{"id":2014,"date":"2021-12-02T19:40:28","date_gmt":"2021-12-03T00:40:28","guid":{"rendered":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/back-matter\/midterm-3-version-e\/"},"modified":"2022-11-02T10:39:08","modified_gmt":"2022-11-02T14:39:08","slug":"midterm-3-version-e","status":"publish","type":"back-matter","link":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/back-matter\/midterm-3-version-e\/","title":{"raw":"Midterm 3: Version E Answer Key","rendered":"Midterm 3: Version E Answer Key"},"content":{"raw":"
    \n \t
  1. [latex]\\dfrac{\\cancel{12}1\\cancel{m^3}}{\\cancel{5}1\\cancel{n^2}}\\cdot \\dfrac{\\cancel{15}\\cancel{3}1\\cancel{n^3}}{\\cancel{36}\\cancel{3}1\\cancel{m^6}}\\cdot \\dfrac{\\cancel{8}4m^{\\cancel{4}}}{\\cancel{6}3n^{\\cancel{2}}}\\Rightarrow \\dfrac{4m}{3n}[\/latex]<\/li>\n \t
  2. [latex]\\dfrac{\\cancel{x}1\\cancel{(x+2)}}{\\cancel{(x+2)}\\cancel{(x+7)}}\\cdot \\dfrac{\\cancel{2}1\\cancel{(x+7)}}{\\cancel{2}1x^{\\cancel{3}2}}=\\dfrac{1}{x^2}[\/latex]<\/li>\n \t
  3. [latex]\\left(\\dfrac{x-3}{7}-\\dfrac{x-15}{28}=\\dfrac{3}{4}\\right)(28) [\/latex]\n[latex]\\begin{array}{rrrrrrrrl}\n4(x&-&3)&-&(x&-&15)&=&3(7) \\\\\n4x&-&12&-&x&+&15&=&21 \\\\\n&+&12&&&-&15&&-15+12 \\\\\n\\hline\n&&&&&&3x&=&18 \\\\\n&&&&&&x&=&6\n\\end{array}[\/latex]<\/li>\n \t
  4. [latex]\\dfrac{\\left(\\dfrac{x^2}{y^2}-36\\right)y^3}{\\left(\\dfrac{x+6y}{y^3}\\right)y^3}\\Rightarrow \\dfrac{x^2y-36y^3}{x+6y}\\Rightarrow \\dfrac{y(x^2-36y^2)}{x+6y}[\/latex]\n[latex]\\Rightarrow \\dfrac{y(x-6y)\\cancel{(x+6y)}}{\\cancel{(x+6y)}}[\/latex]\n[latex]\\Rightarrow y(x-6y)[\/latex]<\/li>\n \t
  5. [latex]\\sqrt{x^6\\cdot x\\cdot y^4\\cdot y}+2xy\\sqrt{36\\cdot x\\cdot y^4\\cdot y}-\\sqrt{x\\cdot y^2\\cdot y}[\/latex]\n[latex]\\begin{array}[t]{l}\nx^3y^2\\sqrt{xy}+2xy\\cdot 6y^2\\sqrt{xy}-y\\sqrt{xy} \\\\ \\\\\nx^3y^2\\sqrt{xy}+12xy^3\\sqrt{xy}-y\\sqrt{xy} \\\\ \\\\\n\\sqrt{xy}(x^3y^2+12xy^3-y)\n\\end{array}[\/latex]<\/li>\n \t
  6. [latex]\\dfrac{\\sqrt{7}}{3-\\sqrt{7}}\\cdot \\dfrac{3+\\sqrt{7}}{3+\\sqrt{7}}\\Rightarrow \\dfrac{3\\sqrt{7}+7}{9-7}\\Rightarrow \\dfrac{3\\sqrt{7}+7}{2}[\/latex]<\/li>\n \t
  7. [latex]\\left(\\dfrac{\\cancel{x^0}1y^4}{z^{-12}}\\right)^{\\frac{1}{4}}\\Rightarrow \\dfrac{y^{4\\cdot \\frac{1}{4}}}{z^{-12\\cdot \\frac{1}{4}}}\\Rightarrow \\dfrac{y^1}{z^{-3}}\\Rightarrow yz^3[\/latex]<\/li>\n \t
  8. [latex](\\sqrt{4x-5})^2=(\\sqrt{2x+3})^2 [\/latex]\n[latex]\\begin{array}{rrrrrrrr}\n&4x&-&5&=&2x&+&3 \\\\\n-&2x&+&5&&-2x&+&5 \\\\\n\\hline\n&&&2x&=&8&& \\\\\n&&&x&=&4&&\n\\end{array}[\/latex]<\/li>\n \t
  9. [latex]\\phantom{a}[\/latex]\n
      \n \t
    1. [latex]\\left(\\dfrac{x^2}{3}=27\\right)(3)\\Rightarrow x^2=81 \\Rightarrow x=\\pm 9[\/latex]<\/li>\n \t
    2. [latex]\\begin{array}[t]{rrl}\\\\\n27x^2+3x&=&0 \\\\\n3x(9x+1)&=&0 \\\\\nx&=&0, -\\dfrac{1}{9}\n\\end{array}[\/latex]<\/li>\n<\/ol>\n<\/li>\n \t
    3. [latex]\\phantom{a}[\/latex]\n
        \n \t
      1. [latex]\\begin{array}[t]{rrl}\\\\\n(x-12)(x+1)&=&0 \\\\\nx&=&12, -1\n\\end{array}[\/latex]<\/li>\n \t
      2. [latex]\\begin{array}[t]{rrl}\\\\\nx^2+13x+12&=&0 \\\\\n(x+12)(x+1)&=&0 \\\\\nx&=&-1, -12\n\\end{array}[\/latex]<\/li>\n<\/ol>\n<\/li>\n \t
      3. [latex]\\left(\\dfrac{2}{x}=\\dfrac{2x}{3x+8}\\right)(x)(3x+8)[\/latex]\n[latex]\\begin{array}[t]{ll}\n\\begin{array}[t]{rrl}\n2(3x+8)&=&2x^2 \\\\\n6x+16&=&2x^2 \\\\\n0&=&2x^2-6x-16 \\\\\n0&=&2(x^2-3x-8)\n\\end{array}\n& \\hspace{0.25in}\n\\begin{array}[t]{l}\n\\dfrac{-(-3)\\pm \\sqrt{(-3)^2-4(1)(-8)}}{2(1)} \\\\ \\\\\n\\dfrac{3\\pm \\sqrt{9+32}}{2} \\\\ \\\\\n\\dfrac{3\\pm \\sqrt{41}}{2}\n\\end{array}\n\\end{array}[\/latex]<\/li>\n \t
      4. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrl}\n(x^2-64)(x^2+1)&=&0 \\\\\n(x-8)(x+8)(x^2+1)&=&0 \\\\\nx&=&\\pm 8\n\\end{array}[\/latex]<\/li>\n \t
      5. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrcrrrrrrrrrr}\n&&&&A&=&20&+&P&&&& \\\\ \\\\\n&&L(L&-&5)&=&20&+&2(L&-&5)&+&2L \\\\\nL^2&-&5L&&&=&20&+&2L&-&10&+&2L \\\\\n&-&4L&-&10&&-10&-&4L&&&& \\\\\n\\hline\nL^2&-&9L&-&10&=&0&&&&&& \\\\\n(L&-&10)(L&+&1)&=&0&&&&&& \\\\\n&&&&L&=&10,&\\cancel{-1}&&&&& \\\\ \\\\\n&&&&W&=&L&-&5&&&& \\\\\n&&&&W&=&10&-&5&&&& \\\\\n&&&&W&=&5&&&&&&\n\\end{array}[\/latex]<\/li>\n \t
      6. [latex]x, x+2, x+4[\/latex]\n[latex]\\begin{array}[t]{rrcrrrrrcrr}\n&&x(x&+&4)&=&35&+&10(x&+&2) \\\\\nx^2&+&4x&&&=&35&+&10x&+&20 \\\\\n&-&10x&-&55&&-55&-&10x&& \\\\\n\\hline\nx^2&-&6x&-&55&=&0&&&& \\\\\n(x&-&11)(x&+&5)&=&0&&&& \\\\\n&&&&x&=&11,&-5&&&\\\\\n\\end{array}[\/latex]\n[latex]\\text{numbers are }11,13,15\\text{ or }-5,-3,-1[\/latex]<\/li>\n \t
      7. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrrrrrcrrr}\n&&&d_{\\text{d}}&=&d_{\\text{u}}&+&9&& \\\\ \\\\\n&3(5&+&r)&=&4(5&-&r)&+&9 \\\\\n&15&+&3r&=&20&-&4r&+&9 \\\\\n-&15&+&4r&&-15&+&4r&& \\\\\n\\hline\n&&&7r&=&14&&&& \\\\\n&&&r&=&2&\\text{km\/h}&&&\n\\end{array}[\/latex]<\/li>\n<\/ol>","rendered":"
          \n
        1. [latex]\\dfrac{\\cancel{12}1\\cancel{m^3}}{\\cancel{5}1\\cancel{n^2}}\\cdot \\dfrac{\\cancel{15}\\cancel{3}1\\cancel{n^3}}{\\cancel{36}\\cancel{3}1\\cancel{m^6}}\\cdot \\dfrac{\\cancel{8}4m^{\\cancel{4}}}{\\cancel{6}3n^{\\cancel{2}}}\\Rightarrow \\dfrac{4m}{3n}[\/latex]<\/li>\n
        2. [latex]\\dfrac{\\cancel{x}1\\cancel{(x+2)}}{\\cancel{(x+2)}\\cancel{(x+7)}}\\cdot \\dfrac{\\cancel{2}1\\cancel{(x+7)}}{\\cancel{2}1x^{\\cancel{3}2}}=\\dfrac{1}{x^2}[\/latex]<\/li>\n
        3. [latex]\\left(\\dfrac{x-3}{7}-\\dfrac{x-15}{28}=\\dfrac{3}{4}\\right)(28)[\/latex]
          \n[latex]\\begin{array}{rrrrrrrrl} 4(x&-&3)&-&(x&-&15)&=&3(7) \\\\ 4x&-&12&-&x&+&15&=&21 \\\\ &+&12&&&-&15&&-15+12 \\\\ \\hline &&&&&&3x&=&18 \\\\ &&&&&&x&=&6 \\end{array}[\/latex]<\/li>\n
        4. [latex]\\dfrac{\\left(\\dfrac{x^2}{y^2}-36\\right)y^3}{\\left(\\dfrac{x+6y}{y^3}\\right)y^3}\\Rightarrow \\dfrac{x^2y-36y^3}{x+6y}\\Rightarrow \\dfrac{y(x^2-36y^2)}{x+6y}[\/latex]
          \n[latex]\\Rightarrow \\dfrac{y(x-6y)\\cancel{(x+6y)}}{\\cancel{(x+6y)}}[\/latex]
          \n[latex]\\Rightarrow y(x-6y)[\/latex]<\/li>\n
        5. [latex]\\sqrt{x^6\\cdot x\\cdot y^4\\cdot y}+2xy\\sqrt{36\\cdot x\\cdot y^4\\cdot y}-\\sqrt{x\\cdot y^2\\cdot y}[\/latex]
          \n[latex]\\begin{array}[t]{l} x^3y^2\\sqrt{xy}+2xy\\cdot 6y^2\\sqrt{xy}-y\\sqrt{xy} \\\\ \\\\ x^3y^2\\sqrt{xy}+12xy^3\\sqrt{xy}-y\\sqrt{xy} \\\\ \\\\ \\sqrt{xy}(x^3y^2+12xy^3-y) \\end{array}[\/latex]<\/li>\n
        6. [latex]\\dfrac{\\sqrt{7}}{3-\\sqrt{7}}\\cdot \\dfrac{3+\\sqrt{7}}{3+\\sqrt{7}}\\Rightarrow \\dfrac{3\\sqrt{7}+7}{9-7}\\Rightarrow \\dfrac{3\\sqrt{7}+7}{2}[\/latex]<\/li>\n
        7. [latex]\\left(\\dfrac{\\cancel{x^0}1y^4}{z^{-12}}\\right)^{\\frac{1}{4}}\\Rightarrow \\dfrac{y^{4\\cdot \\frac{1}{4}}}{z^{-12\\cdot \\frac{1}{4}}}\\Rightarrow \\dfrac{y^1}{z^{-3}}\\Rightarrow yz^3[\/latex]<\/li>\n
        8. [latex](\\sqrt{4x-5})^2=(\\sqrt{2x+3})^2[\/latex]
          \n[latex]\\begin{array}{rrrrrrrr} &4x&-&5&=&2x&+&3 \\\\ -&2x&+&5&&-2x&+&5 \\\\ \\hline &&&2x&=&8&& \\\\ &&&x&=&4&& \\end{array}[\/latex]<\/li>\n
        9. [latex]\\phantom{a}[\/latex]\n
            \n
          1. [latex]\\left(\\dfrac{x^2}{3}=27\\right)(3)\\Rightarrow x^2=81 \\Rightarrow x=\\pm 9[\/latex]<\/li>\n
          2. [latex]\\begin{array}[t]{rrl}\\\\ 27x^2+3x&=&0 \\\\ 3x(9x+1)&=&0 \\\\ x&=&0, -\\dfrac{1}{9} \\end{array}[\/latex]<\/li>\n<\/ol>\n<\/li>\n
          3. [latex]\\phantom{a}[\/latex]\n
              \n
            1. [latex]\\begin{array}[t]{rrl}\\\\ (x-12)(x+1)&=&0 \\\\ x&=&12, -1 \\end{array}[\/latex]<\/li>\n
            2. [latex]\\begin{array}[t]{rrl}\\\\ x^2+13x+12&=&0 \\\\ (x+12)(x+1)&=&0 \\\\ x&=&-1, -12 \\end{array}[\/latex]<\/li>\n<\/ol>\n<\/li>\n
            3. [latex]\\left(\\dfrac{2}{x}=\\dfrac{2x}{3x+8}\\right)(x)(3x+8)[\/latex]
              \n[latex]\\begin{array}[t]{ll} \\begin{array}[t]{rrl} 2(3x+8)&=&2x^2 \\\\ 6x+16&=&2x^2 \\\\ 0&=&2x^2-6x-16 \\\\ 0&=&2(x^2-3x-8) \\end{array} & \\hspace{0.25in} \\begin{array}[t]{l} \\dfrac{-(-3)\\pm \\sqrt{(-3)^2-4(1)(-8)}}{2(1)} \\\\ \\\\ \\dfrac{3\\pm \\sqrt{9+32}}{2} \\\\ \\\\ \\dfrac{3\\pm \\sqrt{41}}{2} \\end{array} \\end{array}[\/latex]<\/li>\n
            4. [latex]\\phantom{a}[\/latex]
              \n[latex]\\begin{array}[t]{rrl} (x^2-64)(x^2+1)&=&0 \\\\ (x-8)(x+8)(x^2+1)&=&0 \\\\ x&=&\\pm 8 \\end{array}[\/latex]<\/li>\n
            5. [latex]\\phantom{a}[\/latex]
              \n[latex]\\begin{array}[t]{rrcrrrrrrrrrr} &&&&A&=&20&+&P&&&& \\\\ \\\\ &&L(L&-&5)&=&20&+&2(L&-&5)&+&2L \\\\ L^2&-&5L&&&=&20&+&2L&-&10&+&2L \\\\ &-&4L&-&10&&-10&-&4L&&&& \\\\ \\hline L^2&-&9L&-&10&=&0&&&&&& \\\\ (L&-&10)(L&+&1)&=&0&&&&&& \\\\ &&&&L&=&10,&\\cancel{-1}&&&&& \\\\ \\\\ &&&&W&=&L&-&5&&&& \\\\ &&&&W&=&10&-&5&&&& \\\\ &&&&W&=&5&&&&&& \\end{array}[\/latex]<\/li>\n
            6. [latex]x, x+2, x+4[\/latex]
              \n[latex]\\begin{array}[t]{rrcrrrrrcrr} &&x(x&+&4)&=&35&+&10(x&+&2) \\\\ x^2&+&4x&&&=&35&+&10x&+&20 \\\\ &-&10x&-&55&&-55&-&10x&& \\\\ \\hline x^2&-&6x&-&55&=&0&&&& \\\\ (x&-&11)(x&+&5)&=&0&&&& \\\\ &&&&x&=&11,&-5&&&\\\\ \\end{array}[\/latex]
              \n[latex]\\text{numbers are }11,13,15\\text{ or }-5,-3,-1[\/latex]<\/li>\n
            7. [latex]\\phantom{a}[\/latex]
              \n[latex]\\begin{array}[t]{rrrrrrcrrr} &&&d_{\\text{d}}&=&d_{\\text{u}}&+&9&& \\\\ \\\\ &3(5&+&r)&=&4(5&-&r)&+&9 \\\\ &15&+&3r&=&20&-&4r&+&9 \\\\ -&15&+&4r&&-15&+&4r&& \\\\ \\hline &&&7r&=&14&&&& \\\\ &&&r&=&2&\\text{km\/h}&&& \\end{array}[\/latex]<\/li>\n<\/ol>\n","protected":false},"author":90,"menu_order":106,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":"cc-by-nc-sa"},"back-matter-type":[],"contributor":[],"license":[56],"class_list":["post-2014","back-matter","type-back-matter","status-publish","hentry","license-cc-by-nc-sa"],"_links":{"self":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter\/2014","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter"}],"about":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/types\/back-matter"}],"author":[{"embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/users\/90"}],"version-history":[{"count":1,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter\/2014\/revisions"}],"predecessor-version":[{"id":2015,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter\/2014\/revisions\/2015"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter\/2014\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/media?parent=2014"}],"wp:term":[{"taxonomy":"back-matter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter-type?post=2014"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/contributor?post=2014"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/license?post=2014"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}