{\frac{1}{4}}\Rightarrow \dfrac{b^{4\cdot \frac{1}{4}}}{c^{8\cdot \frac{1}{4}}d^{-12\cdot \frac{1}{4}}}\Rightarrow \dfrac{b}{c^2d^{-3}}\Rightarrow \dfrac{bd^3}{c^2}\)
- \(\begin{array}{l}
\\
(x-5)(x+1)=0 \\
x=5,-1
\end{array}\) - \(\begin{array}{rrrrrcrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\
&&&(x&-&3)^2&=&(x)^2 \\ \\
&x^2&-&6x&+&9&=&\phantom{-}x^2 \\
-&x^2&&&&&&-x^2 \\
\midrule
&&&-6x&+&9&=&0 \\ \\
&&&&&\dfrac{-6x}{-6}&=&\dfrac{-9}{-6} \\ \\
&&&&&x&=&\dfrac{3}{2}
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
A&=&\dfrac{1}{2}bh \\ \\
20&=&\dfrac{1}{2}(h+6)h \\ \\
40&=&h^2+6h \\ \\
0&=&h^2+6h-40 \\
0&=&h^2+10h-4h-40 \\
0&=&h(h+10)-4(h+10) \\
0&=&(h-4)(h+10) \\ \\
h&=&4, \cancel{-10} \\
b&=&4+6=10
\end{array}\) - \(\phantom{1}\)
\(x, x+2, x+4 \\ \)
\(\begin{array}{rrrrcrrrlrrrr}
&&&&x(x&+&2)&=&\phantom{-}8&+&6(x&+&4) \\
x^2&+&2x&&&&&=&\phantom{-}8&+&6x&+&24 \\
&-&6x&-&8&-&24&&-8&-&6x&-&24 \\
\midrule
&&x^2&-&4x&-&32&=&0&&&& \\ \\
x^2&+&4x&-&8x&-&32&=&0&&&& \\
x(x&+&4)&-&8(x&+&4)&=&0&&&& \\
&&(x&+&4)(x&-&8)&=&0&&&& \\
&&&&&&x&=&-4,8&&&&
\end{array}\)
\(\phantom{1}\)
\(\therefore \text{ numbers are }-4,-2,0 \text{ or } 8,10,12\)
