{"id":2020,"date":"2021-12-02T19:40:30","date_gmt":"2021-12-03T00:40:30","guid":{"rendered":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/back-matter\/answer-key-11-3\/"},"modified":"2022-11-02T10:39:11","modified_gmt":"2022-11-02T14:39:11","slug":"answer-key-11-3","status":"publish","type":"back-matter","link":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/back-matter\/answer-key-11-3\/","title":{"raw":"Answer Key 11.3","rendered":"Answer Key 11.3"},"content":{"raw":"
    \n \t
  1. [latex](g\\circ f)(x)=-(\\sqrt[5]{-x-3})^5-3[\/latex]\n[latex]\\begin{array}[t]{rrl}\n\\phantom{(g\\circ f)x}&=&-(-x-3)-3 \\\\\n&=&x+3-3 \\\\\n&=&x\\hspace{0.75in}\\text{Inverse}\n\\end{array}[\/latex]<\/li>\n \t
  2. [latex](g\\circ f)(x)=4-\\left(\\dfrac{4}{x}\\right)[\/latex]\n[latex]\\phantom{(g\\circ f)(x)}=4-\\dfrac{4}{x}\\hspace{0.5in}\\text{Not inverse}[\/latex]<\/li>\n \t
  3. [latex](g\\circ f)(x)=-10\\left(\\dfrac{x-5}{10}\\right)+5[\/latex]\n[latex]\\begin{array}[t]{rrl}\n\\phantom{(g\\circ f)x}&=& -x+5+5\\\\ \\\\\n\\phantom{(g\\circ f)x}&=&-x+10\\hspace{0.75in}\\text{Not inverse}\n\\end{array}[\/latex]<\/li>\n \t
  4. [latex](f\\circ g)(x)=\\dfrac{(10x+5)-5}{10} [\/latex]\n[latex]\\begin{array}[t]{rrl}\n\\phantom{(f\\circ g)x}&=&\\dfrac{10x+5-5}{10} \\\\ \\\\\n&=&\\dfrac{10x}{10} \\\\ \\\\\n&=&x\\hspace{0.75in}\\text{Inverse}\n\\end{array}[\/latex]<\/li>\n \t
  5. [latex](f\\circ g)(x)=\\dfrac{-2}{\\dfrac{3x+2}{x+2}+3}[\/latex]\n[latex]\\begin{array}[t]{rrl}\n\\phantom{(f\\circ g)x}&=& \\dfrac{-2(x+2)}{3x+2+3(x+2)}\\\\ \\\\\n&=& \\dfrac{-2x-4}{3x+2+3x+6}\\\\ \\\\\n&=& \\dfrac{-2x-4}{6x+8}\\\\ \\\\\n&=& \\dfrac{-x-2}{3x+4}\\hspace{0.75in}\\text{Not inverse}\n\\end{array}[\/latex]<\/li>\n \t
  6. [latex](f\\circ g)=\\dfrac{-\\left(\\dfrac{-2x+1}{-x-1}\\right)-1}{\\dfrac{-2x+1}{-x-1}-2}[\/latex]\n[latex]\\begin{array}[t]{rrl}\n\\phantom{(f\\circ g)}&=&\\dfrac{-(-2x+1)-1(-x-1)}{-2x+1-2(-x-1)} \\\\ \\\\\n&=&\\dfrac{2x-1+x+1}{-2x+1+2x+2} \\\\ \\\\\n&=&\\dfrac{3x}{3} \\\\ \\\\\n&=&x\\hspace{0.75in}\\text{Inverse}\n\\end{array}[\/latex]<\/li>\n \t
  7. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrcrr}\ny&=&(x-2)^5&+&3 \\\\\nx&=&(y-2)^5&+&3 \\\\\n-3&&&-&3 \\\\\n\\hline\nx-3&=&(y-2)^5&& \\\\\n\\sqrt[5]{x-3}&=&y-2&& \\\\ \\\\\ny&=&\\sqrt[5]{x-3}&+&2\n\\end{array}[\/latex]<\/li>\n \t
  8. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrcrr}\ny&=&\\sqrt[3]{x+1}&+&2 \\\\\nx&=&\\sqrt[3]{y+1}&+&2 \\\\\n-2&&&-&2 \\\\\n\\hline\nx-2&=&\\sqrt[3]{y+1}&& \\\\\n(x-2)^3&=&y+1&& \\\\ \\\\\ny&=&(x-2)^3&-&1\n\\end{array}[\/latex]<\/li>\n \t
  9. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrl}\ny&=&\\dfrac{4}{x+2} \\\\ \\\\\nx&=&\\dfrac{4}{y+2} \\\\ \\\\\ny+2&=&\\dfrac{4}{x} \\\\ \\\\\ny&=&\\dfrac{4}{x}-2\n\\end{array}[\/latex]<\/li>\n \t
  10. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrl}\ny&=&\\dfrac{3}{x-3} \\\\ \\\\\nx&=&\\dfrac{3}{y-3} \\\\ \\\\\ny-3&=&\\dfrac{3}{x} \\\\ \\\\\ny&=&\\dfrac{3}{x}+3\n\\end{array}[\/latex]<\/li>\n \t
  11. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrl}\ny&=&\\dfrac{-2x-2}{x+2} \\\\ \\\\\nx&=&\\dfrac{-2y-2}{y+2} \\\\ \\\\\nx(y+2)&=&-2y-2 \\\\\nxy+2x&=&-2y-2 \\\\\n-xy+2&&-xy+2 \\\\\n\\hline\n2x+2&=&-2y-xy \\\\\n2x+2&=&y(-2-x) \\\\ \\\\\ny&=&\\dfrac{2x+2}{-2-x} \\\\ \\\\\ny&=&-\\dfrac{2x+2}{2+x}\n\\end{array}[\/latex]<\/li>\n \t
  12. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrl}\ny&=&\\dfrac{9+x}{3} \\\\ \\\\\nx&=&\\dfrac{9+y}{3} \\\\ \\\\\n3x&=&9+y \\\\\ny&=&3x-9\n\\end{array}[\/latex]<\/li>\n \t
  13. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrl}\ny&=&\\dfrac{10-x}{5} \\\\ \\\\\nx&=&\\dfrac{10-y}{5} \\\\ \\\\\n5x&=&10-y \\\\\ny&=&10-5x\n\\end{array}[\/latex]<\/li>\n \t
  14. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrl}\ny&=&\\dfrac{5x-15}{2} \\\\ \\\\\nx&=&\\dfrac{5y-15}{2} \\\\ \\\\\n5y-15&=&2x \\\\ \\\\\n5y&=&2x+15 \\\\ \\\\\ny&=&\\dfrac{2x+15}{5}\n\\end{array}[\/latex]<\/li>\n \t
  15. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrl}\ny&=&-(x-1)^3 \\\\\nx&=&-(y-1)^3 \\\\\n\\sqrt[3]{x}&=&-(y-1) \\\\\n\\sqrt[3]{x}&=&-y+1 \\\\\n-y&=&\\sqrt[3]{x}-1 \\\\ \\\\\ny&=&1-\\sqrt[3]{x}\n\\end{array}[\/latex]<\/li>\n \t
  16. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrl}\ny&=&\\dfrac{12-3x}{4} \\\\ \\\\\nx&=&\\dfrac{12-3y}{4} \\\\ \\\\\n4x&=&12-3y \\\\ \\\\\n3y&=&12-4x \\\\ \\\\\ny&=&\\dfrac{12-4x}{3} \\\\ \\\\\ny&=&4-\\dfrac{4}{3}x\n\\end{array}[\/latex]<\/li>\n \t
  17. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrl}\ny&=&(x-3)^3 \\\\\nx&=&(y-3)^3 \\\\\n\\sqrt[3]{x}&=&y-3 \\\\ \\\\\ny&=&\\sqrt[3]{x}+3\n\\end{array}[\/latex]<\/li>\n \t
  18. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrl}\ny&=&\\sqrt[5]{-x}+2 \\\\\nx&=&\\sqrt[5]{-y}+2 \\\\\nx-2&=&\\sqrt[5]{-y} \\\\\n(x-2)^5&=&-y \\\\ \\\\\ny&=&-(x-2)^5\n\\end{array}[\/latex]<\/li>\n \t
  19. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrl}\ny&=&\\dfrac{x}{x-1} \\\\ \\\\\nx&=&\\dfrac{y}{y-1} \\\\ \\\\\nx(y-1)&=&y \\\\ \\\\\nxy-x&=&y \\\\ \\\\\ny-xy&=&-x \\\\ \\\\\ny(1-x)&=&-x \\\\ \\\\\ny&=&\\dfrac{-x}{1-x} \\\\ \\\\\ny&=&\\dfrac{x}{x-1}\n\\end{array}[\/latex]<\/li>\n \t
  20. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrl}\ny&=&\\dfrac{-3-2x}{x+3} \\\\ \\\\\nx&=&\\dfrac{-3-2y}{y+3} \\\\ \\\\\nx(y+3)&=&-3-2y \\\\\nxy+3x&=&-3-2y \\\\\n+2y-3x&&-3x+2y \\\\\n\\hline\nxy+2y&=&-3-3x \\\\\ny(x+2)&=&-3-3x \\\\ \\\\\ny&=&\\dfrac{-3-3x}{x+2} \\\\ \\\\\ny&=&-\\dfrac{3x+3}{x+2}\n\\end{array}[\/latex]<\/li>\n \t
  21. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrl}\ny&=&\\dfrac{x-1}{x+1} \\\\ \\\\\nx&=&\\dfrac{y-1}{y+1} \\\\ \\\\\nx(y+1)&=&y-1 \\\\\nxy+x&=&y-1 \\\\\nxy-y&=&-x-1 \\\\\ny(x-1)&=&-x-1 \\\\ \\\\\ny&=&\\dfrac{-x-1}{x-1} \\\\ \\\\\ny&=&-\\dfrac{x+1}{x-1}\n\\end{array}[\/latex]<\/li>\n \t
  22. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrl}\ny&=&\\dfrac{x}{x+2} \\\\ \\\\\nx&=&\\dfrac{y}{y+2} \\\\ \\\\\nx(y+2)&=&y \\\\\nxy+2x&=&y \\\\\n2x&=&y-xy \\\\\n2x&=&y(1-x) \\\\ \\\\\ny&=&\\dfrac{2x}{1-x}\n\\end{array}[\/latex]<\/li>\n<\/ol>","rendered":"
      \n
    1. [latex](g\\circ f)(x)=-(\\sqrt[5]{-x-3})^5-3[\/latex]
      \n[latex]\\begin{array}[t]{rrl} \\phantom{(g\\circ f)x}&=&-(-x-3)-3 \\\\ &=&x+3-3 \\\\ &=&x\\hspace{0.75in}\\text{Inverse} \\end{array}[\/latex]<\/li>\n
    2. [latex](g\\circ f)(x)=4-\\left(\\dfrac{4}{x}\\right)[\/latex]
      \n[latex]\\phantom{(g\\circ f)(x)}=4-\\dfrac{4}{x}\\hspace{0.5in}\\text{Not inverse}[\/latex]<\/li>\n
    3. [latex](g\\circ f)(x)=-10\\left(\\dfrac{x-5}{10}\\right)+5[\/latex]
      \n[latex]\\begin{array}[t]{rrl} \\phantom{(g\\circ f)x}&=& -x+5+5\\\\ \\\\ \\phantom{(g\\circ f)x}&=&-x+10\\hspace{0.75in}\\text{Not inverse} \\end{array}[\/latex]<\/li>\n
    4. [latex](f\\circ g)(x)=\\dfrac{(10x+5)-5}{10}[\/latex]
      \n[latex]\\begin{array}[t]{rrl} \\phantom{(f\\circ g)x}&=&\\dfrac{10x+5-5}{10} \\\\ \\\\ &=&\\dfrac{10x}{10} \\\\ \\\\ &=&x\\hspace{0.75in}\\text{Inverse} \\end{array}[\/latex]<\/li>\n
    5. [latex](f\\circ g)(x)=\\dfrac{-2}{\\dfrac{3x+2}{x+2}+3}[\/latex]
      \n[latex]\\begin{array}[t]{rrl} \\phantom{(f\\circ g)x}&=& \\dfrac{-2(x+2)}{3x+2+3(x+2)}\\\\ \\\\ &=& \\dfrac{-2x-4}{3x+2+3x+6}\\\\ \\\\ &=& \\dfrac{-2x-4}{6x+8}\\\\ \\\\ &=& \\dfrac{-x-2}{3x+4}\\hspace{0.75in}\\text{Not inverse} \\end{array}[\/latex]<\/li>\n
    6. [latex](f\\circ g)=\\dfrac{-\\left(\\dfrac{-2x+1}{-x-1}\\right)-1}{\\dfrac{-2x+1}{-x-1}-2}[\/latex]
      \n[latex]\\begin{array}[t]{rrl} \\phantom{(f\\circ g)}&=&\\dfrac{-(-2x+1)-1(-x-1)}{-2x+1-2(-x-1)} \\\\ \\\\ &=&\\dfrac{2x-1+x+1}{-2x+1+2x+2} \\\\ \\\\ &=&\\dfrac{3x}{3} \\\\ \\\\ &=&x\\hspace{0.75in}\\text{Inverse} \\end{array}[\/latex]<\/li>\n
    7. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrcrr} y&=&(x-2)^5&+&3 \\\\ x&=&(y-2)^5&+&3 \\\\ -3&&&-&3 \\\\ \\hline x-3&=&(y-2)^5&& \\\\ \\sqrt[5]{x-3}&=&y-2&& \\\\ \\\\ y&=&\\sqrt[5]{x-3}&+&2 \\end{array}[\/latex]<\/li>\n
    8. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrcrr} y&=&\\sqrt[3]{x+1}&+&2 \\\\ x&=&\\sqrt[3]{y+1}&+&2 \\\\ -2&&&-&2 \\\\ \\hline x-2&=&\\sqrt[3]{y+1}&& \\\\ (x-2)^3&=&y+1&& \\\\ \\\\ y&=&(x-2)^3&-&1 \\end{array}[\/latex]<\/li>\n
    9. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrl} y&=&\\dfrac{4}{x+2} \\\\ \\\\ x&=&\\dfrac{4}{y+2} \\\\ \\\\ y+2&=&\\dfrac{4}{x} \\\\ \\\\ y&=&\\dfrac{4}{x}-2 \\end{array}[\/latex]<\/li>\n
    10. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrl} y&=&\\dfrac{3}{x-3} \\\\ \\\\ x&=&\\dfrac{3}{y-3} \\\\ \\\\ y-3&=&\\dfrac{3}{x} \\\\ \\\\ y&=&\\dfrac{3}{x}+3 \\end{array}[\/latex]<\/li>\n
    11. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrl} y&=&\\dfrac{-2x-2}{x+2} \\\\ \\\\ x&=&\\dfrac{-2y-2}{y+2} \\\\ \\\\ x(y+2)&=&-2y-2 \\\\ xy+2x&=&-2y-2 \\\\ -xy+2&&-xy+2 \\\\ \\hline 2x+2&=&-2y-xy \\\\ 2x+2&=&y(-2-x) \\\\ \\\\ y&=&\\dfrac{2x+2}{-2-x} \\\\ \\\\ y&=&-\\dfrac{2x+2}{2+x} \\end{array}[\/latex]<\/li>\n
    12. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrl} y&=&\\dfrac{9+x}{3} \\\\ \\\\ x&=&\\dfrac{9+y}{3} \\\\ \\\\ 3x&=&9+y \\\\ y&=&3x-9 \\end{array}[\/latex]<\/li>\n
    13. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrl} y&=&\\dfrac{10-x}{5} \\\\ \\\\ x&=&\\dfrac{10-y}{5} \\\\ \\\\ 5x&=&10-y \\\\ y&=&10-5x \\end{array}[\/latex]<\/li>\n
    14. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrl} y&=&\\dfrac{5x-15}{2} \\\\ \\\\ x&=&\\dfrac{5y-15}{2} \\\\ \\\\ 5y-15&=&2x \\\\ \\\\ 5y&=&2x+15 \\\\ \\\\ y&=&\\dfrac{2x+15}{5} \\end{array}[\/latex]<\/li>\n
    15. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrl} y&=&-(x-1)^3 \\\\ x&=&-(y-1)^3 \\\\ \\sqrt[3]{x}&=&-(y-1) \\\\ \\sqrt[3]{x}&=&-y+1 \\\\ -y&=&\\sqrt[3]{x}-1 \\\\ \\\\ y&=&1-\\sqrt[3]{x} \\end{array}[\/latex]<\/li>\n
    16. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrl} y&=&\\dfrac{12-3x}{4} \\\\ \\\\ x&=&\\dfrac{12-3y}{4} \\\\ \\\\ 4x&=&12-3y \\\\ \\\\ 3y&=&12-4x \\\\ \\\\ y&=&\\dfrac{12-4x}{3} \\\\ \\\\ y&=&4-\\dfrac{4}{3}x \\end{array}[\/latex]<\/li>\n
    17. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrl} y&=&(x-3)^3 \\\\ x&=&(y-3)^3 \\\\ \\sqrt[3]{x}&=&y-3 \\\\ \\\\ y&=&\\sqrt[3]{x}+3 \\end{array}[\/latex]<\/li>\n
    18. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrl} y&=&\\sqrt[5]{-x}+2 \\\\ x&=&\\sqrt[5]{-y}+2 \\\\ x-2&=&\\sqrt[5]{-y} \\\\ (x-2)^5&=&-y \\\\ \\\\ y&=&-(x-2)^5 \\end{array}[\/latex]<\/li>\n
    19. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrl} y&=&\\dfrac{x}{x-1} \\\\ \\\\ x&=&\\dfrac{y}{y-1} \\\\ \\\\ x(y-1)&=&y \\\\ \\\\ xy-x&=&y \\\\ \\\\ y-xy&=&-x \\\\ \\\\ y(1-x)&=&-x \\\\ \\\\ y&=&\\dfrac{-x}{1-x} \\\\ \\\\ y&=&\\dfrac{x}{x-1} \\end{array}[\/latex]<\/li>\n
    20. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrl} y&=&\\dfrac{-3-2x}{x+3} \\\\ \\\\ x&=&\\dfrac{-3-2y}{y+3} \\\\ \\\\ x(y+3)&=&-3-2y \\\\ xy+3x&=&-3-2y \\\\ +2y-3x&&-3x+2y \\\\ \\hline xy+2y&=&-3-3x \\\\ y(x+2)&=&-3-3x \\\\ \\\\ y&=&\\dfrac{-3-3x}{x+2} \\\\ \\\\ y&=&-\\dfrac{3x+3}{x+2} \\end{array}[\/latex]<\/li>\n
    21. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrl} y&=&\\dfrac{x-1}{x+1} \\\\ \\\\ x&=&\\dfrac{y-1}{y+1} \\\\ \\\\ x(y+1)&=&y-1 \\\\ xy+x&=&y-1 \\\\ xy-y&=&-x-1 \\\\ y(x-1)&=&-x-1 \\\\ \\\\ y&=&\\dfrac{-x-1}{x-1} \\\\ \\\\ y&=&-\\dfrac{x+1}{x-1} \\end{array}[\/latex]<\/li>\n
    22. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrl} y&=&\\dfrac{x}{x+2} \\\\ \\\\ x&=&\\dfrac{y}{y+2} \\\\ \\\\ x(y+2)&=&y \\\\ xy+2x&=&y \\\\ 2x&=&y-xy \\\\ 2x&=&y(1-x) \\\\ \\\\ y&=&\\dfrac{2x}{1-x} \\end{array}[\/latex]<\/li>\n<\/ol>\n","protected":false},"author":90,"menu_order":109,"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-2020","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\/2020","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\/2020\/revisions"}],"predecessor-version":[{"id":2021,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter\/2020\/revisions\/2021"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter\/2020\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/media?parent=2020"}],"wp:term":[{"taxonomy":"back-matter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter-type?post=2020"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/contributor?post=2020"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/license?post=2020"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}