{"id":3334,"date":"2019-11-19T15:06:20","date_gmt":"2019-11-19T20:06:20","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/?post_type=back-matter&p=3334"},"modified":"2019-11-19T15:12:36","modified_gmt":"2019-11-19T20:12:36","slug":"midterm-3-version-d","status":"publish","type":"back-matter","link":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/back-matter\/midterm-3-version-d\/","title":{"raw":"Midterm 3: Version D Answer Key","rendered":"Midterm 3: Version D Answer Key"},"content":{"raw":"[latexpage]\r\n
    \r\n \t
  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}\\)<\/li>\r\n \t
  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)}\\)<\/li>\r\n \t
  3. \\(\\phantom{1}\\)\r\n\\(\\left(\\dfrac{2}{x-4}-\\dfrac{6}{x-3}=3\\right)(x+4)(x-3) \\\\ \\)\r\n\\(\\begin{array}{rrrrcrrrcrcrr}\r\n2(x&-&3)&-&6(x&+&4)&=&3(x&+&4)(x&-&3) \\\\\r\n2x&-&6&-&6x&-&24&=&3(x^2&+&x&-&12) \\\\\r\n&&&&-4x&-&30&=&3x^2&+&3x&-&36 \\\\\r\n&&&&+4x&+&30&&&+&4x&+&30 \\\\\r\n\\midrule\r\n&&&&&&0&=&3x^2&+&7x&-&6 \\\\\r\n&&&&&&0&=&(x&+&3)(3x&-&2) \\\\ \\\\\r\n&&&&&&x&=&-3,&\\dfrac{2}{3}&&&\r\n\\end{array}\\)<\/li>\r\n \t
  4. \\(\\begin{array}{l}\r\n\\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)}} \\\\ \\\\\r\n\\Rightarrow y(x-3y)\r\n\\end{array}\\)<\/li>\r\n \t
  5. \\(\\begin{array}{l}\r\n\\\\\r\n5y^2+2\\cdot 7y+\\sqrt{25y^2\\cdot y} \\\\\r\n5y^2+14y+5y\\sqrt{y}\r\n\\end{array}\\)<\/li>\r\n \t
  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}\\)<\/li>\r\n \t
  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}\\)<\/li>\r\n \t
  8. \\(\\begin{array}{ll}\r\n\\begin{array}{rrrrl}\r\n\\\\ \\\\\r\n\\sqrt{2x+9}&-&3&=&x \\\\\r\n&+&3&&\\phantom{x}+3 \\\\\r\n\\midrule\r\n(\\sqrt{2x+9})^2&&&=&(x+3)^2\r\n\\end{array}\r\n& \\hspace{0.25in}\r\n\\begin{array}{rrrrrcrrrr}\r\n\\\\ \\\\ \\\\ \\\\ \\\\\r\n&2x&+&9&=&x^2&+&6x&+&9 \\\\\r\n-&2x&-&9&&&-&2x&-&9 \\\\\r\n\\midrule\r\n&&&0&=&x^2&+&4x&& \\\\\r\n&&&0&=&x(x&+&4)&& \\\\ \\\\\r\n&&&x&=&0,&-4&&&\r\n\\end{array}\r\n\\end{array}\\)<\/li>\r\n \t
  9. \\(\\phantom{1}\\)\r\n
      \r\n \t
    1. \\(\\begin{array}{rrl}\r\n\\\\ \\\\\r\n8x^2-32x&=&0 \\\\\r\n8x(x-4)&=&0 \\\\\r\nx&=&0, 4\r\n\\end{array}\\)<\/li>\r\n \t
    2. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\\r\n\\dfrac{3x^2}{2}&=&\\dfrac{48}{3} \\\\ \\\\\r\n\\sqrt{x^2}&=&\\sqrt{16} \\\\\r\nx&=&\\pm 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\nx^2-5x+4&=&0 \\\\\r\n(x-4)(x-1)&=&0 \\\\\r\nx&=&1, 4\r\n\\end{array}\\)<\/li>\r\n \t
      2. \\(\\begin{array}{rrl}\r\n\\\\\r\n(x-3)(x-1)&=&0 \\\\\r\nx&=&1, 3\r\n\\end{array}\\)<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
      3. \\(\\begin{array}{rrrrcrcrcrr}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n2(x&+&4)&=&x(x)&&&&&& \\\\\r\n2x&+&8&=&x^2&&&&&& \\\\ \\\\\r\n&&0&=&x^2&-&2x&-&8&& \\\\\r\n&&0&=&x^2&-&4x&+&2x&-&8 \\\\\r\n&&0&=&x(x&-&4)&+&2(x&-&4) \\\\\r\n&&0&=&(x&-&4)(x&+&2)&& \\\\ \\\\\r\n&&x&=&-2,&4&&&&&\r\n\\end{array}\\)<\/li>\r\n \t
      4. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n\\text{Let }u&=&x^2 \\\\ \\\\\r\nu^2-48u-49&=&0 \\\\\r\n(u-49)(u+1)&=&0 \\\\ \\\\\r\n(x^2-49)(x^2+1)&=&0 \\\\\r\n(x-7)(x+7)(x^2+1)&=&0 \\\\ \\\\\r\nx^2+1&=&\\text{cannot be factored} \\\\\r\nx&=&\\pm 7\r\n\\end{array}\\)<\/li>\r\n \t
      5. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n40&=&\\dfrac{1}{2}(h-2)(h) \\\\ \\\\\r\n80&=&h^2-2h \\\\\r\n0&=&h^2-2h-80 \\\\\r\n0&=&(h-10)(h+8) \\\\\r\nh&=&10, \\cancel{-8} \\\\ \\\\\r\nb&=&10-2=8\r\n\\end{array}\\)<\/li>\r\n \t
      6. \\(\\phantom{1}\\)\r\n\\(x, x+2, x+4 \\\\ \\)\r\n\\(\\begin{array}{crrrrrcrr}\r\nx(x&+&4)&=&41&+&4(x&+&2) \\\\\r\nx^2&+&4x&=&41&+&4x&+&8 \\\\\r\n&-&4x&&&-&4x&& \\\\\r\n\\midrule\r\n&&\\sqrt{x^2}&=&\\sqrt{49}&&&& \\\\\r\n&&x&=&\\pm 7&&&&\r\n\\end{array}\\)\r\n\\(\\phantom{1}\\)\r\n\\(\\text{numbers are }7, 9, 11\\text{ or }-7,-5,-3\\)<\/li>\r\n \t
      7. \\(\\begin{array}{rrrrrrcrrr}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n&&&d_{\\text{d}}&=&d_{\\text{u}}&+&4&& \\\\ \\\\\r\n&2(6&+&r)&=&3(6&-&r)&+&4 \\\\\r\n&12&+&2r&=&18&-&3r&+&4 \\\\\r\n-&12&+&3r&&-12&+&3r&& \\\\\r\n\\midrule\r\n&&&\\dfrac{5r}{5}&=&\\dfrac{10}{5}&&&& \\\\ \\\\\r\n&&&r&=&2&\\text{km\/h}&&&\r\n\\end{array}\\)<\/li>\r\n<\/ol>\r\n

        <\/h1>","rendered":"
          \n
        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}\"<\/li>\n
        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)}\"<\/li>\n
        3. \"\phantom{1}\"
          \n\"\left(\dfrac{2}{x-4}-\dfrac{6}{x-3}=3\right)(x+4)(x-3) \\\"
          \n\"\begin{array}{rrrrcrrrcrcrr}<\/li>\n
        4. \"\begin{array}{l}<\/li>\n
        5. \"\begin{array}{l}<\/li>\n
        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}\"<\/li>\n
        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}\"<\/li>\n
        8. \"\begin{array}{ll}<\/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. \"\begin{array}{rrrrcrcrcrr}<\/li>\n
            4. \"\begin{array}{rrl}<\/li>\n
            5. \"\begin{array}{rrl}<\/li>\n
            6. \"\phantom{1}\"
              \n\"x, x+2, x+4 \\\"
              \n\"\begin{array}{crrrrrcrr}
              \n\"\phantom{1}\"
              \n\"\text{numbers are }7, 9, 11\text{ or }-7,-5,-3\"<\/li>\n
            7. \"\begin{array}{rrrrrrcrrr}<\/li>\n<\/ol>\n

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