{"id":3328,"date":"2019-11-19T15:05:29","date_gmt":"2019-11-19T20:05:29","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/?post_type=back-matter&p=3328"},"modified":"2019-11-19T15:10:00","modified_gmt":"2019-11-19T20:10:00","slug":"midterm-3-version-a-answer-key","status":"publish","type":"back-matter","link":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/back-matter\/midterm-3-version-a-answer-key\/","title":{"raw":"Midterm 3: Version A Answer Key","rendered":"Midterm 3: Version A Answer Key"},"content":{"raw":"[latexpage]\r\n
    \r\n \t
  1. \\(\\dfrac{15m^3}{4n^2}\\cdot \\dfrac{\\cancel{17}1m^3}{\\cancel{12}4n}\\cdot \\dfrac{\\cancel{3 }1m^4}{\\cancel{34 }2n^2}\\Rightarrow \\dfrac{15m^{10}}{32n^5}\\)<\/li>\r\n \t
  2. \\(\\begin{array}{l}\r\n\\\\ \\\\ \\\\ \\\\\r\n\\dfrac{8x-8y}{x^3+y^3}\\cdot \\dfrac{x^2-xy+y^2}{x^2-y^2} \\\\ \\\\\r\n\\Rightarrow \\dfrac{8\\cancel{(x-y)}}{(x+y)\\cancel{(x^2-xy+y^2)}}\\cdot \\dfrac{\\cancel{x^2-xy+y^2}}{(x+y)\\cancel{(x-y)}}\\Rightarrow \\dfrac{8}{(x+y)^2}\r\n\\end{array}\\)<\/li>\r\n \t
  3. \\(\\begin{array}{l}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n\\text{LCD}=6(n-3) \\\\ \\\\\r\n\\dfrac{5(n-3)-2\\cdot 6(n-3)-5\\cdot 6}{6(n-3)} \\\\ \\\\\r\n\\dfrac{5n-15-12n+36-30}{6(n-3)} \\\\ \\\\\r\n\\dfrac{-7n-9}{6(n-3)}\r\n\\end{array}\\)<\/li>\r\n \t
  4. \\(\\dfrac{\\left(\\dfrac{x^2}{y^2}-4\\right)y^3}{\\left(\\dfrac{x+2y}{y^3}\\right)y^3} \\Rightarrow \\dfrac{x^2y-4y^3}{x+2y}\\Rightarrow \\dfrac{y(x^2-4y^2)}{x+2y}\\Rightarrow \\dfrac{y(x-2y)\\cancel{(x+2y)}}{\\cancel{(x+2y)}} \\\\ \\)\r\n\\(\\Rightarrow y(x-2y)\\)<\/li>\r\n \t
  5. \\(3\\cdot 5+2\\sqrt{36\\cdot 2}-4 \\)\r\n\\(15+2\\cdot 6\\sqrt{2}-4\\)\r\n\\(11+12\\sqrt{2}\\)<\/li>\r\n \t
  6. \\(\\dfrac{\\sqrt{m^7n^{\\cancel{3}2}}}{\\sqrt{2\\cancel{n}}}\\cdot \\dfrac{\\sqrt{2}}{\\sqrt{2}}\\Rightarrow \\dfrac{\\sqrt{m^6\\cdot m\\cdot n^2\\cdot 2}}{\\sqrt{4}}\\Rightarrow \\dfrac{m^3n\\sqrt{2m}}{2}\\)<\/li>\r\n \t
  7. \\(\\begin{array}{l}\r\n\\\\ \\\\ \\\\ \\\\\r\n\\dfrac{2-x}{1-\\sqrt{3}}\\cdot \\dfrac{1+\\sqrt{3}}{1+\\sqrt{3}}\\Rightarrow \\dfrac{2+2\\sqrt{3}-x-x\\sqrt{3}}{1-3} \\\\ \\\\\r\n\\(\\Rightarrow \\dfrac{2+2\\sqrt{3}-x-x\\sqrt{3}}{-2}\\text{ or }\\dfrac{x+x\\sqrt{3}-2-2\\sqrt{3}}{2}\r\n\\end{array}\\)<\/li>\r\n \t
  8. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\\r\n(\\sqrt{7x+8})^2&=&(x)^2 \\\\\r\n7x+8&=&x^2 \\\\\r\n0&=&x^2-7x-8 \\\\\r\n0&=&(x-8)(x+1) \\\\ \\\\\r\nx&=&8, \\cancel{-1}\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{4x^2}{4}&=&\\dfrac{64}{4} \\\\ \\\\\r\nx^2&=&16 \\\\\r\nx&=&\\pm 4\r\n\\end{array}\\)<\/li>\r\n \t
    2. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\\r\n3x^2-12x&=&0 \\\\\r\n3x(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&=&5, 1\r\n\\end{array}\\)<\/li>\r\n \t
      2. \\(\\begin{array}{rrl}\r\n\\\\ \\\\\r\nx^2+10x+9&=&0 \\\\\r\n(x+9)(x+1)&=&0 \\\\\r\nx&=&-9, -1\r\n\\end{array}\\)<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
      3. \\(\\phantom{1}\\)\r\n\\(\\left(\\dfrac{x+4}{-4}=\\dfrac{8}{x}\\right)(-4)(x) \\\\ \\)\r\n\\(\\begin{array}{rrl}\r\nx(x+4)&=&-4(8) \\\\\r\nx^2+4x&=&-32 \\\\\r\n0&=&x^2+4x+32 \\hspace{0.5in} \\text{Does not factor}\r\n\\end{array}\\)<\/li>\r\n \t
      4. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n\\text{Let }u&=&x^2 \\\\ \\\\\r\nu^2-13u+36&=&0 \\\\\r\nu^2-4u-9u+36&=&0 \\\\\r\nu(u-4)-9(u-4)&=&0 \\\\\r\n(u-4)(u-9)&=&0 \\\\ \\\\\r\n(x^2-4)(x^2-9)&=&0 \\\\\r\n(x-2)(x+2)(x-3)(x+3)&=&0 \\\\\r\nx&=& \\pm 2, \\pm 3\r\n\\end{array}\\)<\/li>\r\n \t
      5. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\nA&=&\\dfrac{1}{2}bh \\\\ \\\\\r\n300&=&\\dfrac{1}{2}(h+10)h \\\\ \\\\\r\n600&=&h^2+10h \\\\\r\n0&=&h^2+10h-600 \\\\\r\n0&=&(h-20)(h+30) \\\\ \\\\\r\nh&=& 20, \\cancel{-30} \\\\ \\\\\r\n\\therefore b&=&h+10=30\r\n\\end{array}\\)<\/li>\r\n \t
      6. \\(\\phantom{1}\\)\r\n\\(x, x+2, x+4 \\\\ \\)\r\n\\(\\begin{array}{rrrrrrrrrrr}\r\n&&x(x&+&4)&=&38&+&x&+&2 \\\\\r\nx^2&+&4x&&&=&x&+&40&& \\\\\r\n&-&x&-&40&&-x&-&40&& \\\\\r\n\\midrule\r\nx^2&+&3x&-&40&=&0&&&& \\\\ \\\\\r\n&&&&0&=&(x&+&8)(x&-&5) \\\\\r\n&&&&x&=&\\cancel{-8},&5&&& \\\\\r\n\\end{array}\\)\r\n\u2234 5, 7, 9<\/li>\r\n \t
      7. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\nr_st_s&=&r_ft_f \\\\ \\\\\r\nr(4.5\\text{ h})&=&(r+150)(3.0\\text{ h}) \\\\\r\n4.5r&=&\\phantom{-}3.0r+450 \\\\\r\n-3.0r&&-3.0r \\\\\r\n\\midrule\r\n1.5r&=&450 \\\\ \\\\\r\nr&=&\\dfrac{450}{1.5}\\text{ or }300\\text{ km\/h} \\\\ \\\\\r\nr_f&=&300+150 \\\\\r\nr_f&=&450\\text{ km\/h}\r\n\\end{array}\\)<\/li>\r\n<\/ol>\r\n

        <\/h1>","rendered":"
          \n
        1. \"\dfrac{15m^3}{4n^2}\cdot \dfrac{\cancel{17}1m^3}{\cancel{12}4n}\cdot \dfrac{\cancel{3 }1m^4}{\cancel{34 }2n^2}\Rightarrow \dfrac{15m^{10}}{32n^5}\"<\/li>\n
        2. \"\begin{array}{l}<\/li>\n
        3. \"\begin{array}{l}<\/li>\n
        4. \"\dfrac{\left(\dfrac{x^2}{y^2}-4\right)y^3}{\left(\dfrac{x+2y}{y^3}\right)y^3} \Rightarrow \dfrac{x^2y-4y^3}{x+2y}\Rightarrow \dfrac{y(x^2-4y^2)}{x+2y}\Rightarrow \dfrac{y(x-2y)\cancel{(x+2y)}}{\cancel{(x+2y)}} \\\"
          \n\"\Rightarrow y(x-2y)\"<\/li>\n
        5. \"3\cdot 5+2\sqrt{36\cdot 2}-4\"
          \n\"15+2\cdot 6\sqrt{2}-4\"
          \n\"11+12\sqrt{2}\"<\/li>\n
        6. \"\dfrac{\sqrt{m^7n^{\cancel{3}2}}}{\sqrt{2\cancel{n}}}\cdot \dfrac{\sqrt{2}}{\sqrt{2}}\Rightarrow \dfrac{\sqrt{m^6\cdot m\cdot n^2\cdot 2}}{\sqrt{4}}\Rightarrow \dfrac{m^3n\sqrt{2m}}{2}\"<\/li>\n
        7. \"\begin{array}{l}<\/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+4}{-4}=\dfrac{8}{x}\right)(-4)(x) \\\"
              \n\"\begin{array}{rrl}<\/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}{rrrrrrrrrrr}
              \n\u2234 5, 7, 9<\/li>\n
            7. \"\begin{array}{rrl}<\/li>\n<\/ol>\n

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