{"id":2006,"date":"2021-12-02T19:40:26","date_gmt":"2021-12-03T00:40:26","guid":{"rendered":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/back-matter\/midterm-3-version-a-answer-key\/"},"modified":"2022-11-02T10:39:04","modified_gmt":"2022-11-02T14:39:04","slug":"midterm-3-version-a-answer-key","status":"publish","type":"back-matter","link":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/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":"
    \n \t
  1. [latex]\\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}[\/latex]<\/li>\n \t
  2. [latex]\\dfrac{8x-8y}{x^3+y^3}\\cdot \\dfrac{x^2-xy+y^2}{x^2-y^2}[\/latex]\n[latex]\\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}[\/latex]<\/li>\n \t
  3. [latex]\\text{LCD}=6(n-3)[\/latex]\n[latex]\\begin{array}[t]{l}\n\\dfrac{5(n-3)-2\\cdot 6(n-3)-5\\cdot 6}{6(n-3)} \\\\ \\\\\n\\dfrac{5n-15-12n+36-30}{6(n-3)} \\\\ \\\\\n\\dfrac{-7n-9}{6(n-3)}\n\\end{array}[\/latex]<\/li>\n \t
  4. [latex]\\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)}}[\/latex]\n[latex]\\Rightarrow y(x-2y)[\/latex]<\/li>\n \t
  5. [latex]3\\cdot 5+2\\sqrt{36\\cdot 2}-4 [\/latex]\n[latex]15+2\\cdot 6\\sqrt{2}-4[\/latex]\n[latex]11+12\\sqrt{2}[\/latex]<\/li>\n \t
  6. [latex]\\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}[\/latex]<\/li>\n \t
  7. [latex]\\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}[\/latex]\n[latex]\\Rightarrow \\dfrac{2+2\\sqrt{3}-x-x\\sqrt{3}}{-2}\\text{ or }\\dfrac{x+x\\sqrt{3}-2-2\\sqrt{3}}{2}[\/latex]<\/li>\n \t
  8. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrl}\n(\\sqrt{7x+8})^2&=&(x)^2 \\\\\n7x+8&=&x^2 \\\\\n0&=&x^2-7x-8 \\\\\n0&=&(x-8)(x+1) \\\\ \\\\\nx&=&8, \\cancel{-1}\n\\end{array}[\/latex]<\/li>\n \t
  9. [latex]\\phantom{a}[\/latex]\n
      \n \t
    1. [latex]\\begin{array}{rrl}\\\\\n\\dfrac{4x^2}{4}&=&\\dfrac{64}{4}\nx^2&=&16 \\\\\nx&=&\\pm 4\n\\end{array}[\/latex]<\/li>\n \t
    2. [latex]\\begin{array}{rrl}\\\\ \\\\\n3x^2-12x&=&0 \\\\\n3x(x-4)&=&0 \\\\ \\\\\nx&=&0,4\n\\end{array}[\/latex]<\/li>\n<\/ol>\n<\/li>\n \t
    3. [latex]\\phantom{a}[\/latex]\n
        \n \t
      1. [latex]\\begin{array}{rrl}\\\\\n(x-5)(x-1)&=&0 \\\\\nx&=&5, 1\n\\end{array}[\/latex]<\/li>\n \t
      2. [latex]\\begin{array}{rrl}\\\\ \\\\\nx^2+10x+9&=&0 \\\\\n(x+9)(x+1)&=&0 \\\\\nx&=&-9, -1\n\\end{array}[\/latex]<\/li>\n<\/ol>\n<\/li>\n \t
      3. [latex]\\left(\\dfrac{x+4}{-4}=\\dfrac{8}{x}\\right)(-4)(x)[\/latex]\n[latex]\\begin{array}{rrl}\nx(x+4)&=&-4(8) \\\\\nx^2+4x&=&-32 \\\\\n0&=&x^2+4x+32 \\hspace{0.5in} \\text{Does not factor}\n\\end{array}[\/latex]<\/li>\n \t
      4. [latex]\\text{Let }u=x^2[\/latex]\n[latex]\\begin{array}[t]{rrl}\nu^2-13u+36&=&0 \\\\\nu^2-4u-9u+36&=&0 \\\\\nu(u-4)-9(u-4)&=&0 \\\\\n(u-4)(u-9)&=&0 \\\\ \\\\\n(x^2-4)(x^2-9)&=&0 \\\\\n(x-2)(x+2)(x-3)(x+3)&=&0 \\\\\nx&=& \\pm 2, \\pm 3\n\\end{array}[\/latex]<\/li>\n \t
      5. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrl}\nA&=&\\dfrac{1}{2}bh \\\\ \\\\\n300&=&\\dfrac{1}{2}(h+10)h \\\\ \\\\\n600&=&h^2+10h \\\\\n0&=&h^2+10h-600 \\\\\n0&=&(h-20)(h+30) \\\\ \\\\\nh&=& 20, \\cancel{-30} \\\\ \\\\\n\\therefore b&=&h+10=30\n\\end{array}[\/latex]<\/li>\n \t
      6. [latex]x, x+2, x+4[\/latex]\n[latex]\\begin{array}[t]{rrrrrrrrrrr}\n&&x(x&+&4)&=&38&+&x&+&2 \\\\\nx^2&+&4x&&&=&x&+&40&& \\\\\n&-&x&-&40&&-x&-&40&& \\\\\n\\hline\nx^2&+&3x&-&40&=&0&&&& \\\\ \\\\\n&&&&0&=&(x&+&8)(x&-&5) \\\\\n&&&&x&=&\\cancel{-8},&5&&& \\\\\n\\end{array}[\/latex]\n[latex]\\therefore[\/latex] 5, 7, 9<\/li>\n \t
      7. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrl}\nr_st_s&=&r_ft_f \\\\ \\\\\nr(4.5\\text{ h})&=&(r+150)(3.0\\text{ h}) \\\\\n4.5r&=&\\phantom{-}3.0r+450 \\\\\n-3.0r&&-3.0r \\\\\n\\hline\n1.5r&=&450 \\\\ \\\\\nr&=&\\dfrac{450}{1.5}\\text{ or }300\\text{ km\/h} \\\\ \\\\\nr_f&=&300+150 \\\\\nr_f&=&450\\text{ km\/h}\n\\end{array}[\/latex]<\/li>\n<\/ol>","rendered":"
          \n
        1. [latex]\\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}[\/latex]<\/li>\n
        2. [latex]\\dfrac{8x-8y}{x^3+y^3}\\cdot \\dfrac{x^2-xy+y^2}{x^2-y^2}[\/latex]
          \n[latex]\\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}[\/latex]<\/li>\n
        3. [latex]\\text{LCD}=6(n-3)[\/latex]
          \n[latex]\\begin{array}[t]{l} \\dfrac{5(n-3)-2\\cdot 6(n-3)-5\\cdot 6}{6(n-3)} \\\\ \\\\ \\dfrac{5n-15-12n+36-30}{6(n-3)} \\\\ \\\\ \\dfrac{-7n-9}{6(n-3)} \\end{array}[\/latex]<\/li>\n
        4. [latex]\\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)}}[\/latex]
          \n[latex]\\Rightarrow y(x-2y)[\/latex]<\/li>\n
        5. [latex]3\\cdot 5+2\\sqrt{36\\cdot 2}-4[\/latex]
          \n[latex]15+2\\cdot 6\\sqrt{2}-4[\/latex]
          \n[latex]11+12\\sqrt{2}[\/latex]<\/li>\n
        6. [latex]\\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}[\/latex]<\/li>\n
        7. [latex]\\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}[\/latex]
          \n[latex]\\Rightarrow \\dfrac{2+2\\sqrt{3}-x-x\\sqrt{3}}{-2}\\text{ or }\\dfrac{x+x\\sqrt{3}-2-2\\sqrt{3}}{2}[\/latex]<\/li>\n
        8. [latex]\\phantom{a}[\/latex]
          \n[latex]\\begin{array}[t]{rrl} (\\sqrt{7x+8})^2&=&(x)^2 \\\\ 7x+8&=&x^2 \\\\ 0&=&x^2-7x-8 \\\\ 0&=&(x-8)(x+1) \\\\ \\\\ x&=&8, \\cancel{-1} \\end{array}[\/latex]<\/li>\n
        9. [latex]\\phantom{a}[\/latex]\n
            \n
          1. [latex]\\begin{array}{rrl}\\\\ \\dfrac{4x^2}{4}&=&\\dfrac{64}{4} x^2&=&16 \\\\ x&=&\\pm 4 \\end{array}[\/latex]<\/li>\n
          2. [latex]\\begin{array}{rrl}\\\\ \\\\ 3x^2-12x&=&0 \\\\ 3x(x-4)&=&0 \\\\ \\\\ x&=&0,4 \\end{array}[\/latex]<\/li>\n<\/ol>\n<\/li>\n
          3. [latex]\\phantom{a}[\/latex]\n
              \n
            1. [latex]\\begin{array}{rrl}\\\\ (x-5)(x-1)&=&0 \\\\ x&=&5, 1 \\end{array}[\/latex]<\/li>\n
            2. [latex]\\begin{array}{rrl}\\\\ \\\\ x^2+10x+9&=&0 \\\\ (x+9)(x+1)&=&0 \\\\ x&=&-9, -1 \\end{array}[\/latex]<\/li>\n<\/ol>\n<\/li>\n
            3. [latex]\\left(\\dfrac{x+4}{-4}=\\dfrac{8}{x}\\right)(-4)(x)[\/latex]
              \n[latex]\\begin{array}{rrl} x(x+4)&=&-4(8) \\\\ x^2+4x&=&-32 \\\\ 0&=&x^2+4x+32 \\hspace{0.5in} \\text{Does not factor} \\end{array}[\/latex]<\/li>\n
            4. [latex]\\text{Let }u=x^2[\/latex]
              \n[latex]\\begin{array}[t]{rrl} u^2-13u+36&=&0 \\\\ u^2-4u-9u+36&=&0 \\\\ u(u-4)-9(u-4)&=&0 \\\\ (u-4)(u-9)&=&0 \\\\ \\\\ (x^2-4)(x^2-9)&=&0 \\\\ (x-2)(x+2)(x-3)(x+3)&=&0 \\\\ x&=& \\pm 2, \\pm 3 \\end{array}[\/latex]<\/li>\n
            5. [latex]\\phantom{a}[\/latex]
              \n[latex]\\begin{array}[t]{rrl} A&=&\\dfrac{1}{2}bh \\\\ \\\\ 300&=&\\dfrac{1}{2}(h+10)h \\\\ \\\\ 600&=&h^2+10h \\\\ 0&=&h^2+10h-600 \\\\ 0&=&(h-20)(h+30) \\\\ \\\\ h&=& 20, \\cancel{-30} \\\\ \\\\ \\therefore b&=&h+10=30 \\end{array}[\/latex]<\/li>\n
            6. [latex]x, x+2, x+4[\/latex]
              \n[latex]\\begin{array}[t]{rrrrrrrrrrr} &&x(x&+&4)&=&38&+&x&+&2 \\\\ x^2&+&4x&&&=&x&+&40&& \\\\ &-&x&-&40&&-x&-&40&& \\\\ \\hline x^2&+&3x&-&40&=&0&&&& \\\\ \\\\ &&&&0&=&(x&+&8)(x&-&5) \\\\ &&&&x&=&\\cancel{-8},&5&&& \\\\ \\end{array}[\/latex]
              \n[latex]\\therefore[\/latex] 5, 7, 9<\/li>\n
            7. [latex]\\phantom{a}[\/latex]
              \n[latex]\\begin{array}[t]{rrl} r_st_s&=&r_ft_f \\\\ \\\\ r(4.5\\text{ h})&=&(r+150)(3.0\\text{ h}) \\\\ 4.5r&=&\\phantom{-}3.0r+450 \\\\ -3.0r&&-3.0r \\\\ \\hline 1.5r&=&450 \\\\ \\\\ r&=&\\dfrac{450}{1.5}\\text{ or }300\\text{ km\/h} \\\\ \\\\ r_f&=&300+150 \\\\ r_f&=&450\\text{ km\/h} \\end{array}[\/latex]<\/li>\n<\/ol>\n","protected":false},"author":90,"menu_order":102,"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-2006","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\/2006","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\/2006\/revisions"}],"predecessor-version":[{"id":2007,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter\/2006\/revisions\/2007"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter\/2006\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/media?parent=2006"}],"wp:term":[{"taxonomy":"back-matter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter-type?post=2006"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/contributor?post=2006"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/license?post=2006"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}