{"id":2004,"date":"2021-12-02T19:40:25","date_gmt":"2021-12-03T00:40:25","guid":{"rendered":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/back-matter\/midterm-3-prep-answer-key\/"},"modified":"2022-11-02T10:39:03","modified_gmt":"2022-11-02T14:39:03","slug":"midterm-3-prep-answer-key","status":"publish","type":"back-matter","link":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/back-matter\/midterm-3-prep-answer-key\/","title":{"raw":"Midterm 3 Prep Answer Key","rendered":"Midterm 3 Prep Answer Key"},"content":{"raw":"

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

          Midterm Three Review<\/h1>\n
            \n
          1. [latex]\\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}[\/latex]<\/li>\n
          2. [latex]\\begin{array}[t]{l} \\dfrac{x}{(x+5)(x-5)}-\\dfrac{2}{(x-5)(x-1)} \\\\ \\\\ \\text{LCD}=(x+5)(x-5)(x-1) \\\\ \\\\ \\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)} \\\\ \\\\ \\Rightarrow \\dfrac{\\cancel{(x-5)}(x+2)}{(x+5)\\cancel{(x-5)}(x-1)}\\Rightarrow \\dfrac{x+2}{(x+5)(x-1)} \\end{array}[\/latex]<\/li>\n
          3. [latex]\\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}[\/latex]<\/li>\n
          4. [latex]\\left(\\dfrac{4}{x+4}-\\dfrac{5}{x-2}=5\\right)(x+4)(x-2)[\/latex]
            \n[latex]\\begin{array}{rrrrrrrrrrcrr} 4(x&-&2)&-&5(x&+&4)&=&5(x&+&4)(x&-&2) \\\\ 4x&-&8&-&5x&-&20&=&5(x^2&+&2x&-&8) \\\\ &&&&-x&-&28&=&5x^2&+&10x&-&40 \\\\ &&&&+x&+&28&&&+&x&+&28 \\\\ \\hline &&&&&&0&=&5x^2&+&11x&-&12 \\\\ \\\\ &&&&&&0&=&5x^2&+&15x-4x&-&12 \\\\ &&&&&&0&=&5x(x&+&3)-4(x&+&3) \\\\ &&&&&&0&=&(x&+&3)(5x&-&4) \\\\ \\\\ &&&&&&x&=&-3,&\\dfrac{4}{5}&&& \\\\ \\end{array}[\/latex]<\/li>\n
          5. True<\/li>\n
          6. False<\/li>\n
          7. [latex]4\\cdot 6+3\\sqrt{36\\cdot 2}+4[\/latex]
            \n[latex]24+3\\cdot 6\\sqrt{2}+4[\/latex]
            \n[latex]28+18\\sqrt{2}[\/latex]<\/li>\n
          8. [latex]\\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[\/latex]<\/li>\n
          9. [latex]\\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}[\/latex]<\/li>\n
          10. [latex]\\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}[\/latex]<\/li>\n
          11. [latex]\\phantom{a}[\/latex]
            \n[latex]\\begin{array}[t]{rrl} (\\sqrt{5x-6})^2& =& (x)^2 \\\\ 5x-6&=&x^2 \\\\ 0&=&x^2-5x+6 \\\\ 0&=&(x-3)(x-2) \\\\ \\\\ x&=&3,2 \\end{array}[\/latex]<\/li>\n
          12. [latex]\\phantom{a}[\/latex]
            \n[latex]\\begin{array}[t]{rrcrrrrrr} \\sqrt{2x+9}&+&3&=&x&&&& \\\\ &-&3&&&-&3&& \\\\ \\hline &&\\sqrt{2x+9}&=&x&-&3&& \\\\ \\\\ &&(\\sqrt{2x+9})^2&=&(x&-&3)^2&& \\\\ 2x&+&9&=&x^2&-&6x&+&9 \\\\ -2x&-&9&&&-&2x&-&9 \\\\ \\hline &&0&=&x^2&-&8x&& \\\\ &&0&=&x(x&-&8)&& \\\\ \\\\ &&x&=&0,&8&&& \\end{array}[\/latex]<\/li>\n
          13. [latex](\\sqrt{x-3})^2=(\\sqrt{2x-5})^2[\/latex]
            \n[latex]\\begin{array}{rrrrrrrr} &x&-&3&=&2x&-&5 \\\\ -&x&+&5&&-x&+&5 \\\\ \\hline &&&2&=&x&& \\end{array}[\/latex]<\/li>\n
          14. [latex]\\phantom{a}[\/latex]\n
              \n
            1. [latex]\\phantom{a}[\/latex]
              \n[latex]\\begin{array}[t]{l} b^2-4ac \\\\ =(4)^2-4(2)(3) \\\\ =16-24 \\\\ =-8 \\\\ \\text{2 non-real solutions} \\end{array}[\/latex]<\/li>\n
            2. [latex]\\phantom{a}[\/latex]
              \n[latex]\\begin{array}[t]{l} b^2-4ac \\\\ =(-2)^2-4(3)(-8) \\\\ =4+96 \\\\ =100 \\\\ \\text{2 real solutions} \\end{array}[\/latex]<\/li>\n<\/ol>\n<\/li>\n
            3. [latex]\\phantom{a}[\/latex]\n
                \n
              1. [latex]\\phantom{a}[\/latex]
                \n[latex]\\begin{array}[t]{rrl} \\dfrac{3x^2}{3}&=&\\dfrac{27}{3} \\\\ x^2&=&9 \\\\ x&=&\\pm 3 \\end{array}[\/latex]<\/li>\n
              2. [latex]\\phantom{a}[\/latex]
                \n[latex]\\begin{array}[t]{rrl} 2x^2-16x&=&0 \\\\ 2x(x-8)&=&0 \\\\ x&=&0,8 \\end{array}[\/latex]<\/li>\n<\/ol>\n<\/li>\n
              3. [latex]\\phantom{a}[\/latex]\n
                  \n
                1. [latex](x-4)(x+3)\\Rightarrow x=4,-3[\/latex]<\/li>\n
                2. [latex]\\phantom{a}[\/latex]
                  \n[latex]\\begin{array}[t]{rrl} x^2+9x+8&=&0 \\\\ (x+8)(x+1)&=&0 \\\\ x&=&-1,-8 \\end{array}[\/latex]<\/li>\n<\/ol>\n<\/li>\n
                3. [latex]\\left(\\dfrac{x-3}{2}+\\dfrac{6}{x+3}=1\\right)(2)(x+3)[\/latex]
                  \n[latex]\\begin{array}{rrrrcrrrrrr} (x&-&3)(x&+&3)&+&6(2)&=&2(x&+&3) \\\\ x^2&&&-&9&+&12&=&2x&+&6 \\\\ &-&2x&&&-&6&&-2x&-&6 \\\\ \\hline &&x^2&-&2x&-&3&=&0&& \\\\ &&(x&-&3)(x&+&1)&=&0&& \\\\ \\\\ &&&&&&x&=&3,&-1& \\end{array}[\/latex]<\/li>\n
                4. [latex]\\left(\\dfrac{x-2}{x}=\\dfrac{x}{x+4}\\right)(x)(x+4)[\/latex]
                  \n[latex]\\begin{array}{rrrcrrrl} &(x&-&2)(x&+&4)&=&x^2 \\\\ &x^2&+&2x&-&8&=&x^2 \\\\ -&x^2&&&+&8&&-x^2+8 \\\\ \\hline &&&&&\\dfrac{2x}{2}&=&\\dfrac{8}{2} \\\\ \\\\ &&&&&x&=&4 \\end{array}[\/latex]<\/li>\n
                5. [latex]\\text{width}=W\\hspace{0.5in}\\text{length}=L=3+2W[\/latex]
                  \n[latex]\\begin{array}{rrl} A&=&L\\cdot W \\\\ 65&=&W(3+2W) \\\\ 65&=&3W+2W^2 \\\\ \\\\ 0&=&2W^2+3W-65 \\\\ 0&=&2W^2-10W+13W-65 \\\\ 0&=&2W(W-5)+13(W-5) \\\\ 0&=&(W-5)(2W+13) \\\\ \\\\ W&=&5, \\cancel{-\\dfrac{13}{2}} \\\\ \\\\ L&=&3+2W \\\\ L&=&13 \\end{array}[\/latex]<\/li>\n
                6. [latex]x, x+2, x+4[\/latex]
                  \n[latex]\\begin{array}{rrcrrrrrrrl} &&x(x&+&2)&=&68&+&x&+&4 \\\\ x^2&+&2x&&&=&x&+&72&& \\\\ &-&x&-&72&&-x&-&72&& \\\\ \\hline x^2&+&x&-&72&=&0&&&& \\\\ \\\\ (x&+&9)(x&-&8)&=&0&&&& \\\\ &&&&x&=&\\cancel{-9},&8&&& \\end{array}[\/latex]
                  \n[latex]\\therefore[\/latex] 8, 10, 12<\/li>\n
                7. [latex]d=r\\cdot t\\text{ and }d_{\\text{up}}=d_{\\text{down}}[\/latex]
                  \n[latex]\\begin{array}{rrrrrrcr} &8(r&-&4)&=&6(r&+&4) \\\\ &8r&-&32&=&6r&+&24 \\\\ -&6r&+&32&&-6r&+&32 \\\\ \\hline &&&2r&=&56&& \\\\ &&&r&=&28&\\text{km\/h}& \\\\ \\end{array}[\/latex]<\/li>\n
                8. [latex]\\phantom{a}[\/latex]
                  \n[latex]\\begin{array}[t]{rrl} A&=&\\dfrac{1}{2}bh \\\\ \\\\ (330&=&\\dfrac{1}{2}(h+8)h)(2) \\\\ \\\\ 660&=&h^2+8h \\\\ \\\\ 0&=&h^2+8h-660 \\\\ 0&=&h^2+30h-22h-660 \\\\ 0&=&h(h+30)-22(h+30) \\\\ \\\\ 0&=&(h+30)(h-22) \\\\ h&=&\\cancel{-30}, 22 \\\\ \\\\ \\therefore b&=&h+8 \\\\ &=&22+8 \\\\ &=&30 \\end{array}[\/latex]<\/li>\n<\/ol>\n","protected":false},"author":90,"menu_order":101,"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-2004","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\/2004","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\/2004\/revisions"}],"predecessor-version":[{"id":2005,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter\/2004\/revisions\/2005"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter\/2004\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/media?parent=2004"}],"wp:term":[{"taxonomy":"back-matter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter-type?post=2004"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/contributor?post=2004"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/license?post=2004"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}