{"id":2205,"date":"2019-10-01T11:37:24","date_gmt":"2019-10-01T15:37:24","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/?post_type=back-matter&p=2205"},"modified":"2019-12-29T20:00:23","modified_gmt":"2019-12-30T01:00:23","slug":"final-exam-prep-answer-key","status":"publish","type":"back-matter","link":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/back-matter\/final-exam-prep-answer-key\/","title":{"raw":"Final Exam: Version A Answer Key","rendered":"Final Exam: Version A Answer Key"},"content":{"raw":"[latexpage]\r\n

Questions from Chapters 1 to 3<\/h2>\r\n
    \r\n \t
  1. \\(\\begin{array}{l}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n-(6)-\\sqrt{6^2-4(4)(2)} \\\\ \\\\\r\n-6-\\sqrt{36-32} \\\\ \\\\\r\n-6-\\sqrt{4} \\\\ \\\\\r\n-6-2=-8\r\n\\end{array}\\)<\/li>\r\n \t
  2. \\(\\begin{array}{rrrrrrrrrrr}\r\n\\\\ \\\\ \\\\ \\\\\r\n6x&+&24&=&35&-&5x&-&8&+&12x \\\\\r\n6x&+&24&=&27&+&7x&&&& \\\\\r\n-7x&-&24&&-24&-&7x&&&& \\\\\r\n\\midrule\r\n&&-x&=&3&&&&&& \\\\\r\n&&\\therefore x&=&-3&&&&&& \\\\\r\n\\end{array}\\)<\/li>\r\n \t
  3. \\(\\phantom{1}\\)\r\n\\(\\left(\\dfrac{x+4}{2}-\\dfrac{1}{2}=\\dfrac{x+2}{4}\\right)(4) \\\\ \\)\r\n\\(\\begin{array}{crrrcrrrl}\r\n2(x&+&4)&-&1(2)&=&x&+&2 \\\\\r\n2x&+&8&-&2&=&x&+&2 \\\\\r\n-x&-&8&+&2&&-x&-&8+2 \\\\\r\n\\midrule\r\n&&&&x&=&-4&&\r\n\\end{array}\\)<\/li>\r\n \t
  4. \\(x=-2\\)<\/li>\r\n \t
  5. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\nd^2&=&\\Delta x^2+\\Delta y^2 \\\\\r\n&=&(2--4)^2+(6--2)^2 \\\\\r\n&=&6^2+8^2 \\\\\r\n&=&36+64 \\\\\r\n&=&100 \\\\ \\\\\r\n\\therefore d&=&\\sqrt{100}=10\r\n\\end{array}\\)<\/li>\r\n \t
  6. \r\n\r\n\r\n\r\n\r\n\r\n
    \\(2x-3y=6\\)<\/caption>\r\n
    \\(x\\)<\/th>\r\n\\(y\\)<\/th>\r\n<\/tr>\r\n
    0<\/td>\r\n\u22122<\/td>\r\n<\/tr>\r\n
    3<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
    6<\/td>\r\n2<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n\"Line<\/li>\r\n \t
  7. \\(\\begin{array}{rrrrrrrrr}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\nx&-&2x&+&10&\\le &18&+&3x \\\\\r\n&&-x&+&10&\\le &18&+&3x \\\\\r\n+&&-3x&-&10&&-10&-&3x \\\\\r\n\\midrule\r\n&&&&\\dfrac{-4x}{-4}&\\le &\\dfrac{8}{-4}&& \\\\ \\\\\r\n&&&&x&\\ge &-2&& \\\\\r\n\\end{array}\\)\r\n\\(\\left[-2, \\infty)\\)\r\n\"x<\/li>\r\n \t
  8. \\(\\phantom{1}\\)\r\n\\(\\left(-1 < \\dfrac{3x-2}{7}<1 \\right)(7) \\\\ \\)\r\n\\(\\begin{array}{rrrcrrr}\r\n\\\\ \\\\ \\\\ \\\\\r\n-7&<&3x&-&2&<&7 \\\\\r\n+2&&&+&2&&+2 \\\\\r\n\\midrule\r\n\\dfrac{-5}{3}&<&&\\dfrac{3x}{3}&&<&\\dfrac{9}{3} \\\\ \\\\\r\n-\\dfrac{5}{3}&<&&x&&<&3\r\n\\end{array}\\)\r\n\\(\\phantom{1}\\)\r\n\\(\\left(-\\dfrac{5}{3}, 3\\right)\\)\r\n\"-5<\/li>\r\n \t
  9. \\(\\begin{array}{ll}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\nt=\\dfrac{k}{r} \\\\\r\n\\begin{array}{rrl}\r\n\\\\\r\n&&\\text{1st data} \\\\ \\\\\r\nt&=&45\\text{ min} \\\\\r\nk&=&\\text{find 1st} \\\\\r\nr&=&600\\text{ kL\/min} \\\\ \\\\\r\nt&=&\\dfrac{k}{r} \\\\ \\\\\r\n45&=&\\dfrac{k}{600} \\\\ \\\\\r\nk&=&45(600) \\\\\r\nk&=&27000\\text{ kL}\r\n\\end{array}\r\n& \\hspace{0.25in}\r\n\\begin{array}{rrl}\r\n&&\\text{2nd data} \\\\ \\\\\r\nt&=&\\text{find} \\\\\r\nk&=&27000 \\\\\r\nr&=&1000\\text{ kL\/min} \\\\ \\\\\r\nt&=&\\dfrac{k}{r} \\\\ \\\\\r\nt&=&\\dfrac{27000}{1000} \\\\ \\\\\r\nt&=&27\\text{ min}\r\n\\end{array}\r\n\\end{array}\\)<\/li>\r\n \t
  10. \\(\\phantom{1}\\)\r\n\\(x, x+2 \\\\ \\)\r\n\\(\\begin{array}{rrrrrrrrr}\r\nx&+&x&+&2&=&4(x)&-&12 \\\\\r\n&&2x&+&2&=&4x&-&12 \\\\\r\n&-&2x&+&12&&-2x&+&12 \\\\\r\n\\midrule\r\n&&&&\\dfrac{14}{2}&=&\\dfrac{2x}{2}&& \\\\ \\\\\r\n&&&&x&=&7&&\r\n\\end{array}\\)\r\n\\(\\phantom{1}\\)\r\n\\(\\text{numbers are }7,9\\)<\/li>\r\n<\/ol>\r\n

    Questions from Chapters 4 to 6<\/h2>\r\n
      \r\n \t
    1. \\(\\begin{array}{rrrrrr}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n&2x&+&5y&=&-18 \\\\\r\n+&-2x&+&y&=&6 \\\\\r\n\\midrule\r\n&&&\\dfrac{6y}{6}&=&\\dfrac{-12}{6} \\\\ \\\\\r\n&&&y&=&-2 \\\\ \\\\\r\n&\\therefore y&-&6&=&2x \\\\\r\n&-2&-&6&=&2x \\\\\r\n&&&2x&=&-8 \\\\\r\n&&&x&=&-4\r\n\\end{array}\\)\r\nAnswer: \\((-4, -2)\\)<\/li>\r\n \t
    2. \\(\\begin{array}{ll}\r\n\\begin{array}{rrrrrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n&(8x&+&7y&=&51)(-2) \\\\\r\n&(5x&+&2y&=&20)(7) \\\\ \\\\\r\n&-16x&-&14y&=&-102 \\\\\r\n+&35x&+&14y&=&\\phantom{-}140 \\\\\r\n\\midrule\r\n&&&\\dfrac{19x}{19}&=&\\dfrac{38}{19} \\\\ \\\\\r\n&&&x&=&2 \\\\ \\\\\r\n\\end{array}\r\n& \\hspace{0.25in}\r\n\\begin{array}{rrrrr}\r\n\\\\ \\\\ \\\\ \\\\ \\\\\r\n\\therefore 5x&+&2y&=&20 \\\\\r\n5(2)&+&2y&=&20 \\\\\r\n10&+&2y&=&20 \\\\\r\n-10&&&&-10 \\\\\r\n\\midrule\r\n&&2y&=&10 \\\\\r\n&&y&=&5\r\n\\end{array}\r\n\\end{array}\\)\r\nAnswer: \\((2, 5)\\)<\/li>\r\n \t
    3. \\(\\begin{array}{ll}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n\\begin{array}{rrrrrrrl}\r\n\\\\ \\\\ \\\\ \\\\\r\n&-2x&-&2y&-&12z&=&-10 \\\\\r\n+&2x&&&-&3z&=&\\phantom{-}4 \\\\\r\n\\midrule\r\n&&&(-2y&-&15z&=&-6)(3) \\\\\r\n&&&(3y&+&4z&=&\\phantom{-}9)(2) \\\\ \\\\\r\n&&&-6y&-&45z&=&-18 \\\\\r\n&&+&6y&+&8z&=&\\phantom{-}18 \\\\\r\n\\midrule\r\n&&&&&-37z&=&0 \\\\\r\n&&&&&z&=&0 \\\\ \\\\\r\n\\end{array}\r\n&\\hspace{0.25in}\r\n\\begin{array}{rrrrl}\r\n2x&-&\\cancel{3z}0&=&4 \\\\\r\n&&x&=&\\dfrac{4}{2}\\text{ or }2 \\\\ \\\\\r\n3y&+&\\cancel{4z}0&=&9 \\\\\r\n&&y&=&\\dfrac{9}{3}\\text{ or }3\r\n\\end{array}\r\n\\end{array}\\)\r\nAnswer \\((2, 3, 0)\\)<\/li>\r\n \t
    4. \\(\\begin{array}{l}\r\n\\\\ \\\\\r\n24+\\{-3x-\\cancel{\\left[6x-3(5-2x)\\right]^0}1\\}+3x \\\\\r\n24-3x-1+3x \\\\\r\n23\r\n\\end{array}\\)<\/li>\r\n \t
    5. \\(2ab^3(a^2-16)\\Rightarrow 2a^3b^3-32ab^3\\)<\/li>\r\n \t
    6. \\(\\begin{array}{l}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n(x^{1--2}y^{-3-4})^{-1} \\\\ \\\\\r\n(x^3y^{-7})^{-1} \\\\ \\\\\r\nx^{-3}y^7 \\\\ \\\\\r\n\\dfrac{y^7}{x^3}\r\n\\end{array}\\)<\/li>\r\n \t
    7. \\(\\begin{array}{l}\r\n\\\\ \\\\\r\n3x^2+3x+8x+8 \\\\\r\n3x(x+1)+8(x+1) \\\\\r\n(x+1)(3x+8)\r\n\\end{array}\\)<\/li>\r\n \t
    8. \\((4x)^3-y^3\\Rightarrow (4x-y)(16x^2+4xy+y^2)\\)<\/li>\r\n \t
    9. \\(\\begin{array}{rrrcrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n&(A&+&B&=&\\phantom{191.}50)(-370) \\\\\r\n&(3.95A&+&3.70B&=&191.25)(100) \\\\ \\\\\r\n&-370A&-&370B&=&-18500 \\\\\r\n+&395A&+&370B&=&\\phantom{-}19125 \\\\\r\n\\midrule\r\n&&&25A&=&625 \\\\ \\\\\r\n&&&A&=&\\dfrac{625}{25}\\text{ or }25 \\\\ \\\\\r\n&A&+&B&=&50 \\\\\r\n&25&+&B&=&50 \\\\\r\n&&&B&=&25 \\\\\r\n\\end{array}\\)<\/li>\r\n \t
    10. \\(\\begin{array}{rrrrrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n&(d&+&q&=&16)(-10) \\\\\r\n&10d&+&25q&=&235 \\\\ \\\\\r\n&-10d&-&10q&=&-160 \\\\\r\n+&10d&+&25q&=&\\phantom{-}235 \\\\\r\n\\midrule\r\n&&&\\dfrac{15q}{15}&=&\\dfrac{75}{15} \\\\ \\\\\r\n&&&q&=&5 \\\\\r\n&&&\\therefore d&=&16-5=11 \\\\\r\n\\end{array}\\)<\/li>\r\n<\/ol>\r\n

      Questions from Chapters 7 to 10<\/h2>\r\n
        \r\n \t
      1. \\(\\dfrac{\\cancel{15}3s^{\\cancel{3}2}}{\\cancel{3t^2}1}\\cdot \\dfrac{\\cancel{17}1\\cancel{s^3}}{\\cancel{5}1\\cancel{t}}\\cdot \\dfrac{\\cancel{3t^3}}{\\cancel{34}2\\cancel{s^4}}\\Rightarrow \\dfrac{3s^2}{2}\\)<\/li>\r\n \t
      2. \\(\\begin{array}{l}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n\\text{LCD}=(x+2)(x-2) \\\\ \\\\\r\n\\dfrac{2x(x-2)-4x(x+2)+20}{(x+2)(x-2)} \\\\ \\\\\r\n\\dfrac{2x^2-4x-4x^2-8x+20}{(x+2)(x-2)} \\\\ \\\\\r\n\\dfrac{-2x^2-12x+20}{(x+2)(x-2)} \\\\ \\\\\r\n\\dfrac{-2(x^2+6x-10)}{(x+2)(x-2)}\r\n\\end{array}\\)<\/li>\r\n \t
      3. \\(\\begin{array}{l}\r\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)}} \\\\ \\\\\r\n\\Rightarrow y(x-3y)\r\n\\end{array}\\)<\/li>\r\n \t
      4. \\(\\begin{array}{l}\r\n\\\\ \\\\ \\\\ \\\\\r\n3\\cdot 5\\sqrt{x}-2\\sqrt{36\\cdot 2x}-\\sqrt{16\\cdot x^2\\cdot x} \\\\ \\\\\r\n15\\sqrt{x}-2\\cdot 6\\sqrt{2x}-4x\\sqrt{x} \\\\ \\\\\r\n15\\sqrt{x}-12\\sqrt{2x}-4x\\sqrt{x}\r\n\\end{array}\\)<\/li>\r\n \t
      5. \\(\\dfrac{\\sqrt{m^6\\cancel{n}}}{\\sqrt{3\\cancel{n}}}\\Rightarrow \\dfrac{m^3}{\\sqrt{3}}\\cdot \\dfrac{\\sqrt{3}}{\\sqrt{3}}\\Rightarrow \\dfrac{m^3\\sqrt{3}}{3}\\)<\/li>\r\n \t
      6. \\(\\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}\\)<\/li>\r\n \t
      7. \\(\\begin{array}{l}\r\n\\\\\r\n(x-5)(x+1)=0 \\\\\r\nx=5,-1\r\n\\end{array}\\)<\/li>\r\n \t
      8. \\(\\begin{array}{rrrrrcrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n&&&(x&-&3)^2&=&(x)^2 \\\\ \\\\\r\n&x^2&-&6x&+&9&=&\\phantom{-}x^2 \\\\\r\n-&x^2&&&&&&-x^2 \\\\\r\n\\midrule\r\n&&&-6x&+&9&=&0 \\\\ \\\\\r\n&&&&&\\dfrac{-6x}{-6}&=&\\dfrac{-9}{-6} \\\\ \\\\\r\n&&&&&x&=&\\dfrac{3}{2}\r\n\\end{array}\\)<\/li>\r\n \t
      9. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\nA&=&\\dfrac{1}{2}bh \\\\ \\\\\r\n20&=&\\dfrac{1}{2}(h+6)h \\\\ \\\\\r\n40&=&h^2+6h \\\\ \\\\\r\n0&=&h^2+6h-40 \\\\\r\n0&=&h^2+10h-4h-40 \\\\\r\n0&=&h(h+10)-4(h+10) \\\\\r\n0&=&(h-4)(h+10) \\\\ \\\\\r\nh&=&4, \\cancel{-10} \\\\\r\nb&=&4+6=10\r\n\\end{array}\\)<\/li>\r\n \t
      10. \\(\\phantom{1}\\)\r\n\\(x, x+2, x+4 \\\\ \\)\r\n\\(\\begin{array}{rrrrcrrrlrrrr}\r\n&&&&x(x&+&2)&=&\\phantom{-}8&+&6(x&+&4) \\\\\r\nx^2&+&2x&&&&&=&\\phantom{-}8&+&6x&+&24 \\\\\r\n&-&6x&-&8&-&24&&-8&-&6x&-&24 \\\\\r\n\\midrule\r\n&&x^2&-&4x&-&32&=&0&&&& \\\\ \\\\\r\nx^2&+&4x&-&8x&-&32&=&0&&&& \\\\\r\nx(x&+&4)&-&8(x&+&4)&=&0&&&& \\\\\r\n&&(x&+&4)(x&-&8)&=&0&&&& \\\\\r\n&&&&&&x&=&-4,8&&&&\r\n\\end{array}\\)\r\n\\(\\phantom{1}\\)\r\n\\(\\therefore \\text{ numbers are }-4,-2,0 \\text{ or } 8,10,12\\)<\/li>\r\n<\/ol>","rendered":"

        Questions from Chapters 1 to 3<\/h2>\n
          \n
        1. \"\begin{array}{l}<\/li>\n
        2. \"\begin{array}{rrrrrrrrrrr}<\/li>\n
        3. \"\phantom{1}\"
          \n\"\left(\dfrac{x+4}{2}-\dfrac{1}{2}=\dfrac{x+2}{4}\right)(4) \\\"
          \n\"\begin{array}{crrrcrrrl}<\/li>\n
        4. \"x=-2\"<\/li>\n
        5. \"\begin{array}{rrl}<\/li>\n
        6. \n\n\n\n\n\n\n
          \"2x-3y=6\"<\/caption>\n
          \"x\"<\/th>\n\"y\"<\/th>\n<\/tr>\n
          0<\/td>\n\u22122<\/td>\n<\/tr>\n
          3<\/td>\n0<\/td>\n<\/tr>\n
          6<\/td>\n2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

          \"Line<\/li>\n

        7. \"\begin{array}{rrrrrrrrr}
          \n\"\left[-2, \infty)\"
          \n\"x<\/li>\n
        8. \"\phantom{1}\"
          \n\"\left(-1 < \dfrac{3x-2}{7}<1 \right)(7) \\\"
          \n\"\begin{array}{rrrcrrr}
          \n\"\phantom{1}\"
          \n\"\left(-\dfrac{5}{3}, 3\right)\"
          \n\"-5<\/li>\n
        9. \"\begin{array}{ll}<\/li>\n
        10. \"\phantom{1}\"
          \n\"x, x+2 \\\"
          \n\"\begin{array}{rrrrrrrrr}
          \n\"\phantom{1}\"
          \n\"\text{numbers are }7,9\"<\/li>\n<\/ol>\n

          Questions from Chapters 4 to 6<\/h2>\n
            \n
          1. \"\begin{array}{rrrrrr}
            \nAnswer: \"(-4, -2)\"<\/li>\n
          2. \"\begin{array}{ll}
            \nAnswer: \"(2, 5)\"<\/li>\n
          3. \"\begin{array}{ll}
            \nAnswer \"(2, 3, 0)\"<\/li>\n
          4. \"\begin{array}{l}<\/li>\n
          5. \"2ab^3(a^2-16)\Rightarrow 2a^3b^3-32ab^3\"<\/li>\n
          6. \"\begin{array}{l}<\/li>\n
          7. \"\begin{array}{l}<\/li>\n
          8. \"(4x)^3-y^3\Rightarrow (4x-y)(16x^2+4xy+y^2)\"<\/li>\n
          9. \"\begin{array}{rrrcrl}<\/li>\n
          10. \"\begin{array}{rrrrrl}<\/li>\n<\/ol>\n

            Questions from Chapters 7 to 10<\/h2>\n
              \n
            1. \"\dfrac{\cancel{15}3s^{\cancel{3}2}}{\cancel{3t^2}1}\cdot \dfrac{\cancel{17}1\cancel{s^3}}{\cancel{5}1\cancel{t}}\cdot \dfrac{\cancel{3t^3}}{\cancel{34}2\cancel{s^4}}\Rightarrow \dfrac{3s^2}{2}\"<\/li>\n
            2. \"\begin{array}{l}<\/li>\n
            3. \"\begin{array}{l}<\/li>\n
            4. \"\begin{array}{l}<\/li>\n
            5. \"\dfrac{\sqrt{m^6\cancel{n}}}{\sqrt{3\cancel{n}}}\Rightarrow \dfrac{m^3}{\sqrt{3}}\cdot \dfrac{\sqrt{3}}{\sqrt{3}}\Rightarrow \dfrac{m^3\sqrt{3}}{3}\"<\/li>\n
            6. \"\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}\"<\/li>\n
            7. \"\begin{array}{l}<\/li>\n
            8. \"\begin{array}{rrrrrcrl}<\/li>\n
            9. \"\begin{array}{rrl}<\/li>\n
            10. \"\phantom{1}\"
              \n\"x, x+2, x+4 \\\"
              \n\"\begin{array}{rrrrcrrrlrrrr}
              \n\"\phantom{1}\"
              \n\"\therefore \text{ numbers are }-4,-2,0 \text{ or } 8,10,12\"<\/li>\n<\/ol>\n","protected":false},"author":727,"menu_order":111,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"back-matter-type":[],"contributor":[],"license":[],"class_list":["post-2205","back-matter","type-back-matter","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter\/2205","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":18,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter\/2205\/revisions"}],"predecessor-version":[{"id":3751,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter\/2205\/revisions\/3751"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter\/2205\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/media?parent=2205"}],"wp:term":[{"taxonomy":"back-matter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter-type?post=2205"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/contributor?post=2205"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/license?post=2205"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}