{"id":2177,"date":"2019-09-26T12:41:14","date_gmt":"2019-09-26T16:41:14","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/?post_type=back-matter&p=2177"},"modified":"2019-11-19T15:14:51","modified_gmt":"2019-11-19T20:14:51","slug":"midterm-3-prep-answer-key","status":"publish","type":"back-matter","link":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/back-matter\/midterm-3-prep-answer-key\/","title":{"raw":"Midterm 3 Prep Answer Key","rendered":"Midterm 3 Prep Answer Key"},"content":{"raw":"[latexpage]\r\n

Midterm Three Review<\/h1>\r\n
    \r\n \t
  1. \r\n
      \r\n \t
    1. \\(\\dfrac{6\\cancel{(a-b)}}{(a+b)\\cancel{(a^2-ab+b^2)}}\\cdot \\dfrac{\\cancel{a^2-ab+b^2}}{(a+b)\\cancel{(a-b)}}\\Rightarrow \\dfrac{b}{(a+b)^2}\\)<\/li>\r\n \t
    2. \\(\\begin{array}{l}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n\\dfrac{x}{(x+5)(x-5)}-\\dfrac{2}{(x-5)(x-1)} \\\\ \\\\\r\n\\text{LCD}=(x+5)(x-5)(x-1) \\\\ \\\\\r\n\\therefore \\dfrac{x(x-1)-2(x+5)}{(x+5)(x-5)(x-1)}\\Rightarrow \\dfrac{x^2-x-2x-10}{(x+5)(x-5)(x-1)}\\Rightarrow \\dfrac{x^2-3x-10}{(x+5)(x-5)(x-1)} \\\\ \\\\\r\n\\Rightarrow \\dfrac{\\cancel{(x-5)}(x+2)}{(x+5)\\cancel{(x-5)}(x-1)}\\Rightarrow \\dfrac{x+2}{(x+5)(x-1)}\r\n\\end{array}\\)<\/li>\r\n \t
    3. \\(\\dfrac{\\left(1-\\dfrac{6}{x}\\right)x^2}{\\left(\\dfrac{4}{x}-\\dfrac{24}{x^2}\\right)x^2}\\Rightarrow \\dfrac{x^2-6x}{4x-24}\\Rightarrow \\dfrac{x\\cancel{(x-6)}}{4\\cancel{(x-6)}}\\Rightarrow \\dfrac{x}{4}\\)<\/li>\r\n \t
    4. \\(\\phantom{1}\\)\r\n\\(\\left(\\dfrac{4}{x+4}-\\dfrac{5}{x-2}=5\\right)(x+4)(x-2) \\\\ \\)\r\n\\(\\begin{array}{rrrrrrrrrrcrr}\r\n4(x&-&2)&-&5(x&+&4)&=&5(x&+&4)(x&-&2) \\\\\r\n4x&-&8&-&5x&-&20&=&5(x^2&+&2x&-&8) \\\\\r\n&&&&-x&-&28&=&5x^2&+&10x&-&40 \\\\\r\n&&&&+x&+&28&&&+&x&+&28 \\\\\r\n\\midrule\r\n&&&&&&0&=&5x^2&+&11x&-&12 \\\\ \\\\\r\n&&&&&&0&=&5x^2&+&15x-4x&-&12 \\\\\r\n&&&&&&0&=&5x(x&+&3)-4(x&+&3) \\\\\r\n&&&&&&0&=&(x&+&3)(5x&-&4) \\\\ \\\\\r\n&&&&&&x&=&-3,&\\dfrac{4}{5}&&& \\\\\r\n\\end{array}\\)<\/li>\r\n \t
    5. True<\/li>\r\n \t
    6. False<\/li>\r\n \t
    7. \\(\\begin{array}{l}\r\n\\\\ \\\\\r\n4\\cdot 6+3\\sqrt{36\\cdot 2}+4 \\\\\r\n24+3\\cdot 6\\sqrt{2}+4 \\\\\r\n28+18\\sqrt{2}\r\n\\end{array}\\)<\/li>\r\n \t
    8. \\(\\dfrac{\\sqrt{\\cancel{300}100a^{\\cancel{5}4}\\cancel{b^2}}}{\\sqrt{\\cancel{3}\\cancel{a}\\cancel{b^2}}}\\Rightarrow \\sqrt{100a^4}\\Rightarrow 10a^2\\)<\/li>\r\n \t
    9. \\(\\dfrac{(12)(3+\\sqrt{6})}{(3-\\sqrt{6})(3+\\sqrt{6})}\\Rightarrow \\dfrac{36+12\\sqrt{6}}{9-6}\\Rightarrow \\dfrac{\\cancel{36}12+\\cancel{12}4\\sqrt{6}}{\\cancel{3}1}\\Rightarrow 12+4\\sqrt{6}\\)<\/li>\r\n \t
    10. \\(\\left(\\dfrac{\\cancel{a^0}1b^3}{c^6d^{-12}}\\right)^{\\frac{1}{3}}\\Rightarrow \\left(\\dfrac{b^3d^{12}}{c^6}\\right)^{\\frac{1}{3}}\\Rightarrow \\dfrac{bd^4}{c^2}\\)<\/li>\r\n \t
    11. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\\r\n(\\sqrt{5x-6})^2& =& (x)^2 \\\\\r\n5x-6&=&x^2 \\\\\r\n0&=&x^2-5x+6 \\\\\r\n0&=&(x-3)(x-2) \\\\ \\\\\r\nx&=&3,2\r\n\\end{array}\\)<\/li>\r\n \t
    12. \\(\\begin{array}{rrcrrrrrr}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n\\sqrt{2x+9}&+&3&=&x&&&& \\\\\r\n&-&3&&&-&3&& \\\\\r\n\\midrule\r\n&&\\sqrt{2x+9}&=&x&-&3&& \\\\ \\\\\r\n&&(\\sqrt{2x+9})^2&=&(x&-&3)^2&& \\\\\r\n2x&+&9&=&x^2&-&6x&+&9 \\\\\r\n-2x&-&9&&&-&2x&-&9 \\\\\r\n\\midrule\r\n&&0&=&x^2&-&8x&& \\\\\r\n&&0&=&x(x&-&8)&& \\\\ \\\\\r\n&&x&=&0,&8&&&\r\n\\end{array}\\)<\/li>\r\n \t
    13. \\(\\phantom{1}\\)\r\n\\((\\sqrt{x-3})^2=(\\sqrt{2x-5})^2 \\\\ \\)\r\n\\(\\begin{array}{rrrrrrrr}\r\n&x&-&3&=&2x&-&5 \\\\\r\n-&x&+&5&&-x&+&5 \\\\\r\n\\midrule\r\n&&&2&=&x&&\r\n\\end{array}\\)<\/li>\r\n \t
    14. \\(\\phantom{1}\\)\r\n
        \r\n \t
      1. \\(\\begin{array}{l}\r\n\\\\ \\\\ \\\\ \\\\\r\nb^2-4ac \\\\\r\n=(4)^2-4(2)(3) \\\\\r\n=16-24 \\\\\r\n=-8 \\\\\r\n\\text{2 non-real solutions}\r\n\\end{array}\\)<\/li>\r\n \t
      2. \\(\\begin{array}{l}\r\n\\\\ \\\\ \\\\ \\\\\r\nb^2-4ac \\\\\r\n=(-2)^2-4(3)(-8) \\\\\r\n=4+96 \\\\\r\n=100 \\\\\r\n\\text{2 real solutions}\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\\dfrac{3x^2}{3}&=&\\dfrac{27}{3} \\\\\r\nx^2&=&9 \\\\\r\nx&=&\\pm 3\r\n\\end{array}\\)<\/li>\r\n \t
        2. \\(\\begin{array}{rrl}\r\n\\\\ \\\\\r\n2x^2-16x&=&0 \\\\\r\n2x(x-8)&=&0 \\\\\r\nx&=&0,8\r\n\\end{array}\\)<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
        3. \\(\\phantom{1}\\)\r\n
            \r\n \t
          1. \\((x-4)(x+3)\\Rightarrow x=4,-3\\)<\/li>\r\n \t
          2. \\(\\begin{array}{rrl}\r\n\\\\ \\\\\r\nx^2+9x+8&=&0 \\\\\r\n(x+8)(x+1)&=&0 \\\\\r\nx&=&-1,-8\r\n\\end{array}\\)<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
          3. \\(\\left(\\dfrac{x-3}{2}+\\dfrac{6}{x+3}=1\\right)(2)(x+3) \\\\ \\)\r\n\\(\\begin{array}{rrrrcrrrrrr}\r\n(x&-&3)(x&+&3)&+&6(2)&=&2(x&+&3) \\\\\r\nx^2&&&-&9&+&12&=&2x&+&6 \\\\\r\n&-&2x&&&-&6&&-2x&-&6 \\\\\r\n\\midrule\r\n&&x^2&-&2x&-&3&=&0&& \\\\\r\n&&(x&-&3)(x&+&1)&=&0&& \\\\ \\\\\r\n&&&&&&x&=&3,&-1&\r\n\\end{array}\\)<\/li>\r\n \t
          4. \\(\\left(\\dfrac{x-2}{x}=\\dfrac{x}{x+4}\\right)(x)(x+4) \\\\ \\)\r\n\\(\\begin{array}{rrrcrrrl}\r\n&(x&-&2)(x&+&4)&=&x^2 \\\\\r\n&x^2&+&2x&-&8&=&x^2 \\\\\r\n-&x^2&&&+&8&&-x^2+8 \\\\\r\n\\midrule\r\n&&&&&\\dfrac{2x}{2}&=&\\dfrac{8}{2} \\\\ \\\\\r\n&&&&&x&=&4\r\n\\end{array}\\)<\/li>\r\n \t
          5. \\(\\phantom{1}\\)\r\n\\(\\text{width}=W\\hspace{0.5in}\\text{length}=L=3+2W \\\\ \\)\r\n\\(\\begin{array}{rrl}\r\nA&=&L\\cdot W \\\\\r\n65&=&W(3+2W) \\\\\r\n65&=&3W+2W^2 \\\\ \\\\\r\n0&=&2W^2+3W-65 \\\\\r\n0&=&2W^2-10W+13W-65 \\\\\r\n0&=&2W(W-5)+13(W-5) \\\\\r\n0&=&(W-5)(2W+13) \\\\ \\\\\r\nW&=&5, \\cancel{-\\dfrac{13}{2}} \\\\ \\\\\r\nL&=&3+2W \\\\\r\nL&=&13\r\n\\end{array}\\)<\/li>\r\n \t
          6. \\(\\phantom{1}\\)\r\n\\(x, x+2, x+4 \\\\ \\)\r\n\\(\\begin{array}{rrcrrrrrrrl}\r\n&&x(x&+&2)&=&68&+&x&+&4 \\\\\r\nx^2&+&2x&&&=&x&+&72&& \\\\\r\n&-&x&-&72&&-x&-&72&& \\\\\r\n\\midrule\r\nx^2&+&x&-&72&=&0&&&& \\\\ \\\\\r\n(x&+&9)(x&-&8)&=&0&&&& \\\\\r\n&&&&x&=&\\cancel{-9},&8&&&\r\n\\end{array}\\)\r\n\u2234 8, 10, 12<\/li>\r\n \t
          7. \\(\\phantom{1}\\)\r\n\\(d=r\\cdot t\\text{ and }d_{\\text{up}}=d_{\\text{down}} \\\\ \\)\r\n\\(\\begin{array}{rrrrrrcr}\r\n&8(r&-&4)&=&6(r&+&4) \\\\\r\n&8r&-&32&=&6r&+&24 \\\\\r\n-&6r&+&32&&-6r&+&32 \\\\\r\n\\midrule\r\n&&&2r&=&56&& \\\\\r\n&&&r&=&28&\\text{km\/h}& \\\\\r\n\\end{array}\\)<\/li>\r\n \t
          8. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\nA&=&\\dfrac{1}{2}bh \\\\ \\\\\r\n(330&=&\\dfrac{1}{2}(h+8)h)(2) \\\\ \\\\\r\n660&=&h^2+8h \\\\ \\\\\r\n0&=&h^2+8h-660 \\\\\r\n0&=&h^2+30h-22h-660 \\\\\r\n0&=&h(h+30)-22(h+30) \\\\ \\\\\r\n0&=&(h+30)(h-22) \\\\\r\nh&=&\\cancel{-30}, 22 \\\\ \\\\\r\n\\therefore b&=&h+8 \\\\\r\n&=&22+8 \\\\\r\n&=&30\r\n\\end{array}\\)<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>","rendered":"

            Midterm Three Review<\/h1>\n
              \n
            1. \n
                \n
              1. \"\dfrac{6\cancel{(a-b)}}{(a+b)\cancel{(a^2-ab+b^2)}}\cdot \dfrac{\cancel{a^2-ab+b^2}}{(a+b)\cancel{(a-b)}}\Rightarrow \dfrac{b}{(a+b)^2}\"<\/li>\n
              2. \"\begin{array}{l}<\/li>\n
              3. \"\dfrac{\left(1-\dfrac{6}{x}\right)x^2}{\left(\dfrac{4}{x}-\dfrac{24}{x^2}\right)x^2}\Rightarrow \dfrac{x^2-6x}{4x-24}\Rightarrow \dfrac{x\cancel{(x-6)}}{4\cancel{(x-6)}}\Rightarrow \dfrac{x}{4}\"<\/li>\n
              4. \"\phantom{1}\"
                \n\"\left(\dfrac{4}{x+4}-\dfrac{5}{x-2}=5\right)(x+4)(x-2) \\\"
                \n\"\begin{array}{rrrrrrrrrrcrr}<\/li>\n
              5. True<\/li>\n
              6. False<\/li>\n
              7. \"\begin{array}{l}<\/li>\n
              8. \"\dfrac{\sqrt{\cancel{300}100a^{\cancel{5}4}\cancel{b^2}}}{\sqrt{\cancel{3}\cancel{a}\cancel{b^2}}}\Rightarrow \sqrt{100a^4}\Rightarrow 10a^2\"<\/li>\n
              9. \"\dfrac{(12)(3+\sqrt{6})}{(3-\sqrt{6})(3+\sqrt{6})}\Rightarrow \dfrac{36+12\sqrt{6}}{9-6}\Rightarrow \dfrac{\cancel{36}12+\cancel{12}4\sqrt{6}}{\cancel{3}1}\Rightarrow 12+4\sqrt{6}\"<\/li>\n
              10. \"\left(\dfrac{\cancel{a^0}1b^3}{c^6d^{-12}}\right)^{\frac{1}{3}}\Rightarrow \left(\dfrac{b^3d^{12}}{c^6}\right)^{\frac{1}{3}}\Rightarrow \dfrac{bd^4}{c^2}\"<\/li>\n
              11. \"\begin{array}{rrl}<\/li>\n
              12. \"\begin{array}{rrcrrrrrr}<\/li>\n
              13. \"\phantom{1}\"
                \n\"(\sqrt{x-3})^2=(\sqrt{2x-5})^2 \\\"
                \n\"\begin{array}{rrrrrrrr}<\/li>\n
              14. \"\phantom{1}\"\n
                  \n
                1. \"\begin{array}{l}<\/li>\n
                2. \"\begin{array}{l}<\/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
                      \n
                    1. \"(x-4)(x+3)\Rightarrow x=4,-3\"<\/li>\n
                    2. \"\begin{array}{rrl}<\/li>\n<\/ol>\n<\/li>\n
                    3. \"\left(\dfrac{x-3}{2}+\dfrac{6}{x+3}=1\right)(2)(x+3) \\\"
                      \n\"\begin{array}{rrrrcrrrrrr}<\/li>\n
                    4. \"\left(\dfrac{x-2}{x}=\dfrac{x}{x+4}\right)(x)(x+4) \\\"
                      \n\"\begin{array}{rrrcrrrl}<\/li>\n
                    5. \"\phantom{1}\"
                      \n\"\text{width}=W\hspace{0.5in}\text{length}=L=3+2W \\\"
                      \n\"\begin{array}{rrl}<\/li>\n
                    6. \"\phantom{1}\"
                      \n\"x, x+2, x+4 \\\"
                      \n\"\begin{array}{rrcrrrrrrrl}
                      \n\u2234 8, 10, 12<\/li>\n
                    7. \"\phantom{1}\"
                      \n\"d=r\cdot t\text{ and }d_{\text{up}}=d_{\text{down}} \\\"
                      \n\"\begin{array}{rrrrrrcr}<\/li>\n
                    8. \"\begin{array}{rrl}<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n","protected":false},"author":727,"menu_order":96,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"back-matter-type":[],"contributor":[],"license":[],"class_list":["post-2177","back-matter","type-back-matter","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter\/2177","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/types\/back-matter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/users\/727"}],"version-history":[{"count":17,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter\/2177\/revisions"}],"predecessor-version":[{"id":3351,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter\/2177\/revisions\/3351"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter\/2177\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/media?parent=2177"}],"wp:term":[{"taxonomy":"back-matter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter-type?post=2177"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/contributor?post=2177"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/license?post=2177"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}