{"id":2008,"date":"2021-12-02T19:40:26","date_gmt":"2021-12-03T00:40:26","guid":{"rendered":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/back-matter\/midterm-3-version-b\/"},"modified":"2022-11-02T10:39:05","modified_gmt":"2022-11-02T14:39:05","slug":"midterm-3-version-b","status":"publish","type":"back-matter","link":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/back-matter\/midterm-3-version-b\/","title":{"raw":"Midterm 3: Version B Answer Key","rendered":"Midterm 3: Version B Answer Key"},"content":{"raw":"
    \n \t
  1. [latex]\\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}[\/latex]<\/li>\n \t
  2. [latex]\\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)}[\/latex]<\/li>\n \t
  3. [latex]\\left(\\dfrac{5x}{x+3}-\\dfrac{5x}{x-3}+\\dfrac{90}{x^2-9}\\right)(x-3)(x+3)[\/latex]\n[latex]\\begin{array}[t]{l}\n\\Rightarrow \\dfrac{5x(x-3)-5x(x+3)+90}{(x+3)(x-3)} \\\\ \\\\\n\\Rightarrow \\dfrac{5x^2-15x-5x^2-15x+90}{(x+3)(x-3)} \\\\ \\\\\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}\n\\end{array}[\/latex]<\/li>\n \t
  4. [latex]\\dfrac{(\\dfrac{9{a}^{2}}{{b}^{2}}-25)(b^2)}{(\\dfrac{3a}{b}+5)(b^2)} \\Rightarrow \\dfrac{9a^2-25b^2}{3ab+5b^2}[\/latex]\n[latex]\\Rightarrow \\dfrac{(3a-5b)\\cancel{(3a+5b)}}{b\\cancel{(3a+5b)}}[\/latex]\n[latex]\\Rightarrow \\dfrac{3a-5b}{b}[\/latex]<\/li>\n \t
  5. [latex]\\sqrt{2\\cdot 36\\cdot d^2\\cdot d}+4\\sqrt{2\\cdot 9\\cdot d^2\\cdot d}-2(7d^2)[\/latex]\n[latex]\\begin{array}[t]{l}\n6d\\sqrt{2d}+4\\cdot 3d\\sqrt{2d}-14d^2 \\\\\n6d\\sqrt{2d}+12d\\sqrt{2d}-14d^2 \\\\\n18d\\sqrt{2d}-14d^2\n\\end{array}[\/latex]<\/li>\n \t
  6. [latex]\\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}[\/latex]<\/li>\n \t
  7. [latex]\\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}[\/latex]<\/li>\n \t
  8. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrl}\n(\\sqrt{4x+12})^2&=&(x)^2 \\\\\n4x+12&=&x^2 \\\\\n0&=&x^2-4x-12 \\\\\n0&=&(x-6)(x+2) \\\\ \\\\\nx&=&6, \\cancel{-2}\n\\end{array}[\/latex]<\/li>\n \t
  9. [latex]\\phantom{a}[\/latex]\n
      \n \t
    1. [latex]\\begin{array}{rrl}\\\\ \\\\\n\\dfrac{2x^2}{2}&=&\\dfrac{98}{2} \\\\ \\\\\nx^2&=&49 \\\\\nx&=& \\pm 7\n\\end{array}[\/latex]<\/li>\n \t
    2. [latex]\\begin{array}{rrl}\\\\ \\\\\n4x^2-12x&=&0 \\\\\n4x(x-3)&=&0 \\\\\nx&=&0, 3\n\\end{array}[\/latex]<\/li>\n<\/ol>\n<\/li>\n \t
    3. [latex]\\phantom{a}[\/latex]\n
        \n \t
      1. [latex]\\begin{array}{rrl}\\\\\n(x-5)(x+4)&=&0 \\\\\nx&=&5, -4\n\\end{array}[\/latex]<\/li>\n \t
      2. [latex]\\begin{array}{rrl}\\\\\nx^2-2x-35&=&0 \\\\\n(x-7)(x+5)&=&0 \\\\\nx&=&7, -5\n\\end{array}[\/latex]<\/li>\n<\/ol>\n<\/li>\n \t
      3. [latex]\\left(\\dfrac{x-3}{x+2}+\\dfrac{6}{x+3}=1\\right)(x+2)(x+3)[\/latex]\n[latex]\\begin{array}{rcccrcrrrrrcrr}\n&(x-3)&\\cdot &(x+3)&+&6(x&+&2)&=&(x&+&2)(x&+&3) \\\\\n&x^2&-&9&+&6x&+&12&=&x^2&+&5x&+&6 \\\\\n-&x^2&+&9&-&6x&-&12&&-x^2&-&6x&-&12 \\\\\n\\hline\n&&&&&&&0&=&-x&+&3&& \\\\\n&&&&&&+&x&&+x&&&& \\\\\n\\hline\n&&&&&&&x&=&3&&&&\n\\end{array}[\/latex]<\/li>\n \t
      4. [latex]\\text{Let }u=x^2[\/latex]\n[latex]\\begin{array}[t]{rrl}\nu^2-5u+4&=&0 \\\\\n(u-4)(u-1)&=&0 \\\\ \\\\\n(x^2-4)(x^2-1)&=&0 \\\\\n(x-2)(x+2)(x-1)(x+1)&=&0 \\\\\nx&=&\\pm 2, \\pm 1\n\\end{array}[\/latex]<\/li>\n \t
      5. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrrrrrr}\nL&=&3&+&W&& \\\\ \\\\\nP&=&2L&+&2W&& \\\\\n46&=&2(3&+&W)&+&2W \\\\\n46&=&6&+&2W&+&2W \\\\\n-6&&-6&&&& \\\\\n\\hline\n40&=&4W&&&& \\\\ \\\\\nW&=&\\dfrac{40}{4}&=&10&& \\\\ \\\\\n\\therefore L&=&W&+&3&& \\\\\nL&=&10&+&3&=&13\n\\end{array}[\/latex]<\/li>\n \t
      6. [latex]x, x+2, x+4[\/latex]\n[latex]\\begin{array}{rrcrrrrrcrr}\n&&x(x&+&2)&=&16&+&x&+&4 \\\\\nx^2&+&2x&&&=&20&+&x&& \\\\\n&-&x&-&20&&-20&-&x&& \\\\\n\\hline\nx^2&+&x&-&20&=&0&&&& \\\\ \\\\\n&&&&0&=&(x&+&5)(x&-&4) \\\\\n&&&&x&=&\\cancel{-5},&4&&& \\\\\n\\end{array}[\/latex]\n[latex]\\therefore[\/latex] 4, 6, 8<\/li>\n \t
      7. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrl}\nd&=&r\\cdot t \\\\\nd_{\\text{up}}&=&d_{\\text{return}} \\\\ \\\\\n4(r-5)&=&2(r+5) \\\\\n4r-20&=&\\phantom{-}2r+10 \\\\\n-2r+20&&-2r+20 \\\\\n\\hline\n\\dfrac{2r}{2}&=&\\dfrac{30}{2} \\\\ \\\\\nr&=&15\n\\end{array}[\/latex]<\/li>\n<\/ol>","rendered":"
          \n
        1. [latex]\\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}[\/latex]<\/li>\n
        2. [latex]\\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)}[\/latex]<\/li>\n
        3. [latex]\\left(\\dfrac{5x}{x+3}-\\dfrac{5x}{x-3}+\\dfrac{90}{x^2-9}\\right)(x-3)(x+3)[\/latex]
          \n[latex]\\begin{array}[t]{l} \\Rightarrow \\dfrac{5x(x-3)-5x(x+3)+90}{(x+3)(x-3)} \\\\ \\\\ \\Rightarrow \\dfrac{5x^2-15x-5x^2-15x+90}{(x+3)(x-3)} \\\\ \\\\ \\Rightarrow \\dfrac{-30x+90}{(x+3)(x-3)}\\Rightarrow \\dfrac{-30\\cancel{(x-3)}}{(x+3)\\cancel{(x-3)}}\\Rightarrow \\dfrac{-30}{x+3} \\end{array}[\/latex]<\/li>\n
        4. [latex]\\dfrac{(\\dfrac{9{a}^{2}}{{b}^{2}}-25)(b^2)}{(\\dfrac{3a}{b}+5)(b^2)} \\Rightarrow \\dfrac{9a^2-25b^2}{3ab+5b^2}[\/latex]
          \n[latex]\\Rightarrow \\dfrac{(3a-5b)\\cancel{(3a+5b)}}{b\\cancel{(3a+5b)}}[\/latex]
          \n[latex]\\Rightarrow \\dfrac{3a-5b}{b}[\/latex]<\/li>\n
        5. [latex]\\sqrt{2\\cdot 36\\cdot d^2\\cdot d}+4\\sqrt{2\\cdot 9\\cdot d^2\\cdot d}-2(7d^2)[\/latex]
          \n[latex]\\begin{array}[t]{l} 6d\\sqrt{2d}+4\\cdot 3d\\sqrt{2d}-14d^2 \\\\ 6d\\sqrt{2d}+12d\\sqrt{2d}-14d^2 \\\\ 18d\\sqrt{2d}-14d^2 \\end{array}[\/latex]<\/li>\n
        6. [latex]\\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}[\/latex]<\/li>\n
        7. [latex]\\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}[\/latex]<\/li>\n
        8. [latex]\\phantom{a}[\/latex]
          \n[latex]\\begin{array}[t]{rrl} (\\sqrt{4x+12})^2&=&(x)^2 \\\\ 4x+12&=&x^2 \\\\ 0&=&x^2-4x-12 \\\\ 0&=&(x-6)(x+2) \\\\ \\\\ x&=&6, \\cancel{-2} \\end{array}[\/latex]<\/li>\n
        9. [latex]\\phantom{a}[\/latex]\n
            \n
          1. [latex]\\begin{array}{rrl}\\\\ \\\\ \\dfrac{2x^2}{2}&=&\\dfrac{98}{2} \\\\ \\\\ x^2&=&49 \\\\ x&=& \\pm 7 \\end{array}[\/latex]<\/li>\n
          2. [latex]\\begin{array}{rrl}\\\\ \\\\ 4x^2-12x&=&0 \\\\ 4x(x-3)&=&0 \\\\ x&=&0, 3 \\end{array}[\/latex]<\/li>\n<\/ol>\n<\/li>\n
          3. [latex]\\phantom{a}[\/latex]\n
              \n
            1. [latex]\\begin{array}{rrl}\\\\ (x-5)(x+4)&=&0 \\\\ x&=&5, -4 \\end{array}[\/latex]<\/li>\n
            2. [latex]\\begin{array}{rrl}\\\\ x^2-2x-35&=&0 \\\\ (x-7)(x+5)&=&0 \\\\ x&=&7, -5 \\end{array}[\/latex]<\/li>\n<\/ol>\n<\/li>\n
            3. [latex]\\left(\\dfrac{x-3}{x+2}+\\dfrac{6}{x+3}=1\\right)(x+2)(x+3)[\/latex]
              \n[latex]\\begin{array}{rcccrcrrrrrcrr} &(x-3)&\\cdot &(x+3)&+&6(x&+&2)&=&(x&+&2)(x&+&3) \\\\ &x^2&-&9&+&6x&+&12&=&x^2&+&5x&+&6 \\\\ -&x^2&+&9&-&6x&-&12&&-x^2&-&6x&-&12 \\\\ \\hline &&&&&&&0&=&-x&+&3&& \\\\ &&&&&&+&x&&+x&&&& \\\\ \\hline &&&&&&&x&=&3&&&& \\end{array}[\/latex]<\/li>\n
            4. [latex]\\text{Let }u=x^2[\/latex]
              \n[latex]\\begin{array}[t]{rrl} u^2-5u+4&=&0 \\\\ (u-4)(u-1)&=&0 \\\\ \\\\ (x^2-4)(x^2-1)&=&0 \\\\ (x-2)(x+2)(x-1)(x+1)&=&0 \\\\ x&=&\\pm 2, \\pm 1 \\end{array}[\/latex]<\/li>\n
            5. [latex]\\phantom{a}[\/latex]
              \n[latex]\\begin{array}[t]{rrrrrrr} L&=&3&+&W&& \\\\ \\\\ P&=&2L&+&2W&& \\\\ 46&=&2(3&+&W)&+&2W \\\\ 46&=&6&+&2W&+&2W \\\\ -6&&-6&&&& \\\\ \\hline 40&=&4W&&&& \\\\ \\\\ W&=&\\dfrac{40}{4}&=&10&& \\\\ \\\\ \\therefore L&=&W&+&3&& \\\\ L&=&10&+&3&=&13 \\end{array}[\/latex]<\/li>\n
            6. [latex]x, x+2, x+4[\/latex]
              \n[latex]\\begin{array}{rrcrrrrrcrr} &&x(x&+&2)&=&16&+&x&+&4 \\\\ x^2&+&2x&&&=&20&+&x&& \\\\ &-&x&-&20&&-20&-&x&& \\\\ \\hline x^2&+&x&-&20&=&0&&&& \\\\ \\\\ &&&&0&=&(x&+&5)(x&-&4) \\\\ &&&&x&=&\\cancel{-5},&4&&& \\\\ \\end{array}[\/latex]
              \n[latex]\\therefore[\/latex] 4, 6, 8<\/li>\n
            7. [latex]\\phantom{a}[\/latex]
              \n[latex]\\begin{array}[t]{rrl} d&=&r\\cdot t \\\\ d_{\\text{up}}&=&d_{\\text{return}} \\\\ \\\\ 4(r-5)&=&2(r+5) \\\\ 4r-20&=&\\phantom{-}2r+10 \\\\ -2r+20&&-2r+20 \\\\ \\hline \\dfrac{2r}{2}&=&\\dfrac{30}{2} \\\\ \\\\ r&=&15 \\end{array}[\/latex]<\/li>\n<\/ol>\n","protected":false},"author":90,"menu_order":103,"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-2008","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\/2008","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\/2008\/revisions"}],"predecessor-version":[{"id":2009,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter\/2008\/revisions\/2009"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter\/2008\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/media?parent=2008"}],"wp:term":[{"taxonomy":"back-matter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter-type?post=2008"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/contributor?post=2008"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/license?post=2008"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}