Midterm 1: Version D Answer Key

  1. [latex]3(4)-\sqrt{4^2-4(4)(1)}[/latex]
    [latex]\begin{array}[t]{l}\\ 12-\sqrt{16-16} \\ \\ 12 \end{array}[/latex]
  2. [latex]\phantom{a}[/latex]
    [latex]\begin{array}[t]{rrrrrrrrrrr} 2x&-&8&+&8&=&-6&+&3x&+&9 \\ &&&&2x&=&3x&+&3&& \\ &&&&-3x&&-3x&&&& \\ \hline &&&&-x&=&3&&&& \\ &&&&x&=&-3&&&& \end{array}[/latex]
  3. [latex]\left(\dfrac{1}{R}=\dfrac{1}{r_1}+\dfrac{1}{r_2}\right)(Rr_1r_2)[/latex]
    [latex]\begin{array}[t]{rrrrlrr} &&r_1r_2&=&\phantom{-}Rr_2&+&Rr_1 \\ &&-Rr_2&&-Rr_2&& \\ \hline r_1r_2&-&Rr_2&=&Rr_1&& \\ r_2(r_1&-&R)&=&Rr_1&& \\ \\ &&r_2&=&\dfrac{Rr_1}{r_1-R}&& \end{array}[/latex]
  4. [latex]\left(\dfrac{x}{15}-\dfrac{x-3}{3}=\dfrac{1}{3}\right)(15)[/latex]
    [latex]\begin{array}[t]{rrcrrrr} x&-&5(x&-&3)&=&5 \\ x&-&5x&+&15&=&5 \\ &&&-&15&&-15 \\ \hline &&&&\dfrac{-4x}{-4}&=&\dfrac{-10}{-4} \\ \\ &&&&x&=&\dfrac{5}{2} \end{array}[/latex]
  5. [latex]y=5[/latex]
    point at (-2,5) y=5
  6. [latex]\phantom{a}[/latex]
    [latex]\begin{array}[t]{ll} \begin{array}[t]{rrl} m&=&\dfrac{\Delta y}{\Delta x} \\ \\ \dfrac{2}{3}&=&\dfrac{y-4}{x--2} \end{array} & \hspace{0.25in} \begin{array}[t]{rrrrrrlrr} &&2(x&+&2)&=&\phantom{-}3(y&-&4) \\ &&2x&+&4&=&\phantom{-}3y&-&12 \\ &-&3y&+&12&&-3y&+&12 \\ \hline 2x&-&3y&+&16&=&0&& \\ \\ &&&&y&=&\dfrac{2}{3}x&+&\dfrac{16}{3} \end{array} \end{array}[/latex]
  7. [latex]\phantom{a}[/latex]
    [latex]\begin{array}[t]{ll} \begin{array}[t]{rrl} &&\textbf{1st slope:} \\ m&=&\dfrac{\Delta y}{\Delta x} \\ \\ m&=&\dfrac{-9--7}{8-12} \\ \\ m&=&\dfrac{-2}{-4} \\ \\ m&=&\dfrac{1}{2} \\ \\ &&\textbf{Now:} \\ m&=&\dfrac{\Delta y}{\Delta x} \\ \\ \dfrac{1}{2}&=&\dfrac{y--9}{x-8} \end{array} & \hspace{0.25in} \begin{array}[t]{rrrrrrrrr} &&x&-&8&=&2(y&+&9) \\ &&x&-&8&=&2y&+&18 \\ &-&2y&-&18&&-2y&-&18 \\ \hline x&-&2y&-&26&=&0&& \end{array} \end{array}[/latex]
  8. [latex]\phantom{a}[/latex]
    Line on graph passes through (0,-2) and (3,0)
  9. [latex]\phantom{a}[/latex]
    [latex]\begin{array}[t]{rrrrrrr} -27&\le &6x&-&9&\le &3 \\ +9&&&+&9&&+9 \\ \hline \dfrac{-18}{6}&\le &&\dfrac{6x}{6}&&\le &\dfrac{12}{6} \\ \\ -3&\le &&x&&\le &2 \end{array}[/latex]
    -3 , or equal to x < or equal to 2
  10. [latex]\phantom{a}[/latex]
    [latex]\begin{array}[t]{ll} \dfrac{2x+2}{6}=2& \hspace{0.75in} \dfrac{2x+2}{6}=-2 \\ \\ \begin{array}[t]{rrrrl} 2x&+&2&=&2\cdot 6 \\ 2x&+&2&=&12 \\ &-&2&&-2 \\ \hline &&\dfrac{2x}{2}&=&\dfrac{10}{2} \\ \\ &&x&=&5 \end{array} & \hspace{0.25in} \begin{array}[t]{rrrrl} 2x&+&2&=&-2\cdot 6 \\ 2x&+&2&=&-12 \\ &-&2&&-2 \\ \hline &&\dfrac{2x}{2}&=&\dfrac{-14}{2} \\ \\ &&x&=&-7 \end{array} \end{array}[/latex]
    x=5, x=-7
  11. [latex]\phantom{a}[/latex]
    [latex]\begin{array}[t]{rrrrrrrrrrr} 2x&-&1&<&-6&\hspace{0.25in}&6&<&2x&-&1 \\ &+&1&&+1&&+1&&&+&1 \\ \hline &&\dfrac{2x}{2}&<&\dfrac{-5}{2}&&\dfrac{7}{2}&<&\dfrac{2x}{2}&& \\ \\ &&x&<&-\dfrac{5}{2}&&x&>&\dfrac{7}{2}&& \end{array}[/latex]
  12. [latex]y=|2x|-1[/latex]
    [latex]x[/latex] [latex]y[/latex]
    3 5
    2 3
    1 1
    0 −1
    −1 1
    −2 3
    −3 5

    V line with point at (0,-1)

  13. [latex]\phantom{a}[/latex]
    [latex]\begin{array}[t]{ll} \begin{array}[t]{rrl} A_1&=&A_2 \\ A_3&=&2A_1-10^{\circ} \\ \\ A_1+A_2+A_3&=&180^{\circ} \\ A_1+A_1+2A_1-10^{\circ}&=&180^{\circ} \\ +10^{\circ}&&+10^{\circ} \\ \hline \dfrac{4A_1}{4}&=&\dfrac{190^{\circ}}{4} \end{array} & \hspace{0.25in} \begin{array}[t]{rrl} A_1&=&47.5^{\circ} \\ \\ A_2&=&47.5^{\circ} \\ \\ A_3&=&2A_1-10^{\circ} \\ A_3&=&2(47.5^{\circ})-10^{\circ} \\ A_3&=&95^{\circ}-10^{\circ} \\ A_3&=&85^{\circ} \end{array} \end{array}[/latex]
  14. [latex]x, x+2[/latex]
    [latex]\begin{array}[t]{rrrrrrrrr} x&+&x&+&2&=&x&-&20 \\ &&2x&+&2&=&x&-&20 \\ &&-x&-&2&&-x&-&2 \\ \hline &&&&x&=&-22&&\\ \end{array}[/latex]
    numbers are −22, −20
  15. [latex]\phantom{a}[/latex]
    [latex]\begin{array}[t]{ll} \begin{array}[t]{rrl} y&=&\dfrac{kmn^2}{d} \\ \\ &&\textbf{1st data} \\ y&=&16 \\ k&=&\text{find 1st} \\ m&=&3 \\ n&=&4 \\ d&=&6 \\ \\ y&=&\dfrac{kmn^2}{d} \\ \\ 16&=&\dfrac{k(3)(4)^2}{6} \\ \\ k&=&\dfrac{16\cdot 6}{3\cdot (4)^2} \\ \\ k&=&2 \end{array} & \hspace{0.25in} \begin{array}[t]{rrl}\\ \\ &&\textbf{2nd data} \\ y&=&\text{find 2nd} \\ k&=&2 \\ m&=&-2 \\ n&=&4 \\ d&=&8 \\ \\ y&=&\dfrac{kmn^2}{d} \\ \\ y&=&\dfrac{(2)(-2)(4)^2}{8} \\ \\ y&=&-8 \end{array} \end{array}[/latex]

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