{"id":2012,"date":"2021-12-02T19:40:28","date_gmt":"2021-12-03T00:40:28","guid":{"rendered":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/back-matter\/midterm-3-version-d\/"},"modified":"2022-11-02T10:39:07","modified_gmt":"2022-11-02T14:39:07","slug":"midterm-3-version-d","status":"publish","type":"back-matter","link":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/back-matter\/midterm-3-version-d\/","title":{"raw":"Midterm 3: Version D Answer Key","rendered":"Midterm 3: Version D Answer Key"},"content":{"raw":"
    \n \t
  1. [latex]\\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}[\/latex]<\/li>\n \t
  2. [latex]\\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)}[\/latex]<\/li>\n \t
  3. [latex]\\left(\\dfrac{2}{x-4}-\\dfrac{6}{x-3}=3\\right)(x+4)(x-3)[\/latex]\n[latex]\\begin{array}{rrrrcrrrcrcrr}\n2(x&-&3)&-&6(x&+&4)&=&3(x&+&4)(x&-&3) \\\\\n2x&-&6&-&6x&-&24&=&3(x^2&+&x&-&12) \\\\\n&&&&-4x&-&30&=&3x^2&+&3x&-&36 \\\\\n&&&&+4x&+&30&&&+&4x&+&30 \\\\\n\\hline\n&&&&&&0&=&3x^2&+&7x&-&6 \\\\\n&&&&&&0&=&(x&+&3)(3x&-&2) \\\\ \\\\\n&&&&&&x&=&-3,&\\dfrac{2}{3}&&&\n\\end{array}[\/latex]<\/li>\n \t
  4. [latex]\\begin{array}[t]{l}\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)}} \\\\ \\\\\n\\Rightarrow y(x-3y)\n\\end{array}[\/latex]<\/li>\n \t
  5. [latex]5y^2+2\\cdot 7y+\\sqrt{25y^2\\cdot y}[\/latex]\n[latex]5y^2+14y+5y\\sqrt{y}[\/latex]<\/li>\n \t
  6. [latex]\\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}[\/latex]<\/li>\n \t
  7. [latex]\\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}[\/latex]<\/li>\n \t
  8. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{ll}\n\\begin{array}[t]{rrrrl}\n\\sqrt{2x+9}&-&3&=&x \\\\\n&+&3&&\\phantom{x}+3 \\\\\n\\hline\n(\\sqrt{2x+9})^2&&&=&(x+3)^2\n\\end{array}\n& \\hspace{0.25in}\n\\begin{array}[t]{rrrrrcrrrr}\n&2x&+&9&=&x^2&+&6x&+&9 \\\\\n-&2x&-&9&&&-&2x&-&9 \\\\\n\\hline\n&&&0&=&x^2&+&4x&& \\\\\n&&&0&=&x(x&+&4)&& \\\\ \\\\\n&&&x&=&0,&-4&&&\n\\end{array}\n\\end{array}[\/latex]<\/li>\n \t
  9. [latex]\\phantom{a}[\/latex]\n
      \n \t
    1. [latex]\\begin{array}{rrl}\\\\\n8x^2-32x&=&0 \\\\\n8x(x-4)&=&0 \\\\\nx&=&0, 4\n\\end{array}[\/latex]<\/li>\n \t
    2. [latex]\\begin{array}{rrl}\\\\\n\\dfrac{3x^2}{2}&=&\\dfrac{48}{3} \\\\ \\\\\n\\sqrt{x^2}&=&\\sqrt{16} \\\\\nx&=&\\pm 4\n\\end{array}[\/latex]<\/li>\n<\/ol>\n<\/li>\n \t
    3. [latex]\\phantom{a}[\/latex]\n
        \n \t
      1. [latex]\\begin{array}{rrl}\\\\\nx^2-5x+4&=&0 \\\\\n(x-4)(x-1)&=&0 \\\\\nx&=&1, 4\n\\end{array}[\/latex]<\/li>\n \t
      2. [latex]\\begin{array}{rrl}\\\\\n(x-3)(x-1)&=&0 \\\\\nx&=&1, 3\n\\end{array}[\/latex]<\/li>\n<\/ol>\n<\/li>\n \t
      3. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrrrcrcrcrr}\n2(x&+&4)&=&x(x)&&&&&& \\\\\n2x&+&8&=&x^2&&&&&& \\\\ \\\\\n&&0&=&x^2&-&2x&-&8&& \\\\\n&&0&=&x^2&-&4x&+&2x&-&8 \\\\\n&&0&=&x(x&-&4)&+&2(x&-&4) \\\\\n&&0&=&(x&-&4)(x&+&2)&& \\\\ \\\\\n&&x&=&-2,&4&&&&&\n\\end{array}[\/latex]<\/li>\n \t
      4. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrl}\n\\text{Let }u&=&x^2 \\\\ \\\\\nu^2-48u-49&=&0 \\\\\n(u-49)(u+1)&=&0 \\\\ \\\\\n(x^2-49)(x^2+1)&=&0 \\\\\n(x-7)(x+7)(x^2+1)&=&0 \\\\ \\\\\nx^2+1&=&\\text{cannot be factored} \\\\\nx&=&\\pm 7\n\\end{array}[\/latex]<\/li>\n \t
      5. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrl}\n40&=&\\dfrac{1}{2}(h-2)(h) \\\\ \\\\\n80&=&h^2-2h \\\\\n0&=&h^2-2h-80 \\\\\n0&=&(h-10)(h+8) \\\\\nh&=&10, \\cancel{-8} \\\\ \\\\\nb&=&10-2=8\n\\end{array}[\/latex]<\/li>\n \t
      6. [latex]x, x+2, x+4[\/latex]\n[latex]\\begin{array}{crrrrrcrr}\nx(x&+&4)&=&41&+&4(x&+&2) \\\\\nx^2&+&4x&=&41&+&4x&+&8 \\\\\n&-&4x&&&-&4x&& \\\\\n\\hline\n&&\\sqrt{x^2}&=&\\sqrt{49}&&&& \\\\\n&&x&=&\\pm 7&&&& \\\\\n\\end{array}[\/latex]\n[latex]\\text{numbers are }7, 9, 11\\text{ or }-7,-5,-3[\/latex]<\/li>\n \t
      7. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrrrrrcrrr}\n&&&d_{\\text{d}}&=&d_{\\text{u}}&+&4&& \\\\ \\\\\n&2(6&+&r)&=&3(6&-&r)&+&4 \\\\\n&12&+&2r&=&18&-&3r&+&4 \\\\\n-&12&+&3r&&-12&+&3r&& \\\\\n\\hline\n&&&\\dfrac{5r}{5}&=&\\dfrac{10}{5}&&&& \\\\ \\\\\n&&&r&=&2&\\text{km\/h}&&&\n\\end{array}[\/latex]<\/li>\n<\/ol>","rendered":"
          \n
        1. [latex]\\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}[\/latex]<\/li>\n
        2. [latex]\\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)}[\/latex]<\/li>\n
        3. [latex]\\left(\\dfrac{2}{x-4}-\\dfrac{6}{x-3}=3\\right)(x+4)(x-3)[\/latex]
          \n[latex]\\begin{array}{rrrrcrrrcrcrr} 2(x&-&3)&-&6(x&+&4)&=&3(x&+&4)(x&-&3) \\\\ 2x&-&6&-&6x&-&24&=&3(x^2&+&x&-&12) \\\\ &&&&-4x&-&30&=&3x^2&+&3x&-&36 \\\\ &&&&+4x&+&30&&&+&4x&+&30 \\\\ \\hline &&&&&&0&=&3x^2&+&7x&-&6 \\\\ &&&&&&0&=&(x&+&3)(3x&-&2) \\\\ \\\\ &&&&&&x&=&-3,&\\dfrac{2}{3}&&& \\end{array}[\/latex]<\/li>\n
        4. [latex]\\begin{array}[t]{l} \\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)}} \\\\ \\\\ \\Rightarrow y(x-3y) \\end{array}[\/latex]<\/li>\n
        5. [latex]5y^2+2\\cdot 7y+\\sqrt{25y^2\\cdot y}[\/latex]
          \n[latex]5y^2+14y+5y\\sqrt{y}[\/latex]<\/li>\n
        6. [latex]\\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}[\/latex]<\/li>\n
        7. [latex]\\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}[\/latex]<\/li>\n
        8. [latex]\\phantom{a}[\/latex]
          \n[latex]\\begin{array}[t]{ll} \\begin{array}[t]{rrrrl} \\sqrt{2x+9}&-&3&=&x \\\\ &+&3&&\\phantom{x}+3 \\\\ \\hline (\\sqrt{2x+9})^2&&&=&(x+3)^2 \\end{array} & \\hspace{0.25in} \\begin{array}[t]{rrrrrcrrrr} &2x&+&9&=&x^2&+&6x&+&9 \\\\ -&2x&-&9&&&-&2x&-&9 \\\\ \\hline &&&0&=&x^2&+&4x&& \\\\ &&&0&=&x(x&+&4)&& \\\\ \\\\ &&&x&=&0,&-4&&& \\end{array} \\end{array}[\/latex]<\/li>\n
        9. [latex]\\phantom{a}[\/latex]\n
            \n
          1. [latex]\\begin{array}{rrl}\\\\ 8x^2-32x&=&0 \\\\ 8x(x-4)&=&0 \\\\ x&=&0, 4 \\end{array}[\/latex]<\/li>\n
          2. [latex]\\begin{array}{rrl}\\\\ \\dfrac{3x^2}{2}&=&\\dfrac{48}{3} \\\\ \\\\ \\sqrt{x^2}&=&\\sqrt{16} \\\\ x&=&\\pm 4 \\end{array}[\/latex]<\/li>\n<\/ol>\n<\/li>\n
          3. [latex]\\phantom{a}[\/latex]\n
              \n
            1. [latex]\\begin{array}{rrl}\\\\ x^2-5x+4&=&0 \\\\ (x-4)(x-1)&=&0 \\\\ x&=&1, 4 \\end{array}[\/latex]<\/li>\n
            2. [latex]\\begin{array}{rrl}\\\\ (x-3)(x-1)&=&0 \\\\ x&=&1, 3 \\end{array}[\/latex]<\/li>\n<\/ol>\n<\/li>\n
            3. [latex]\\phantom{a}[\/latex]
              \n[latex]\\begin{array}[t]{rrrrcrcrcrr} 2(x&+&4)&=&x(x)&&&&&& \\\\ 2x&+&8&=&x^2&&&&&& \\\\ \\\\ &&0&=&x^2&-&2x&-&8&& \\\\ &&0&=&x^2&-&4x&+&2x&-&8 \\\\ &&0&=&x(x&-&4)&+&2(x&-&4) \\\\ &&0&=&(x&-&4)(x&+&2)&& \\\\ \\\\ &&x&=&-2,&4&&&&& \\end{array}[\/latex]<\/li>\n
            4. [latex]\\phantom{a}[\/latex]
              \n[latex]\\begin{array}[t]{rrl} \\text{Let }u&=&x^2 \\\\ \\\\ u^2-48u-49&=&0 \\\\ (u-49)(u+1)&=&0 \\\\ \\\\ (x^2-49)(x^2+1)&=&0 \\\\ (x-7)(x+7)(x^2+1)&=&0 \\\\ \\\\ x^2+1&=&\\text{cannot be factored} \\\\ x&=&\\pm 7 \\end{array}[\/latex]<\/li>\n
            5. [latex]\\phantom{a}[\/latex]
              \n[latex]\\begin{array}[t]{rrl} 40&=&\\dfrac{1}{2}(h-2)(h) \\\\ \\\\ 80&=&h^2-2h \\\\ 0&=&h^2-2h-80 \\\\ 0&=&(h-10)(h+8) \\\\ h&=&10, \\cancel{-8} \\\\ \\\\ b&=&10-2=8 \\end{array}[\/latex]<\/li>\n
            6. [latex]x, x+2, x+4[\/latex]
              \n[latex]\\begin{array}{crrrrrcrr} x(x&+&4)&=&41&+&4(x&+&2) \\\\ x^2&+&4x&=&41&+&4x&+&8 \\\\ &-&4x&&&-&4x&& \\\\ \\hline &&\\sqrt{x^2}&=&\\sqrt{49}&&&& \\\\ &&x&=&\\pm 7&&&& \\\\ \\end{array}[\/latex]
              \n[latex]\\text{numbers are }7, 9, 11\\text{ or }-7,-5,-3[\/latex]<\/li>\n
            7. [latex]\\phantom{a}[\/latex]
              \n[latex]\\begin{array}[t]{rrrrrrcrrr} &&&d_{\\text{d}}&=&d_{\\text{u}}&+&4&& \\\\ \\\\ &2(6&+&r)&=&3(6&-&r)&+&4 \\\\ &12&+&2r&=&18&-&3r&+&4 \\\\ -&12&+&3r&&-12&+&3r&& \\\\ \\hline &&&\\dfrac{5r}{5}&=&\\dfrac{10}{5}&&&& \\\\ \\\\ &&&r&=&2&\\text{km\/h}&&& \\end{array}[\/latex]<\/li>\n<\/ol>\n","protected":false},"author":90,"menu_order":105,"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-2012","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\/2012","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\/2012\/revisions"}],"predecessor-version":[{"id":2013,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter\/2012\/revisions\/2013"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter\/2012\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/media?parent=2012"}],"wp:term":[{"taxonomy":"back-matter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter-type?post=2012"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/contributor?post=2012"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/license?post=2012"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}