{"id":3330,"date":"2019-11-19T15:05:47","date_gmt":"2019-11-19T20:05:47","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/?post_type=back-matter&p=3330"},"modified":"2019-11-19T15:11:17","modified_gmt":"2019-11-19T20:11:17","slug":"midterm-3-version-b","status":"publish","type":"back-matter","link":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/back-matter\/midterm-3-version-b\/","title":{"raw":"Midterm 3: Version B Answer Key","rendered":"Midterm 3: Version B Answer Key"},"content":{"raw":"[latexpage]\r\n
    \r\n \t
  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}\\)<\/li>\r\n \t
  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)}\\)<\/li>\r\n \t
  3. \\(\\begin{array}{l}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n\\left(\\dfrac{5x}{x+3}-\\dfrac{5x}{x-3}+\\dfrac{90}{x^2-9}\\right)(x-3)(x+3) \\\\ \\\\\r\n\\Rightarrow \\dfrac{5x(x-3)-5x(x+3)+90}{(x+3)(x-3)} \\\\ \\\\\r\n\\Rightarrow \\dfrac{5x^2-15x-5x^2-15x+90}{(x+3)(x-3)} \\\\ \\\\\r\n\\Rightarrow \\dfrac{-30x+90}{(x+3)(x-3)}\\Rightarrow \\dfrac{-30\\cancel{(x-3)}}{(x+3)\\cancel{(x-3)}}\\Rightarrow \\dfrac{-30}{x+3}\r\n\\end{array}\\)<\/li>\r\n \t
  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}\\)<\/li>\r\n \t
  5. \\(\\begin{array}{l}\r\n\\\\ \\\\ \\\\\r\n\\sqrt{2\\cdot 36\\cdot d^2\\cdot d}+4\\sqrt{2\\cdot 9\\cdot d^2\\cdot d}-2(7d^2) \\\\\r\n6d\\sqrt{2d}+4\\cdot 3d\\sqrt{2d}-14d^2 \\\\\r\n6d\\sqrt{2d}+12d\\sqrt{2d}-14d^2 \\\\\r\n18d\\sqrt{2d}-14d^2\r\n\\end{array}\\)<\/li>\r\n \t
  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}\\)<\/li>\r\n \t
  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}\\)<\/li>\r\n \t
  8. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\\r\n(\\sqrt{4x+12})^2&=&(x)^2 \\\\\r\n4x+12&=&x^2 \\\\\r\n0&=&x^2-4x-12 \\\\\r\n0&=&(x-6)(x+2) \\\\ \\\\\r\nx&=&6, \\cancel{-2}\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{98}{2} \\\\ \\\\\r\nx^2&=&49 \\\\\r\nx&=& \\pm 7\r\n\\end{array}\\)<\/li>\r\n \t
    2. \\(\\begin{array}{rrl}\r\n\\\\ \\\\\r\n4x^2-12x&=&0 \\\\\r\n4x(x-3)&=&0 \\\\\r\nx&=&0, 3\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+4)&=&0 \\\\\r\nx&=&5, -4\r\n\\end{array}\\)<\/li>\r\n \t
      2. \\(\\begin{array}{rrl}\r\n\\\\ \\\\\r\nx^2-2x-35&=&0 \\\\\r\n(x-7)(x+5)&=&0 \\\\\r\nx&=&7, -5\r\n\\end{array}\\)<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
      3. \\(\\phantom{1}\\)\r\n\\(\\left(\\dfrac{x-3}{x+2}+\\dfrac{6}{x+3}=1\\right)(x+2)(x+3) \\\\ \\\\ \\)\r\n\\(\\begin{array}{rcccrcrrrrrcrr}\r\n&(x-3)&\\cdot &(x+3)&+&6(x&+&2)&=&(x&+&2)(x&+&3) \\\\\r\n&x^2&-&9&+&6x&+&12&=&x^2&+&5x&+&6 \\\\\r\n-&x^2&+&9&-&6x&-&12&&-x^2&-&6x&-&12 \\\\\r\n\\midrule\r\n&&&&&&&0&=&-x&+&3&& \\\\\r\n&&&&&&+&x&&+x&&&& \\\\\r\n\\midrule\r\n&&&&&&&x&=&3&&&&\r\n\\end{array}\\)<\/li>\r\n \t
      4. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\\r\n\\text{Let }u&=&x^2 \\\\ \\\\\r\nu^2-5u+4&=&0 \\\\\r\n(u-4)(u-1)&=&0 \\\\ \\\\\r\n(x^2-4)(x^2-1)&=&0 \\\\\r\n(x-2)(x+2)(x-1)(x+1)&=&0 \\\\\r\nx&=&\\pm 2, \\pm 1\r\n\\end{array}\\)<\/li>\r\n \t
      5. \\(\\begin{array}{rrrrrrr}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\nL&=&3&+&W&& \\\\ \\\\\r\nP&=&2L&+&2W&& \\\\\r\n46&=&2(3&+&W)&+&2W \\\\\r\n46&=&6&+&2W&+&2W \\\\\r\n-6&&-6&&&& \\\\\r\n\\midrule\r\n40&=&4W&&&& \\\\ \\\\\r\nW&=&\\dfrac{40}{4}&=&10&& \\\\ \\\\\r\n\\therefore L&=&W&+&3&& \\\\\r\nL&=&10&+&3&=&13\r\n\\end{array}\\)<\/li>\r\n \t
      6. \\(\\phantom{1}\\)\r\n\\(x, x+2, x+4 \\\\ \\)\r\n\\(\\begin{array}{rrcrrrrrcrr}\r\n&&x(x&+&2)&=&16&+&x&+&4 \\\\\r\nx^2&+&2x&&&=&20&+&x&& \\\\\r\n&-&x&-&20&&-20&-&x&& \\\\\r\n\\midrule\r\nx^2&+&x&-&20&=&0&&&& \\\\ \\\\\r\n&&&&0&=&(x&+&5)(x&-&4) \\\\\r\n&&&&x&=&\\cancel{-5},&4&&& \\\\\r\n\\end{array}\\)\r\n\u2234 4, 6, 8<\/li>\r\n \t
      7. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\nd&=&r\\cdot t \\\\\r\nd_{\\text{up}}&=&d_{\\text{return}} \\\\ \\\\\r\n4(r-5)&=&2(r+5) \\\\\r\n4r-20&=&\\phantom{-}2r+10 \\\\\r\n-2r+20&&-2r+20 \\\\\r\n\\midrule\r\n\\dfrac{2r}{2}&=&\\dfrac{30}{2} \\\\ \\\\\r\nr&=&15\r\n\\end{array}\\)<\/li>\r\n<\/ol>\r\n

        <\/h1>","rendered":"
          \n
        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}\"<\/li>\n
        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)}\"<\/li>\n
        3. \"\begin{array}{l}<\/li>\n
        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}\"<\/li>\n
        5. \"\begin{array}{l}<\/li>\n
        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}\"<\/li>\n
        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}\"<\/li>\n
        8. \"\begin{array}{rrl}<\/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}{rrl}<\/li>\n<\/ol>\n<\/li>\n
            3. \"\phantom{1}\"
              \n\"\left(\dfrac{x-3}{x+2}+\dfrac{6}{x+3}=1\right)(x+2)(x+3) \\ \\\"
              \n\"\begin{array}{rcccrcrrrrrcrr}<\/li>\n
            4. \"\begin{array}{rrl}<\/li>\n
            5. \"\begin{array}{rrrrrrr}<\/li>\n
            6. \"\phantom{1}\"
              \n\"x, x+2, x+4 \\\"
              \n\"\begin{array}{rrcrrrrrcrr}
              \n\u2234 4, 6, 8<\/li>\n
            7. \"\begin{array}{rrl}<\/li>\n<\/ol>\n

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