{"id":1816,"date":"2019-08-14T12:54:09","date_gmt":"2019-08-14T16:54:09","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/?post_type=back-matter&p=1816"},"modified":"2019-08-14T14:49:04","modified_gmt":"2019-08-14T18:49:04","slug":"answer-key-11-2","status":"publish","type":"back-matter","link":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/back-matter\/answer-key-11-2\/","title":{"raw":"Answer Key 11.2","rendered":"Answer Key 11.2"},"content":{"raw":"[latexpage]\r\n
    \r\n \t
  1. \\(\\begin{array}{ll}\r\n\\\\ \\\\\r\n\\begin{array}{rrl}\r\ng(3)&=&(3)^3+5(3)^2 \\\\\r\n&=&27+45 \\\\\r\n&=&72\r\n\\end{array}\r\n&\\hspace{0.25in}\r\n\\begin{array}{rrl}\r\nf(3)&=&2(3)+4 \\\\\r\n&=&6+4 \\\\\r\n&=&10 \\\\\r\n\\end{array}\r\n\\end{array}\\)\r\n\r\n\\(g(3)+f(3)=72+10=82\\)<\/li>\r\n \t
  2. \\(\\begin{array}{ll}\r\n\\\\ \\\\\r\n\\begin{array}{rrl}\r\n\\\\\r\nf(-4)&=&-3(-4)^2+3(-4) \\\\\r\n&=&-3(16)-12 \\\\\r\n&=&-48-12 \\\\\r\n&=&-60\r\n\\end{array}\r\n&\\hspace{0.25in}\r\n\\begin{array}{rrl}\r\ng(-4)&=&2(-4)+5 \\\\\r\n&=&-8+5 \\\\\r\n&=&-3\r\n\\end{array}\r\n\\end{array}\\)\r\n\r\n\\(\\dfrac{f(-4)}{g(-4)}=\\dfrac{-60}{-3}=20\\)<\/li>\r\n \t
  3. \\(\\begin{array}{ll}\r\n\\\\ \\\\\r\n\\begin{array}{rrl}\r\ng(5)&=&-4(5)+1 \\\\\r\n&=&-20+1 \\\\\r\n&=&-19\r\n\\end{array}\r\n& \\hspace{0.25in}\r\n\\begin{array}{rrl}\r\nh(5)&=&-2(5)-1 \\\\\r\n&=&-10-1 \\\\\r\n&=& -11\r\n\\end{array}\r\n\\end{array}\\)\r\n\r\n\\(g(5)+h(5)=-19-11=-30\\)<\/li>\r\n \t
  4. \\(\\begin{array}{ll}\r\n\\\\ \\\\\r\n\\begin{array}{rrl}\r\ng(2)&=&3(2)+1 \\\\\r\n&=&6+1 \\\\\r\n&=&7\r\n\\end{array}\r\n& \\hspace{0.25in}\r\n\\begin{array}{rrl}\r\n\\\\\r\nf(2)&=&(2)^3+3(2)^2 \\\\\r\n&=&8+3\\cdot 4 \\\\\r\n&=&8+12 \\\\\r\n&=&20\r\n\\end{array}\r\n\\end{array}\\)\r\n\r\n\\(g(2)\\cdot f(2)=7\\cdot 20=140\\)<\/li>\r\n \t
  5. \\(\\begin{array}{ll}\r\n\\\\\r\n\\begin{array}{rrl}\r\ng(1)&=&1-3 \\\\\r\n&=&-2\r\n\\end{array}\r\n&\\hspace{0.25in}\r\n\\begin{array}{rrl}\r\n\\\\\r\nh(1)&=&-3(1)^3+6(1) \\\\\r\n&=&-3+6 \\\\\r\n&=&3\r\n\\end{array}\r\n\\end{array}\\)\r\n\r\n\\(g(1)+h(1)=-2+3=1\\)<\/li>\r\n \t
  6. \\(\\begin{array}{ll}\r\n\\\\ \\\\\r\n\\begin{array}{rrl}\r\ng(-6)&=&(-6)^2-2 \\\\\r\n&=&36-2 \\\\\r\n&=&34\r\n\\end{array}\r\n& \\hspace{0.25in}\r\n\\begin{array}{rrl}\r\nh(-6)&=&2(-6)+5 \\\\\r\n&=&-12+5 \\\\\r\n&=&-7\r\n\\end{array}\r\n\\end{array}\\)\r\n\r\n\\(g(-6)+h(-6)=34-7=27\\)<\/li>\r\n \t
  7. \\(\\begin{array}{ll}\r\n\\\\\r\n\\begin{array}{rrl}\r\nh(0)&=&\\cancel{2(0)}-1 \\\\\r\n&=&-1\r\n\\end{array}\r\n& \\hspace{0.25in}\r\n\\begin{array}{rrl}\r\ng(0)&=&\\cancel{3(0)}-5 \\\\\r\n&=&-5\r\n\\end{array}\r\n\\end{array}\\)\r\n\r\n\\(\\dfrac{h(0)}{g(0)}=\\dfrac{-1}{-5}=\\dfrac{1}{5}\\)<\/li>\r\n \t
  8. \\((g+h)=\r\n\\begin{array}{rrrr}\r\n\\\\ \\\\\r\n&3a&-&2 \\\\\r\n+&4a&-&2 \\\\\r\n\\midrule\r\n&7a&-&4\r\n\\end{array}\\hspace{0.25in}\r\n\\begin{array}{rrl}\r\n\\\\ \\\\\r\n(g+h)(10)&=&7(10)-4 \\\\\r\n&=&70-4 \\\\\r\n&=&66\r\n\\end{array}\\)<\/li>\r\n \t
  9. \\((g+f)=\r\n\\begin{array}{rrrr}\r\n\\\\ \\\\\r\n&3a&+&3 \\\\\r\n+&2a&-&2 \\\\\r\n\\midrule\r\n&5a&+&1\r\n\\end{array}\\hspace{0.25in}\r\n\\begin{array}{rrl}\r\n\\\\ \\\\\r\n(g+f)(9)&=&5(9)+1 \\\\\r\n&=&45+1 \\\\\r\n&=&46\r\n\\end{array}\\)<\/li>\r\n \t
  10. \\((g-h)=\r\n\\begin{array}{r}\r\n\\\\ \\\\\r\n4x+3 \\\\\r\n- \\hspace{0.42in} (x^3-2x^2) \\\\\r\n\\midrule\r\n-x^3+2x^2+4x+3\r\n\\end{array}\\hspace{0.25in}\r\n\\begin{array}{rrl}\r\n\\\\ \\\\\r\n(g-h)(-1)&=&-(-1)^3+2(-1)^2+4(-1)+3 \\\\\r\n&=&1+2-4+3 \\\\\r\n&=&2\r\n\\end{array}\\)<\/li>\r\n \t
  11. \\((g-f)=\r\n\\begin{array}{rrrr}\r\n\\\\ \\\\\r\n&x&+&3 \\\\\r\n-&(-x&+&4) \\\\\r\n\\midrule\r\n&2x&-&1\r\n\\end{array}\\hspace{0.25in}\r\n\\begin{array}{rrl}\r\n\\\\ \\\\\r\n(g-f)(3)&=&2(3)-1 \\\\\r\n&=&6-1 \\\\\r\n&=&5\r\n\\end{array}\\)<\/li>\r\n \t
  12. \\((g-f)=\r\n\\begin{array}{rrrrrr}\r\n\\\\ \\\\\r\n&x^2&&&+&2 \\\\\r\n-&&&(2x&+&5) \\\\\r\n\\midrule\r\n&x^2&-&2x&-&3\r\n\\end{array}\\hspace{0.25in}\r\n\\begin{array}{rrl}\r\n\\\\ \\\\\r\n(g-f)(0)&=&\\cancel{(0)^2}-\\cancel{2(0)}-3 \\\\\r\n&=&-3\r\n\\end{array}\\)<\/li>\r\n \t
  13. \\((f+g)=\r\n\\begin{array}{rrrr}\r\n\\\\ \\\\\r\n&n&-&5 \\\\\r\n+&4n&+&2 \\\\\r\n\\midrule\r\n&5n&-&3\r\n\\end{array}\\hspace{0.25in}\r\n\\begin{array}{rrl}\r\n\\\\ \\\\\r\n(f+g)(-8)&=&5(-8)-3 \\\\\r\n&=&-40-3 \\\\\r\n&=&-43\r\n\\end{array}\\)<\/li>\r\n \t
  14. \\((h\\cdot g)=\r\n\\begin{array}{rrrrrr}\r\n\\\\ \\\\ \\\\ \\\\ \\\\\r\n&&&t&+&5 \\\\\r\n\\times &&&3t&-&5 \\\\\r\n\\midrule\r\n&3t^2&+&15t&& \\\\\r\n&&-&5t&-&25 \\\\\r\n\\midrule\r\n&3t^2&+&10t&-&25\r\n\\end{array}\\hspace{0.25in}\r\n\\begin{array}{rrl}\r\n\\\\ \\\\\r\n(h\\cdot g)(5)&=&3(5)^2+10(5)-25 \\\\\r\n&=&75+50-25 \\\\\r\n&=&100\r\n\\end{array}\\)<\/li>\r\n \t
  15. \\((g\\cdot h)=\r\n\\begin{array}{rrrr}\r\n\\\\ \\\\\r\n&t&-&4 \\\\\r\n\\times &&&2t \\\\\r\n\\midrule\r\n&2t^2&-&8t\r\n\\end{array}\\hspace{0.25in}\r\n\\begin{array}{rrl}\r\n\\\\ \\\\\r\n(g\\cdot h)(3t)&=&2(3t)^2-8(3t) \\\\\r\n&=&2(9t^2)-24t \\\\\r\n&=&18t^2-24t\r\n\\end{array}\\)<\/li>\r\n \t
  16. \\(\\dfrac{g(n)}{f(n)}=\\dfrac{n^2+5}{2n+5}\\hspace{0.25in} \\text{Does not reduce}\\)<\/li>\r\n \t
  17. \\(\\dfrac{g}{f}=\\dfrac{-2a+5}{3a+5}\\hspace{0.3in}\\left(\\dfrac{g}{f}\\right)(a^2)=\\dfrac{-2a^2+5}{3a^2+5}\\hspace{0.25in} \\text{Does not reduce}\\)<\/li>\r\n \t
  18. \\(h(n)+g(n)=\r\n\\begin{array}{rrrrrr}\r\n\\\\ \\\\\r\n&n^3&+&4n&& \\\\\r\n+&&&4n&+&5 \\\\\r\n\\midrule\r\n&n^3&+&8n&+&5\r\n\\end{array}\\)<\/li>\r\n \t
  19. \\(\\begin{array}{rrl}\r\n\\\\\r\ng(n^2)&=&(n^2)^2-4(n^2) \\\\\r\n&=&n^4-4n^2\r\n\\end{array}\\hspace{0.25in}\r\nh(n^2)=n^2-5\\)\r\n\r\n\\(g(n^2)\\cdot h(n^2)=\r\n\\begin{array}{rrrrrr}\r\n\\\\ \\\\ \\\\\r\n&&&n^4&-&4n^2 \\\\\r\n\\times&&&n^2&-&5 \\\\\r\n\\midrule\r\n&n^6&-&4n^4&& \\\\\r\n&&-&5n^4&+&20n^2 \\\\\r\n\\midrule\r\n&n^6&-&9n^4&+&20n^2 \\\\\r\n\\end{array}\\)<\/li>\r\n \t
  20. \\((g\\cdot h)=\r\n\\begin{array}{rrrrrr}\r\n\\\\ \\\\ \\\\\r\n&&&n&+&5 \\\\\r\n\\times &&&2n&-&5 \\\\\r\n\\midrule\r\n&2n^2&+&10n&& \\\\\r\n&&-&5n&-&25 \\\\\r\n\\midrule\r\n&2n^2&+&5n&-&25\r\n\\end{array}\\hspace{0.25in}\r\n\\begin{array}{rrl}\r\n(g\\cdot h)(-3n)&=&2(-3n)^2+5(-3n)-25 \\\\\r\n&=&2(9n^2)-15n-25 \\\\\r\n&=&18n^2-15n-25\r\n\\end{array}\\)<\/li>\r\n \t
  21. \\(\\begin{array}{ll}\r\n\\\\ \\\\\r\n\\begin{array}{rrl}\r\n(f\\circ g)&=&-4(4x+3)+1 \\\\\r\n&=&-16x-12+1 \\\\\r\n&=&-16x-11\r\n\\end{array}\r\n&\\hspace{0.25in}\r\n\\begin{array}{rrl}\r\n(f\\circ g)(9)&=&-16(9)-11 \\\\\r\n&=&-144-11 \\\\\r\n&=&-155\r\n\\end{array}\r\n\\end{array}\\)<\/li>\r\n \t
  22. \\(\\begin{array}{ll}\r\n\\\\ \\\\\r\n\\begin{array}{rrl}\r\n(h\\circ g)&=&3(a+1)+3 \\\\\r\n&=&3a+3+3 \\\\\r\n&=&3a+6\r\n\\end{array}\r\n&\\hspace{0.25in}\r\n\\begin{array}{rrl}\r\n(h\\circ g)(5)&=&3(5)+6 \\\\\r\n&=&15+6 \\\\\r\n&=&21\r\n\\end{array}\r\n\\end{array}\\)<\/li>\r\n \t
  23. \\(\\begin{array}{ll}\r\n\\\\ \\\\\r\n\\begin{array}{rrl}\r\n(g\\circ h)&=&(x^2-1)+4 \\\\\r\n&=&x^2-1+4 \\\\\r\n&=&x^2+3\r\n\\end{array}\r\n&\\hspace{0.25in}\r\n\\begin{array}{rrl}\r\n(g\\circ h)(10)&=&(10)^2+3 \\\\\r\n&=&100+3 \\\\\r\n&=&103\r\n\\end{array}\r\n\\end{array}\\)<\/li>\r\n \t
  24. \\(\\begin{array}{ll}\r\n\\\\ \\\\\r\n\\begin{array}{rrl}\r\n(f\\circ g)&=&-4(n+4)+2 \\\\\r\n&=&-4n-16+2 \\\\\r\n&=&-4n-14\r\n\\end{array}\r\n&\\hspace{0.25in}\r\n\\begin{array}{rrl}\r\n(f\\circ g)(9)&=&-4(9)-14 \\\\\r\n&=&-36-14 \\\\\r\n&=&-50\r\n\\end{array}\r\n\\end{array}\\)<\/li>\r\n \t
  25. \\(\\begin{array}{ll}\r\n\\begin{array}{rrl}\r\n\\\\\r\n(g\\circ h)&=&2(2x^3+4x^2)-4 \\\\\r\n&=&4x^3+8x^2-4\r\n\\end{array}\r\n&\\hspace{0.25in}\r\n\\begin{array}{rrl}\r\n\\\\ \\\\\r\n(g\\circ h)(3)&=&4(3)^3+8(3)^2-4 \\\\\r\n&=&108+72-4 \\\\\r\n&=&176\r\n\\end{array}\r\n\\end{array}\\)<\/li>\r\n \t
  26. \\(\\begin{array}{rrl}\r\n\\\\ \\\\\r\n(g\\circ h)&=&(4x+4)^2-5(4x+4) \\\\\r\n&=&16x^2+32x+16-20x-20 \\\\\r\n&=&16x^2+12x-4\r\n\\end{array}\\)<\/li>\r\n \t
  27. \\(\\begin{array}{rrl}\r\n\\\\\r\n(f\\circ g)&=&-2(4a)+2 \\\\\r\n&=&-8a+2\r\n\\end{array}\\)<\/li>\r\n \t
  28. \\(\\begin{array}{rrl}\r\n\\\\ \\\\\r\n(g\\circ f)&=&4(x^3-1)+4 \\\\\r\n&=&4x^3-4+4 \\\\\r\n& =&4x^3\r\n\\end{array}\\)<\/li>\r\n \t
  29. \\(\\begin{array}{rrl}\r\n\\\\ \\\\\r\n(g\\circ f)&=&-(2x-3)+5 \\\\\r\n&=&-2x+6+5 \\\\\r\n&=&-2x+11\r\n\\end{array}\\)<\/li>\r\n \t
  30. \\(\\begin{array}{rrl}\r\n\\\\ \\\\\r\n(f\\circ g)&=&4(-4t-2)+3 \\\\\r\n&=&-16t-8+3 \\\\\r\n&=&-16t-5\r\n\\end{array}\\)<\/li>\r\n<\/ol>","rendered":"
      \n
    1. \"\begin{array}{ll}\n

      \"g(3)+f(3)=72+10=82\"<\/li>\n

    2. \"\begin{array}{ll}\n

      \"\dfrac{f(-4)}{g(-4)}=\dfrac{-60}{-3}=20\"<\/li>\n

    3. \"\begin{array}{ll}\n

      \"g(5)+h(5)=-19-11=-30\"<\/li>\n

    4. \"\begin{array}{ll}\n

      \"g(2)\cdot f(2)=7\cdot 20=140\"<\/li>\n

    5. \"\begin{array}{ll}\n

      \"g(1)+h(1)=-2+3=1\"<\/li>\n

    6. \"\begin{array}{ll}\n

      \"g(-6)+h(-6)=34-7=27\"<\/li>\n

    7. \"\begin{array}{ll}\n

      \"\dfrac{h(0)}{g(0)}=\dfrac{-1}{-5}=\dfrac{1}{5}\"<\/li>\n

    8. \"(g+h)=<\/li>\n
    9. \"(g+f)=<\/li>\n
    10. \"(g-h)=<\/li>\n
    11. \"(g-f)=<\/li>\n
    12. \"(g-f)=<\/li>\n
    13. \"(f+g)=<\/li>\n
    14. \"(h\cdot g)=<\/li>\n
    15. \"(g\cdot h)=<\/li>\n
    16. \"\dfrac{g(n)}{f(n)}=\dfrac{n^2+5}{2n+5}\hspace{0.25in} \text{Does not reduce}\"<\/li>\n
    17. \"\dfrac{g}{f}=\dfrac{-2a+5}{3a+5}\hspace{0.3in}\left(\dfrac{g}{f}\right)(a^2)=\dfrac{-2a^2+5}{3a^2+5}\hspace{0.25in} \text{Does not reduce}\"<\/li>\n
    18. \"h(n)+g(n)=<\/li>\n
    19. \"\begin{array}{rrl}\n

      \"g(n^2)\cdot h(n^2)=<\/li>\n

    20. \"(g\cdot h)=<\/li>\n
    21. \"\begin{array}{ll}<\/li>\n
    22. \"\begin{array}{ll}<\/li>\n
    23. \"\begin{array}{ll}<\/li>\n
    24. \"\begin{array}{ll}<\/li>\n
    25. \"\begin{array}{ll}<\/li>\n
    26. \"\begin{array}{rrl}<\/li>\n
    27. \"\begin{array}{rrl}<\/li>\n
    28. \"\begin{array}{rrl}<\/li>\n
    29. \"\begin{array}{rrl}<\/li>\n
    30. \"\begin{array}{rrl}<\/li>\n<\/ol>\n","protected":false},"author":727,"menu_order":103,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"back-matter-type":[],"contributor":[],"license":[],"class_list":["post-1816","back-matter","type-back-matter","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter\/1816","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":4,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter\/1816\/revisions"}],"predecessor-version":[{"id":2006,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter\/1816\/revisions\/2006"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter\/1816\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/media?parent=1816"}],"wp:term":[{"taxonomy":"back-matter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter-type?post=1816"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/contributor?post=1816"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/license?post=1816"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}