{"id":3332,"date":"2019-11-19T15:06:05","date_gmt":"2019-11-19T20:06:05","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/?post_type=back-matter&p=3332"},"modified":"2019-11-19T15:11:55","modified_gmt":"2019-11-19T20:11:55","slug":"midterm-3-version-c","status":"publish","type":"back-matter","link":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/back-matter\/midterm-3-version-c\/","title":{"raw":"Midterm 3: Version C Answer Key","rendered":"Midterm 3: Version C Answer Key"},"content":{"raw":"[latexpage]\r\n
    \r\n \t
  1. \\(\\dfrac{\\cancel{15}3\\cancel{m^3}}{4\\cancel{n^2}}\\cdot \\dfrac{\\cancel{17}\\cancel{n^3}}{\\cancel{30}\\cancel{10}2\\cancel{m^3}}\\cdot \\dfrac{\\cancel{3}1m^4}{\\cancel{34}2n^{\\cancel{2}}}\\Rightarrow \\dfrac{3m^4}{16n}\\)<\/li>\r\n \t
  2. \\(\\dfrac{5v^2-25v}{5v+25}\\cdot \\dfrac{10v}{v^2-11v+30}\\Rightarrow \\dfrac{\\cancel{5}v\\cancel{(v-5)}}{\\cancel{5}(v+5)}\\cdot \\dfrac{10v}{\\cancel{(v-5)}(v-6)}\\Rightarrow \\dfrac{10v^2}{(v+5)(v-6)}\\)<\/li>\r\n \t
  3. \\(\\phantom{1}\\)\r\n\\(\\left(\\dfrac{8}{2x}=\\dfrac{2}{x}+1\\right)(2x) \\\\ \\)\r\n\\(\\begin{array}{rrl}\r\n8&=&2\\cdot 2+1(2x) \\\\\r\n8&=&\\phantom{-}4+2x \\\\\r\n-4&&-4 \\\\\r\n\\midrule\r\n\\dfrac{4}{2}&=&\\dfrac{2x}{2} \\\\ \\\\\r\nx&=&2\r\n\\end{array}\\)<\/li>\r\n \t
  4. \\(\\begin{array}{l}\r\n\\dfrac{\\left(\\dfrac{x^2}{y^2}-16\\right)y^3}{\\left(\\dfrac{x+4y}{y^3}\\right)y^3}\\Rightarrow \\dfrac{x^2y-16y^3}{x+4y}\\Rightarrow \\dfrac{y(x^2-16y^2)}{x+4y}\\Rightarrow\\dfrac{y(x-4y)\\cancel{(x+4y)}}{\\cancel{x+4y}} \\\\ \\\\\r\n\\Rightarrow y(x-4y)\r\n\\end{array}\\)<\/li>\r\n \t
  5. \\(\\begin{array}{l}\r\n\\\\\r\n5y+2\\cdot 9y+6y\\sqrt{y} \\\\\r\n23y+6y\\sqrt{y}\r\n\\end{array}\\)<\/li>\r\n \t
  6. \\(\\dfrac{(28)(7+3\\sqrt{5})}{(7-3\\sqrt{5})(7+3\\sqrt{5})}\\Rightarrow \\dfrac{196+84\\sqrt{5}}{49-9\\cdot 5}\\Rightarrow \\dfrac{196+84\\sqrt{5}}{4}\\Rightarrow 49+21\\sqrt{5}\\)<\/li>\r\n \t
  7. \\(\\begin{array}{l}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n(27a^{-\\frac{3}{8}})^{\\frac{1}{3}} \\\\ \\\\\r\n27^{\\frac{1}{3}}a^{-\\frac{3}{8}\\cdot \\frac{1}{3}} \\\\ \\\\\r\n3a^{-\\frac{1}{8}} \\\\ \\\\\r\n\\dfrac{3}{a^{\\frac{1}{8}}}\\Rightarrow \\dfrac{3}{\\sqrt[8]{a}}\r\n\\end{array}\\)<\/li>\r\n \t
  8. \\(\\phantom{1}\\)\r\n\\((\\sqrt{3x-2})^2=(\\sqrt{5x+4})^2 \\\\ \\)\r\n\\(\\begin{array}{rrrrrrrr}\r\n&3x&-&2&=&5x&+&4 \\\\\r\n-&3x&-&4&&-3x&-&4 \\\\\r\n\\midrule\r\n&&&-6&=&2x&& \\\\ \\\\\r\n&&&x&=&\\dfrac{-6}{2}&=&-3\r\n\\end{array}\\)<\/li>\r\n \t
  9. \\(\\phantom{1}\\)\r\n
      \r\n \t
    1. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\\r\n\\dfrac{2x^2}{2}&=&\\dfrac{72}{2} \\\\ \\\\\r\nx^2&=&36 \\\\\r\nx&=&\\pm 6\r\n\\end{array}\\)<\/li>\r\n \t
    2. \\(\\begin{array}{rrl}\r\n\\\\ \\\\\r\n2x^2-8x&=&0 \\\\\r\n2x(x-4)&=&0 \\\\\r\nx&=&0, 4\r\n\\end{array}\\)<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
    3. \\(\\phantom{1}\\)\r\n
        \r\n \t
      1. \\(\\begin{array}{rrl}\r\n\\\\\r\n(x+5)(x+1)&=&0 \\\\\r\nx&=&-1, -5\r\n\\end{array}\\)<\/li>\r\n \t
      2. \\(\\begin{array}{l}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n\\text{Quadratic:} \\\\ \\\\\r\nx^2-10x+4=0 \\\\ \\\\\r\n\\dfrac{-(-10)\\pm \\sqrt{(-10)^2-4(1)(4)}}{2} \\\\ \\\\\r\n\\dfrac{-10\\pm \\sqrt{100-16}}{2} \\\\ \\\\\r\n\\dfrac{10\\pm \\sqrt{84}}{2} \\\\ \\\\\r\n\\dfrac{10\\pm 2\\sqrt{21}}{2}\\Rightarrow 5 \\pm \\sqrt{21}\r\n\\end{array}\\)<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
      3. \\(\\phantom{1}\\)\r\n\\(\\left(\\dfrac{8}{4x}=\\dfrac{2}{x}+3\\right)(4x) \\\\ \\)\r\n\\(\\begin{array}{rrrrl}\r\n8&=&8&+&3(4x) \\\\\r\n-8&&-8&& \\\\\r\n\\midrule\r\n\\dfrac{0}{12}&=&\\dfrac{12x}{12}&& \\\\ \\\\\r\nx&=&0&&\\therefore \\text{Undefined. No solution}\r\n\\end{array}\\)<\/li>\r\n \t
      4. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n\\text{Let }u&=&x^2 \\\\ \\\\\r\nu^2-17u+16&=&0 \\\\\r\n(u-16)(u-1)&=&0 \\\\ \\\\\r\n(x^2-16)(x^2-1)&=&0 \\\\\r\n(x-4)(x+4)(x-1)(x+1)&=&0 \\\\\r\nx&=& \\pm 1, \\pm 4\r\n\\end{array}\\)<\/li>\r\n \t
      5. \\(\\begin{array}{ll}\r\n\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\\r\nL&=&W+6 \\\\\r\n\\text{Area}&=&12+\\text{Perimeter} \\\\ \\\\\r\nL\\cdot W&=& 12+2L+2W \\\\\r\n(W+6)W&=&12+2(W+6)+2W \\\\\r\nW^2+6W&=&12+2W+12+2W\r\n\\end{array}\r\n& \\hspace{0.25in}\r\n\\begin{array}{rrrrcrr}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n0&=&W^2&+&6W&& \\\\\r\n&&&-&4W&-&24 \\\\\r\n\\midrule\r\n0&=&W^2&+&2W&-&24 \\\\\r\n0&=&(W&+&6)(W&-&4) \\\\\r\nW&=&\\cancel{-6},&4&&& \\\\ \\\\\r\nL&=&W&+&6&& \\\\\r\nL&=&4&+&6&& \\\\\r\nL&=&10&&&& \\\\\r\n\\end{array}\r\n\\end{array}\\)<\/li>\r\n \t
      6. \\(\\phantom{1}\\)\r\n\\(x, x+2, x+4 \\\\ \\)\r\n\\(\\begin{array}{ll}\r\n\\begin{array}{rrrrrrrrrrr}\r\n&&x(x&+&4)&=&31&+&x&+&2 \\\\\r\nx^2&+&4x&&&=&33&+&x&& \\\\\r\n&-&x&-&33&&-33&-&x&& \\\\\r\n\\midrule\r\nx^2&+&3x&-&33&=&0&&&&\r\n\\end{array}\r\n& \\hspace{0.25in}\r\n\\begin{array}{l}\r\n\\dfrac{-b\\pm \\sqrt{b^2-4ac}}{2a} \\\\ \\\\\r\n\\dfrac{-3\\pm \\sqrt{3^2-4(1)(-33)}}{2} \\\\ \\\\\r\n\\dfrac{-3\\pm \\sqrt{141}}{2}\r\n\\end{array}\r\n\\end{array}\\)<\/li>\r\n \t
      7. \\(\\begin{array}{rrrrcl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n&d&=&r&\\cdot &t \\\\\r\n\\text{To outpost:}&60&=&(B&-&C)5 \\\\\r\n\\text{Back:}&60&=&(B&+&C)3 \\\\ \\\\\r\n&60&=&5B&-&5C \\\\\r\n&60&=&3B&+&3C \\\\ \\\\\r\n&12&=&B&-&C \\\\\r\n+&20&=&B&+&C \\\\\r\n\\midrule\r\n&32&=&2B&& \\\\\r\n&\\therefore B&=&16&\\text{ km\/h}& \\\\ \\\\\r\n&\\therefore B&+&C&=&\\phantom{-}20 \\\\\r\n&16&+&C&=&\\phantom{-}20 \\\\\r\n-&16&&&&-16 \\\\\r\n\\midrule\r\n&&&C&=&4\\text{ km\/h}\r\n\\end{array}\\)<\/li>\r\n<\/ol>\r\n

        <\/h1>","rendered":"
          \n
        1. \"\dfrac{\cancel{15}3\cancel{m^3}}{4\cancel{n^2}}\cdot \dfrac{\cancel{17}\cancel{n^3}}{\cancel{30}\cancel{10}2\cancel{m^3}}\cdot \dfrac{\cancel{3}1m^4}{\cancel{34}2n^{\cancel{2}}}\Rightarrow \dfrac{3m^4}{16n}\"<\/li>\n
        2. \"\dfrac{5v^2-25v}{5v+25}\cdot \dfrac{10v}{v^2-11v+30}\Rightarrow \dfrac{\cancel{5}v\cancel{(v-5)}}{\cancel{5}(v+5)}\cdot \dfrac{10v}{\cancel{(v-5)}(v-6)}\Rightarrow \dfrac{10v^2}{(v+5)(v-6)}\"<\/li>\n
        3. \"\phantom{1}\"
          \n\"\left(\dfrac{8}{2x}=\dfrac{2}{x}+1\right)(2x) \\\"
          \n\"\begin{array}{rrl}<\/li>\n
        4. \"\begin{array}{l}<\/li>\n
        5. \"\begin{array}{l}<\/li>\n
        6. \"\dfrac{(28)(7+3\sqrt{5})}{(7-3\sqrt{5})(7+3\sqrt{5})}\Rightarrow \dfrac{196+84\sqrt{5}}{49-9\cdot 5}\Rightarrow \dfrac{196+84\sqrt{5}}{4}\Rightarrow 49+21\sqrt{5}\"<\/li>\n
        7. \"\begin{array}{l}<\/li>\n
        8. \"\phantom{1}\"
          \n\"(\sqrt{3x-2})^2=(\sqrt{5x+4})^2 \\\"
          \n\"\begin{array}{rrrrrrrr}<\/li>\n
        9. \"\phantom{1}\"\n
            \n
          1. \"\begin{array}{rrl}<\/li>\n
          2. \"\begin{array}{rrl}<\/li>\n<\/ol>\n<\/li>\n
          3. \"\phantom{1}\"\n
              \n
            1. \"\begin{array}{rrl}<\/li>\n
            2. \"\begin{array}{l}<\/li>\n<\/ol>\n<\/li>\n
            3. \"\phantom{1}\"
              \n\"\left(\dfrac{8}{4x}=\dfrac{2}{x}+3\right)(4x) \\\"
              \n\"\begin{array}{rrrrl}<\/li>\n
            4. \"\begin{array}{rrl}<\/li>\n
            5. \"\begin{array}{ll}<\/li>\n
            6. \"\phantom{1}\"
              \n\"x, x+2, x+4 \\\"
              \n\"\begin{array}{ll}<\/li>\n
            7. \"\begin{array}{rrrrcl}<\/li>\n<\/ol>\n

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