{"id":1596,"date":"2021-12-02T19:38:41","date_gmt":"2021-12-03T00:38:41","guid":{"rendered":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/back-matter\/answer-key-2-6\/"},"modified":"2023-08-31T19:59:10","modified_gmt":"2023-08-31T23:59:10","slug":"answer-key-2-6","status":"publish","type":"back-matter","link":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/back-matter\/answer-key-2-6\/","title":{"raw":"Answer Key 2.6","rendered":"Answer Key 2.6"},"content":{"raw":"
    \r\n \t
  1. (3) + 1 + (4) \u2212 (1)\r\n3 + 1 + 4 \u2212 1\r\n7<\/li>\r\n \t
  2. (5)2<\/sup> + (5) \u2212 (1)\r\n25 + 5 \u2212 1\r\n29<\/li>\r\n \t
  3. (6) \u2212 [(6)(5)\u00a0\u00f7 6]\r\n6 \u2212 [30 \u00f7 6]\r\n6 \u2212 5\r\n1<\/li>\r\n \t
  4. [6 + (4) \u2212 (1)] \u00f7 3\r\n[6 + 4 \u2212 1] \u00f7 3\r\n9 \u00f7 3\r\n3<\/li>\r\n \t
  5. (5)2<\/sup> \u2212 ((3) \u2212 1)\r\n25 \u2212 (3 \u2212 1)\r\n25 \u2212 2\r\n23<\/li>\r\n \t
  6. (6) + 6(4) \u2212 4(4)\r\n6 + 24 \u2212 16\r\n14<\/li>\r\n \t
  7. 5(4) + (2)(5) \u00f7 2\r\n20 + 10 \u00f7 2\r\n20 + 5\r\n25<\/li>\r\n \t
  8. 5((6) + (2)) + 1 + (5)\r\n5(6 + 2) + 1 + 5\r\n5(8) + 6\r\n40 + 6\r\n46<\/li>\r\n \t
  9. [4 \u2212 ((6) \u2212 (4))] \u00f7 2 + (6)\r\n[4 \u2212 (6\u00a0\u2212 4)] \u00f7 2 + 6\r\n[4 \u2212 2] \u00f7 2 + 6\r\n2 \u00f7 2 + 6\r\n1 + 6\r\n7<\/li>\r\n \t
  10. (4) + (5) \u2212 1\r\n4 + 5 \u2212 1\r\n8<\/li>\r\n \t
  11. [latex]\\phantom{a}[\/latex]\r\n[latex]\\begin{array}[t]{rrl}\r\n\\dfrac{ab}{a}&=&\\dfrac{c}{a} \\\\ \\\\\r\nb&=&\\dfrac{c}{a}\r\n\\end{array}[\/latex]<\/li>\r\n \t
  12. [latex]\\phantom{a}[\/latex]\r\n[latex]\\begin{array}[t]{l}\r\n\\left(g=\\dfrac{h}{i}\\right)(i) \\\\ \\\\\r\nh=gi\r\n\\end{array}[\/latex]<\/li>\r\n \t
  13. [latex]\\phantom{a}[\/latex]\r\n[latex]\\begin{array}[t]{l}\r\n\\left(\\left(\\dfrac{f}{g}\\right)x=b\\right)\\dfrac{g}{f} \\\\ \\\\\r\nx=\\dfrac{bg}{f}\r\n\\end{array}[\/latex]<\/li>\r\n \t
  14. [latex]\\phantom{a}[\/latex]\r\n[latex]\\begin{array}[t]{l}\r\n\\left(p=\\dfrac{3y}{q}\\right)\\left(\\dfrac{q}{3}\\right) \\\\ \\\\\r\ny=\\dfrac{pq}{3}\r\n\\end{array}[\/latex]<\/li>\r\n \t
  15. [latex]\\phantom{a}[\/latex]\r\n[latex]\\begin{array}[t]{l}\r\n\\left(3x=\\dfrac{a}{b}\\right)\\left(\\dfrac{1}{3}\\right) \\\\ \\\\\r\nx=\\dfrac{a}{3b}\r\n\\end{array}[\/latex]<\/li>\r\n \t
  16. [latex]\\phantom{a}[\/latex]\r\n[latex]\\begin{array}[t]{l}\r\n\\left(\\dfrac{ym}{b}=\\dfrac{c}{d}\\right)\\left(\\dfrac{b}{m}\\right) \\\\ \\\\\r\ny=\\dfrac{bc}{dm}\r\n\\end{array}[\/latex]<\/li>\r\n \t
  17. [latex]\\phantom{a}[\/latex]\r\n[latex]\\begin{array}[t]{l}\r\n\\left(V=\\dfrac{4}{3}\\pi r^3\\right)\\left(\\dfrac{3}{4r^3}\\right) \\\\ \\\\\r\n\\pi=\\dfrac{3V}{4r^3}\r\n\\end{array}[\/latex]<\/li>\r\n \t
  18. [latex]\\phantom{a}[\/latex]\r\n[latex]\\begin{array}[t]{l}\r\n\\left(E=mv^2\\right)\\left(\\dfrac{1}{v^2}\\right) \\\\ \\\\\r\nm=\\dfrac{E}{v^2}\r\n\\end{array}[\/latex]<\/li>\r\n \t
  19. [latex]\\phantom{a}[\/latex]\r\n[latex]\\begin{array}[t]{l}\r\n\\left(c=\\dfrac{4y}{m+n}\\right)\\left(\\dfrac{m+n}{4}\\right) \\\\ \\\\\r\ny=\\dfrac{c(m+n)}{4}\r\n\\end{array}[\/latex]<\/li>\r\n \t
  20. [latex]\\phantom{a}[\/latex]\r\n[latex]\\begin{array}[t]{l}\r\n\\left(\\dfrac{rs}{a-3}=k\\right)\\left(\\dfrac{a-3}{s}\\right) \\\\ \\\\\r\nr=\\dfrac{k(a-3)}{s}\r\n\\end{array}[\/latex]<\/li>\r\n \t
  21. [latex]\\phantom{a}[\/latex]\r\n[latex]\\begin{array}[t]{l}\r\n\\left(V=\\dfrac{\\pi Dn}{12}\\right)\\left(\\dfrac{12}{\\pi n}\\right) \\\\ \\\\\r\nD=\\dfrac{12V}{\\pi n}\r\n\\end{array}[\/latex]<\/li>\r\n \t
  22. [latex]\\phantom{a}[\/latex]\r\n[latex]\\begin{array}[t]{rrl}\r\nF\\phantom{+kL}&=&kR-kL \\\\\r\n+kL&&\\phantom{kR}+kL \\\\\r\n\\hline\r\n\\dfrac{F+kL}{k}&=&\\dfrac{kR}{k} \\\\ \\\\\r\nR&=&\\dfrac{F+kL}{k}\r\n\\end{array}[\/latex]<\/li>\r\n \t
  23. [latex]\\phantom{a}[\/latex]\r\n[latex]\\begin{array}[t]{rrl}\r\nP\\phantom{-np}&=&\\phantom{-}np-nc \\\\\r\n-np&&-np \\\\\r\n\\hline\r\n\\dfrac{P-np}{-n}&=&\\dfrac{-nc}{-n} \\\\ \\\\\r\nc&=&\\dfrac{P-np}{-n}\r\n\\end{array}[\/latex]<\/li>\r\n \t
  24. [latex]\\phantom{a}[\/latex]\r\n[latex]\\begin{array}[t]{rrrrr}\r\nS\\phantom{-2B}&=&L&+&2B \\\\\r\n-2B&&&-&2B \\\\\r\n\\hline\r\nL&=&S&-&2B\r\n\\end{array}[\/latex]<\/li>\r\n \t
  25. [latex]\\left(T=\\dfrac{D-d}{L}\\right)(L)[\/latex]\r\n[latex]\\begin{array}{rrrrr}\r\nTL\\phantom{+d}&=&D&-&d \\\\\r\n+d&&&+&d \\\\\r\n\\hline\r\nD&=&TL&+&d\r\n\\end{array}[\/latex]<\/li>\r\n \t
  26. [latex]\\left(I=\\dfrac{E_a-E_q}{R}\\right)(R) [\/latex]\r\n[latex]\\begin{array}{rrrrr}\r\nIR&=&E_a&-&E_q \\\\\r\n+E_q&&&+&E_q \\\\\r\n\\hline\r\nE_a&=&IR&+&E_q\r\n\\end{array}[\/latex]<\/li>\r\n \t
  27. [latex]\\phantom{a}[\/latex]\r\n[latex]\\begin{array}[t]{rrl}\r\n\\dfrac{L}{1+at}&=&\\dfrac{L_o(1+at)}{1+at} \\\\ \\\\\r\nL_o&=&\\dfrac{L}{1+at}\r\n\\end{array}[\/latex]<\/li>\r\n \t
  28. [latex]\\phantom{a}[\/latex]\r\n[latex]\\begin{array}[t]{rrl}\r\n2m+p&=&\\phantom{-}4m+q \\\\\r\n-2m-q&&-2m-q \\\\\r\n\\hline\r\n\\dfrac{p-q}{2}&=&\\dfrac{2m}{2} \\\\ \\\\\r\nm&=&\\dfrac{p-q}{2}\r\n\\end{array}[\/latex]<\/li>\r\n \t
  29. [latex]\\left(\\dfrac{k-m}{r}=q\\right)(r) [\/latex]\r\n[latex]\\begin{array}{rrl}\r\nk-m&=&qr \\\\\r\n+m&&\\phantom{qr}+m \\\\\r\n\\hline\r\nk&=&qr+m\r\n\\end{array}[\/latex]<\/li>\r\n \t
  30. [latex]\\phantom{a}[\/latex]\r\n[latex]\\begin{array}[t]{rrrrr}\r\nR\\phantom{-b}&=&aT&+&b \\\\\r\n-b&&&-&b \\\\\r\n\\hline\r\n\\dfrac{R-b}{a}&=&\\dfrac{aT}{a}&& \\\\ \\\\\r\nT&=&\\dfrac{R-b}{a}&&\r\n\\end{array}[\/latex]<\/li>\r\n \t
  31. [latex]\\phantom{a}[\/latex]\r\n[latex]\\begin{array}[t]{rrl}\r\nQ_1\\phantom{+PQ_1}&=&PQ_2-PQ_1 \\\\\r\n+PQ_1&&\\phantom{PQ_2}+PQ_1 \\\\\r\n\\hline\r\n\\dfrac{Q_1+PQ_1}{P}&=&\\dfrac{PQ_2}{P} \\\\ \\\\\r\nQ_2&=&\\dfrac{Q_1+PQ_1}{P}\r\n\\end{array}[\/latex]<\/li>\r\n \t
  32. [latex]\\phantom{a}[\/latex]\r\n[latex]\\begin{array}[t]{rrl}\r\nL\\phantom{-\\pi r_2-2d}&=&\\pi r_1+\\pi r_2+2d \\\\\r\n-\\pi r_2-2d&&\\phantom{\\pi r_1}-\\pi r_2-2d \\\\\r\n\\hline\r\n\\dfrac{L-\\pi r_2-2d}{\\pi}&=&\\dfrac{\\pi r_1}{\\pi} \\\\ \\\\\r\nr_1&=&\\dfrac{L-\\pi r_2-2d}{\\pi}\r\n\\end{array}[\/latex]<\/li>\r\n \t
  33. [latex]\\left(R=\\dfrac{kA(T+T_1)}{d}\\right)\\left(\\dfrac{d}{kA}\\right)[\/latex]\r\n[latex]\\begin{array}{rrl}\r\n\\dfrac{Rd}{kA}\\phantom{-T}&=&\\phantom{-}T+T_1 \\\\\r\n-T&&-T \\\\\r\n\\hline\r\nT_1&=&\\dfrac{Rd}{kA}-T\r\n\\end{array}[\/latex]<\/li>\r\n \t
  34. [latex]\\left(P=\\dfrac{V_1(V_2-V_1)}{g}\\right)\\left(\\dfrac{g}{V_1}\\right)[\/latex]\r\n[latex]\\begin{array}{rrl}\r\n\\dfrac{Pg}{V_1}\\phantom{+V_1}&=&V_2-V_1 \\\\\r\n+V_1&&\\phantom{V_2}+V_1 \\\\\r\n\\hline\r\nV_2&=&\\dfrac{Pg}{V_1}+V_1\r\n\\end{array}[\/latex]<\/li>\r\n<\/ol>","rendered":"
      \n
    1. (3) + 1 + (4) \u2212 (1)
      \n3 + 1 + 4 \u2212 1
      \n7<\/li>\n
    2. (5)2<\/sup> + (5) \u2212 (1)
      \n25 + 5 \u2212 1
      \n29<\/li>\n
    3. (6) \u2212 [(6)(5)\u00a0\u00f7 6]
      \n6 \u2212 [30 \u00f7 6]
      \n6 \u2212 5
      \n1<\/li>\n
    4. [6 + (4) \u2212 (1)] \u00f7 3
      \n[6 + 4 \u2212 1] \u00f7 3
      \n9 \u00f7 3
      \n3<\/li>\n
    5. (5)2<\/sup> \u2212 ((3) \u2212 1)
      \n25 \u2212 (3 \u2212 1)
      \n25 \u2212 2
      \n23<\/li>\n
    6. (6) + 6(4) \u2212 4(4)
      \n6 + 24 \u2212 16
      \n14<\/li>\n
    7. 5(4) + (2)(5) \u00f7 2
      \n20 + 10 \u00f7 2
      \n20 + 5
      \n25<\/li>\n
    8. 5((6) + (2)) + 1 + (5)
      \n5(6 + 2) + 1 + 5
      \n5(8) + 6
      \n40 + 6
      \n46<\/li>\n
    9. [4 \u2212 ((6) \u2212 (4))] \u00f7 2 + (6)
      \n[4 \u2212 (6\u00a0\u2212 4)] \u00f7 2 + 6
      \n[4 \u2212 2] \u00f7 2 + 6
      \n2 \u00f7 2 + 6
      \n1 + 6
      \n7<\/li>\n
    10. (4) + (5) \u2212 1
      \n4 + 5 \u2212 1
      \n8<\/li>\n
    11. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrl} \\dfrac{ab}{a}&=&\\dfrac{c}{a} \\\\ \\\\ b&=&\\dfrac{c}{a} \\end{array}[\/latex]<\/li>\n
    12. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{l} \\left(g=\\dfrac{h}{i}\\right)(i) \\\\ \\\\ h=gi \\end{array}[\/latex]<\/li>\n
    13. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{l} \\left(\\left(\\dfrac{f}{g}\\right)x=b\\right)\\dfrac{g}{f} \\\\ \\\\ x=\\dfrac{bg}{f} \\end{array}[\/latex]<\/li>\n
    14. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{l} \\left(p=\\dfrac{3y}{q}\\right)\\left(\\dfrac{q}{3}\\right) \\\\ \\\\ y=\\dfrac{pq}{3} \\end{array}[\/latex]<\/li>\n
    15. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{l} \\left(3x=\\dfrac{a}{b}\\right)\\left(\\dfrac{1}{3}\\right) \\\\ \\\\ x=\\dfrac{a}{3b} \\end{array}[\/latex]<\/li>\n
    16. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{l} \\left(\\dfrac{ym}{b}=\\dfrac{c}{d}\\right)\\left(\\dfrac{b}{m}\\right) \\\\ \\\\ y=\\dfrac{bc}{dm} \\end{array}[\/latex]<\/li>\n
    17. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{l} \\left(V=\\dfrac{4}{3}\\pi r^3\\right)\\left(\\dfrac{3}{4r^3}\\right) \\\\ \\\\ \\pi=\\dfrac{3V}{4r^3} \\end{array}[\/latex]<\/li>\n
    18. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{l} \\left(E=mv^2\\right)\\left(\\dfrac{1}{v^2}\\right) \\\\ \\\\ m=\\dfrac{E}{v^2} \\end{array}[\/latex]<\/li>\n
    19. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{l} \\left(c=\\dfrac{4y}{m+n}\\right)\\left(\\dfrac{m+n}{4}\\right) \\\\ \\\\ y=\\dfrac{c(m+n)}{4} \\end{array}[\/latex]<\/li>\n
    20. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{l} \\left(\\dfrac{rs}{a-3}=k\\right)\\left(\\dfrac{a-3}{s}\\right) \\\\ \\\\ r=\\dfrac{k(a-3)}{s} \\end{array}[\/latex]<\/li>\n
    21. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{l} \\left(V=\\dfrac{\\pi Dn}{12}\\right)\\left(\\dfrac{12}{\\pi n}\\right) \\\\ \\\\ D=\\dfrac{12V}{\\pi n} \\end{array}[\/latex]<\/li>\n
    22. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrl} F\\phantom{+kL}&=&kR-kL \\\\ +kL&&\\phantom{kR}+kL \\\\ \\hline \\dfrac{F+kL}{k}&=&\\dfrac{kR}{k} \\\\ \\\\ R&=&\\dfrac{F+kL}{k} \\end{array}[\/latex]<\/li>\n
    23. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrl} P\\phantom{-np}&=&\\phantom{-}np-nc \\\\ -np&&-np \\\\ \\hline \\dfrac{P-np}{-n}&=&\\dfrac{-nc}{-n} \\\\ \\\\ c&=&\\dfrac{P-np}{-n} \\end{array}[\/latex]<\/li>\n
    24. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrrrr} S\\phantom{-2B}&=&L&+&2B \\\\ -2B&&&-&2B \\\\ \\hline L&=&S&-&2B \\end{array}[\/latex]<\/li>\n
    25. [latex]\\left(T=\\dfrac{D-d}{L}\\right)(L)[\/latex]
      \n[latex]\\begin{array}{rrrrr} TL\\phantom{+d}&=&D&-&d \\\\ +d&&&+&d \\\\ \\hline D&=&TL&+&d \\end{array}[\/latex]<\/li>\n
    26. [latex]\\left(I=\\dfrac{E_a-E_q}{R}\\right)(R)[\/latex]
      \n[latex]\\begin{array}{rrrrr} IR&=&E_a&-&E_q \\\\ +E_q&&&+&E_q \\\\ \\hline E_a&=&IR&+&E_q \\end{array}[\/latex]<\/li>\n
    27. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrl} \\dfrac{L}{1+at}&=&\\dfrac{L_o(1+at)}{1+at} \\\\ \\\\ L_o&=&\\dfrac{L}{1+at} \\end{array}[\/latex]<\/li>\n
    28. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrl} 2m+p&=&\\phantom{-}4m+q \\\\ -2m-q&&-2m-q \\\\ \\hline \\dfrac{p-q}{2}&=&\\dfrac{2m}{2} \\\\ \\\\ m&=&\\dfrac{p-q}{2} \\end{array}[\/latex]<\/li>\n
    29. [latex]\\left(\\dfrac{k-m}{r}=q\\right)(r)[\/latex]
      \n[latex]\\begin{array}{rrl} k-m&=&qr \\\\ +m&&\\phantom{qr}+m \\\\ \\hline k&=&qr+m \\end{array}[\/latex]<\/li>\n
    30. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrrrr} R\\phantom{-b}&=&aT&+&b \\\\ -b&&&-&b \\\\ \\hline \\dfrac{R-b}{a}&=&\\dfrac{aT}{a}&& \\\\ \\\\ T&=&\\dfrac{R-b}{a}&& \\end{array}[\/latex]<\/li>\n
    31. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrl} Q_1\\phantom{+PQ_1}&=&PQ_2-PQ_1 \\\\ +PQ_1&&\\phantom{PQ_2}+PQ_1 \\\\ \\hline \\dfrac{Q_1+PQ_1}{P}&=&\\dfrac{PQ_2}{P} \\\\ \\\\ Q_2&=&\\dfrac{Q_1+PQ_1}{P} \\end{array}[\/latex]<\/li>\n
    32. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrl} L\\phantom{-\\pi r_2-2d}&=&\\pi r_1+\\pi r_2+2d \\\\ -\\pi r_2-2d&&\\phantom{\\pi r_1}-\\pi r_2-2d \\\\ \\hline \\dfrac{L-\\pi r_2-2d}{\\pi}&=&\\dfrac{\\pi r_1}{\\pi} \\\\ \\\\ r_1&=&\\dfrac{L-\\pi r_2-2d}{\\pi} \\end{array}[\/latex]<\/li>\n
    33. [latex]\\left(R=\\dfrac{kA(T+T_1)}{d}\\right)\\left(\\dfrac{d}{kA}\\right)[\/latex]
      \n[latex]\\begin{array}{rrl} \\dfrac{Rd}{kA}\\phantom{-T}&=&\\phantom{-}T+T_1 \\\\ -T&&-T \\\\ \\hline T_1&=&\\dfrac{Rd}{kA}-T \\end{array}[\/latex]<\/li>\n
    34. [latex]\\left(P=\\dfrac{V_1(V_2-V_1)}{g}\\right)\\left(\\dfrac{g}{V_1}\\right)[\/latex]
      \n[latex]\\begin{array}{rrl} \\dfrac{Pg}{V_1}\\phantom{+V_1}&=&V_2-V_1 \\\\ +V_1&&\\phantom{V_2}+V_1 \\\\ \\hline V_2&=&\\dfrac{Pg}{V_1}+V_1 \\end{array}[\/latex]<\/li>\n<\/ol>\n","protected":false},"author":90,"menu_order":20,"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-1596","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\/1596","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":3,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter\/1596\/revisions"}],"predecessor-version":[{"id":2197,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter\/1596\/revisions\/2197"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter\/1596\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/media?parent=1596"}],"wp:term":[{"taxonomy":"back-matter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter-type?post=1596"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/contributor?post=1596"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/license?post=1596"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}