Midterm 3: Version B Answer Key

  1. \dfrac{5\cancel{m^3}}{\cancel{4}n^{\cancel{2}}}\cdot \dfrac{\cancel{13}\cancel{n^3}}{\cancel{3}\cancel{m^3}}\cdot \dfrac{\cancel{12}m^4}{\cancel{26}2\cancel{n^2}}\Rightarrow \dfrac{5m^4}{2n}
  2. \dfrac{\cancel{3x}\cancel{(x+3)}}{\cancel{3}\cancel{(x+3)}}\cdot \dfrac{6x(x+3)}{(x+6)(x-3)}\Rightarrow \dfrac{6x^2(x+3)}{(x+6)(x-3)}
  3. \begin{array}{l} \\ \\ \\ \\ \\ \\ \\ \\ \left(\dfrac{5x}{x+3}-\dfrac{5x}{x-3}+\dfrac{90}{x^2-9}\right)(x-3)(x+3) \\ \\ \Rightarrow \dfrac{5x(x-3)-5x(x+3)+90}{(x+3)(x-3)} \\ \\ \Rightarrow \dfrac{5x^2-15x-5x^2-15x+90}{(x+3)(x-3)} \\ \\ \Rightarrow \dfrac{-30x+90}{(x+3)(x-3)}\Rightarrow \dfrac{-30\cancel{(x-3)}}{(x+3)\cancel{(x-3)}}\Rightarrow \dfrac{-30}{x+3} \end{array}
  4. \dfrac{\left(\dfrac{9a^2}{b^2}-25\right)(b^2)}{\left(\dfrac{3a}{b}+5}\right)(b^2)}\Rightarrow \dfrac{9a^2-25b^2}{3ab+5b^2}\Rightarrow \dfrac{(3a-5b)\cancel{(3a+5b)}}{b\cancel{(3a+5b)}}\Rightarrow \dfrac{3a-5b}{b}
  5. \begin{array}{l} \\ \\ \\ \sqrt{2\cdot 36\cdot d^2\cdot d}+4\sqrt{2\cdot 9\cdot d^2\cdot d}-2(7d^2) \\ 6d\sqrt{2d}+4\cdot 3d\sqrt{2d}-14d^2 \\ 6d\sqrt{2d}+12d\sqrt{2d}-14d^2 \\ 18d\sqrt{2d}-14d^2 \end{array}
  6. \dfrac{\sqrt{a^{\cancel{6}5}b^3}}{\sqrt{5\cancel{a}}}\cdot \dfrac{\sqrt{5}}{\sqrt{5}}\Rightarrow \dfrac{\sqrt{5a^5b^3}}{\sqrt{25}}\Rightarrow \dfrac{\sqrt{5\cdot a^4\cdot a\cdot b^2\cdot b}}{5}\Rightarrow \dfrac{a^2b\sqrt{5ab}}{5}
  7. \dfrac{\sqrt{5}}{3+\sqrt{5}}\cdot \dfrac{3-\sqrt{5}}{3-\sqrt{5}}\Rightarrow \dfrac{3\sqrt{5}-5}{9-5}\Rightarrow \dfrac{3\sqrt{5}-5}{4}
  8. \begin{array}{rrl} \\ \\ \\ \\ \\ (\sqrt{4x+12})^2&=&(x)^2 \\ 4x+12&=&x^2 \\ 0&=&x^2-4x-12 \\ 0&=&(x-6)(x+2) \\ \\ x&=&6, \cancel{-2} \end{array}
  9. \phantom{1}
    1. \begin{array}{rrl} \\ \\ \\ \dfrac{2x^2}{2}&=&\dfrac{98}{2} \\ \\ x^2&=&49 \\ x&=& \pm 7 \end{array}
    2. \begin{array}{rrl} \\ \\ 4x^2-12x&=&0 \\ 4x(x-3)&=&0 \\ x&=&0, 3 \end{array}
  10. \phantom{1}
    1. \begin{array}{rrl} \\ (x-5)(x+4)&=&0 \\ x&=&5, -4 \end{array}
    2. \begin{array}{rrl} \\ \\ x^2-2x-35&=&0 \\ (x-7)(x+5)&=&0 \\ x&=&7, -5 \end{array}
  11. \phantom{1}
    \left(\dfrac{x-3}{x+2}+\dfrac{6}{x+3}=1\right)(x+2)(x+3) \\ \\
    \begin{array}{rcccrcrrrrrcrr} &(x-3)&\cdot &(x+3)&+&6(x&+&2)&=&(x&+&2)(x&+&3) \\ &x^2&-&9&+&6x&+&12&=&x^2&+&5x&+&6 \\ -&x^2&+&9&-&6x&-&12&&-x^2&-&6x&-&12 \\ \midrule &&&&&&&0&=&-x&+&3&& \\ &&&&&&+&x&&+x&&&& \\ \midrule &&&&&&&x&=&3&&&& \end{array}
  12. \begin{array}{rrl} \\ \\ \\ \\ \\ \text{Let }u&=&x^2 \\ \\ u^2-5u+4&=&0 \\ (u-4)(u-1)&=&0 \\ \\ (x^2-4)(x^2-1)&=&0 \\ (x-2)(x+2)(x-1)(x+1)&=&0 \\ x&=&\pm 2, \pm 1 \end{array}
  13. \begin{array}{rrrrrrr} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ L&=&3&+&W&& \\ \\ P&=&2L&+&2W&& \\ 46&=&2(3&+&W)&+&2W \\ 46&=&6&+&2W&+&2W \\ -6&&-6&&&& \\ \midrule 40&=&4W&&&& \\ \\ W&=&\dfrac{40}{4}&=&10&& \\ \\ \therefore L&=&W&+&3&& \\ L&=&10&+&3&=&13 \end{array}
  14. \phantom{1}
    x, x+2, x+4 \\
    \begin{array}{rrcrrrrrcrr} &&x(x&+&2)&=&16&+&x&+&4 \\ x^2&+&2x&&&=&20&+&x&& \\ &-&x&-&20&&-20&-&x&& \\ \midrule x^2&+&x&-&20&=&0&&&& \\ \\ &&&&0&=&(x&+&5)(x&-&4) \\ &&&&x&=&\cancel{-5},&4&&& \\ \end{array}
    ∴ 4, 6, 8
  15. \begin{array}{rrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ d&=&r\cdot t \\ d_{\text{up}}&=&d_{\text{return}} \\ \\ 4(r-5)&=&2(r+5) \\ 4r-20&=&\phantom{-}2r+10 \\ -2r+20&&-2r+20 \\ \midrule \dfrac{2r}{2}&=&\dfrac{30}{2} \\ \\ r&=&15 \end{array}

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