Midterm 3: Version D Answer Key

  1. \dfrac{\cancel{15}1\cancel{m^3}}{4\cancel{n^2}}\cdot \dfrac{\cancel{13}1\cancel{n^3}}{\cancel{45}3\cancel{m^6}}\cdot \dfrac{\cancel{3}1m^{\cancel{4}}}{\cancel{39}\cancel{3}1n^{\cancel{2}}}\Rightarrow \dfrac{m}{12n}
  2. \dfrac{3x^2-9x}{3x+9}\cdot \dfrac{12x}{x^2+2x-15}\Rightarrow \dfrac{3x\cancel{(x-3)}}{\cancel{3}(x+3)}\cdot \dfrac{\cancel{12}4x}{(x+5)\cancel{(x-3)}}\Rightarrow \dfrac{12x^2}{(x+3)(x+5)}
  3. \phantom{1}
    \left(\dfrac{2}{x-4}-\dfrac{6}{x-3}=3\right)(x+4)(x-3) \\
    \begin{array}{rrrrcrrrcrcrr} 2(x&-&3)&-&6(x&+&4)&=&3(x&+&4)(x&-&3) \\ 2x&-&6&-&6x&-&24&=&3(x^2&+&x&-&12) \\ &&&&-4x&-&30&=&3x^2&+&3x&-&36 \\ &&&&+4x&+&30&&&+&4x&+&30 \\ \midrule &&&&&&0&=&3x^2&+&7x&-&6 \\ &&&&&&0&=&(x&+&3)(3x&-&2) \\ \\ &&&&&&x&=&-3,&\dfrac{2}{3}&&& \end{array}
  4. \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}
  5. \begin{array}{l} \\ 5y^2+2\cdot 7y+\sqrt{25y^2\cdot y} \\ 5y^2+14y+5y\sqrt{y} \end{array}
  6. \dfrac{15}{3-\sqrt{5}}\cdot \dfrac{3+\sqrt{5}}{3+\sqrt{5}}\Rightarrow \dfrac{45+15\sqrt{5}}{9-5}\Rightarrow \dfrac{45+15\sqrt{5}}{4}
  7. \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}
  8. \begin{array}{ll} \begin{array}{rrrrl} \\ \\ \sqrt{2x+9}&-&3&=&x \\ &+&3&&\phantom{x}+3 \\ \midrule (\sqrt{2x+9})^2&&&=&(x+3)^2 \end{array} & \hspace{0.25in} \begin{array}{rrrrrcrrrr} \\ \\ \\ \\ \\ &2x&+&9&=&x^2&+&6x&+&9 \\ -&2x&-&9&&&-&2x&-&9 \\ \midrule &&&0&=&x^2&+&4x&& \\ &&&0&=&x(x&+&4)&& \\ \\ &&&x&=&0,&-4&&& \end{array} \end{array}
  9. \phantom{1}
    1. \begin{array}{rrl} \\ \\ 8x^2-32x&=&0 \\ 8x(x-4)&=&0 \\ x&=&0, 4 \end{array}
    2. \begin{array}{rrl} \\ \\ \\ \dfrac{3x^2}{2}&=&\dfrac{48}{3} \\ \\ \sqrt{x^2}&=&\sqrt{16} \\ x&=&\pm 4 \end{array}
  10. \phantom{1}
    1. \begin{array}{rrl} \\ \\ x^2-5x+4&=&0 \\ (x-4)(x-1)&=&0 \\ x&=&1, 4 \end{array}
    2. \begin{array}{rrl} \\ (x-3)(x-1)&=&0 \\ x&=&1, 3 \end{array}
  11. \begin{array}{rrrrcrcrcrr} \\ \\ \\ \\ \\ \\ \\ \\ 2(x&+&4)&=&x(x)&&&&&& \\ 2x&+&8&=&x^2&&&&&& \\ \\ &&0&=&x^2&-&2x&-&8&& \\ &&0&=&x^2&-&4x&+&2x&-&8 \\ &&0&=&x(x&-&4)&+&2(x&-&4) \\ &&0&=&(x&-&4)(x&+&2)&& \\ \\ &&x&=&-2,&4&&&&& \end{array}
  12. \begin{array}{rrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ \text{Let }u&=&x^2 \\ \\ u^2-48u-49&=&0 \\ (u-49)(u+1)&=&0 \\ \\ (x^2-49)(x^2+1)&=&0 \\ (x-7)(x+7)(x^2+1)&=&0 \\ \\ x^2+1&=&\text{cannot be factored} \\ x&=&\pm 7 \end{array}
  13. \begin{array}{rrl} \\ \\ \\ \\ \\ \\ \\ 40&=&\dfrac{1}{2}(h-2)(h) \\ \\ 80&=&h^2-2h \\ 0&=&h^2-2h-80 \\ 0&=&(h-10)(h+8) \\ h&=&10, \cancel{-8} \\ \\ b&=&10-2=8 \end{array}
  14. \phantom{1}
    x, x+2, x+4 \\
    \begin{array}{crrrrrcrr} x(x&+&4)&=&41&+&4(x&+&2) \\ x^2&+&4x&=&41&+&4x&+&8 \\ &-&4x&&&-&4x&& \\ \midrule &&\sqrt{x^2}&=&\sqrt{49}&&&& \\ &&x&=&\pm 7&&&& \end{array}
    \phantom{1}
    \text{numbers are }7, 9, 11\text{ or }-7,-5,-3
  15. \begin{array}{rrrrrrcrrr} \\ \\ \\ \\ \\ \\ \\ \\ &&&d_{\text{d}}&=&d_{\text{u}}&+&4&& \\ \\ &2(6&+&r)&=&3(6&-&r)&+&4 \\ &12&+&2r&=&18&-&3r&+&4 \\ -&12&+&3r&&-12&+&3r&& \\ \midrule &&&\dfrac{5r}{5}&=&\dfrac{10}{5}&&&& \\ \\ &&&r&=&2&\text{km/h}&&& \end{array}

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Intermediate Algebra by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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