{"id":2439,"date":"2019-10-16T13:18:32","date_gmt":"2019-10-16T17:18:32","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/?post_type=back-matter&p=2439"},"modified":"2019-10-17T13:48:48","modified_gmt":"2019-10-17T17:48:48","slug":"answer-key-2-7","status":"publish","type":"back-matter","link":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/back-matter\/answer-key-2-7\/","title":{"raw":"Answer Key 2.7","rendered":"Answer Key 2.7"},"content":{"raw":"[latexpage]\r\n
    \r\n \t
  1. \\(x=ky\\)<\/li>\r\n \t
  2. \\(x=kyz\\)<\/li>\r\n \t
  3. \\(x=\\dfrac{k}{y}\\)<\/li>\r\n \t
  4. \\(x=ky^2\\)<\/li>\r\n \t
  5. \\(x=kzy\\)<\/li>\r\n \t
  6. \\(x=\\dfrac{k}{y^3}\\)<\/li>\r\n \t
  7. \\(x=ky^2\\sqrt{z}\\)<\/li>\r\n \t
  8. \\(x=\\dfrac{k}{y^6}\\)<\/li>\r\n \t
  9. \\(x=\\dfrac{ky^3}{\\sqrt{z}}\\)<\/li>\r\n \t
  10. \\(x=\\dfrac{k}{y^2\\sqrt{z}}\\)<\/li>\r\n \t
  11. \\(x=\\dfrac{kzy}{p^3}\\)<\/li>\r\n \t
  12. \\(x=\\dfrac{k}{y^3z^2}\\)<\/li>\r\n \t
  13. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\nA&=&kB \\\\\r\n(15)&=&k(5) \\\\ \\\\\r\n\\dfrac{15}{5}&=&\\dfrac{k(5)}{5} \\\\ \\\\\r\nk&=&3\r\n\\end{array}\\)<\/li>\r\n \t
  14. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\nP&=&kQR \\\\\r\n(12)&=&k(8)(3) \\\\ \\\\\r\n\\dfrac{12}{24}&=&\\dfrac{k(8)(3)}{24} \\\\ \\\\\r\nk&=&\\dfrac{1}{2}\r\n\\end{array}\\)<\/li>\r\n \t
  15. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\nA&=&\\dfrac{k}{B} \\\\ \\\\\r\n(7)&=&\\dfrac{k}{(4)} \\\\ \\\\\r\n(4)7&=&\\dfrac{k}{\\cancel{4}}\\cancel{(4)} \\\\ \\\\\r\nk&=&28\r\n\\end{array}\\)<\/li>\r\n \t
  16. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\nA&=&kB^2 \\\\\r\n(6)&=&k(3)^2 \\\\ \\\\\r\n\\dfrac{6}{9}&=&\\dfrac{k(3)^2}{9} \\\\ \\\\\r\nk&=&\\dfrac{6}{9}\\text{ or }\\dfrac{2}{3}\r\n\\end{array}\\)<\/li>\r\n \t
  17. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\nC&=&kAB \\\\\r\n(24)&=&k(3)(2) \\\\ \\\\\r\n\\dfrac{24}{6}&=&\\dfrac{k(3)(2)}{6} \\\\ \\\\\r\nk&=&4\r\n\\end{array}\\)<\/li>\r\n \t
  18. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\ny&=&\\dfrac{k}{x^3} \\\\ \\\\\r\n(54)&=&\\dfrac{k}{(3)^3} \\\\ \\\\\r\n54&=&\\dfrac{k}{27} \\\\ \\\\\r\n27\\cdot 54&=&\\dfrac{k}{\\cancel{27}}\\cdot \\cancel{27} \\\\ \\\\\r\nk&=&1458\r\n\\end{array}\\)<\/li>\r\n \t
  19. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\nx&=&kY \\\\\r\n(12)&=&k(8) \\\\ \\\\\r\n\\dfrac{12}{8}&=&\\dfrac{k(8)}{8} \\\\ \\\\\r\nk&=&\\dfrac{12}{8}\\text{ or }\\dfrac{3}{2}\r\n\\end{array}\\)<\/li>\r\n \t
  20. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\nA&=&kB^2\\sqrt{C} \\\\\r\n(25)&=&k(5)^2\\sqrt{(9)} \\\\\r\n25&=&k(75) \\\\ \\\\\r\nk&=&\\dfrac{25}{75} \\\\ \\\\\r\nk&=&\\dfrac{1}{3}\r\n\\end{array}\\)<\/li>\r\n \t
  21. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\ny&=&\\dfrac{kmn^2}{d} \\\\ \\\\\r\n(10)&=&\\dfrac{k(4)(5)^2}{(6)} \\\\ \\\\\r\nk&=&\\dfrac{\\cancel{10}\\cancel{5}\\cdot \\cancel{6}3}{\\cancel{(4)}(5)^{\\cancel{2}}} \\\\ \\\\\r\nk&=&\\dfrac{3}{5}\r\n\\end{array}\\)<\/li>\r\n \t
  22. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\nP&=&\\dfrac{kT}{V} \\\\ \\\\\r\n(10)&=&\\dfrac{k(250)}{(400)} \\\\ \\\\\r\nk&=&\\dfrac{10(400)}{250} \\\\ \\\\\r\nk&=&16\r\n\\end{array}\\)<\/li>\r\n \t
  23. \\(\\phantom{1}\\)\r\n\\(I=kV \\\\ \\)\r\n\\(\\begin{array}{ll}\r\n\\begin{array}{rrl}\r\n&&\\textbf{1st Data} \\\\\r\nI&=&5 \\text{ A} \\\\\r\nV&=&15\\text{ V} \\\\\r\nk&=&\\text{find} \\\\ \\\\\r\nI&=&kV \\\\\r\n5\\text{ A}&=&k(\\text{15 V}) \\\\ \\\\\r\nk&=&\\dfrac{\\text{5 A}}{\\text{15 V}} \\\\ \\\\\r\nk&=&\\dfrac{1}{3}\\text{ A\/V}\r\n\\end{array}\r\n& \\hspace{0.5in}\r\n\\begin{array}{rrl}\r\n&&\\textbf{2nd Data} \\\\\r\nI&=&\\text{find} \\\\\r\nk&=&\\dfrac{1}{3} \\\\ \\\\\r\nV&=&\\text{25 V} \\\\ \\\\\r\nI&=&kV \\\\\r\nI&=&\\left(\\dfrac{1}{3}\\right)(25) \\\\ \\\\\r\nI&=&8\\dfrac{1}{3}\\text{ A}\r\n\\end{array}\r\n\\end{array}\\)<\/li>\r\n \t
  24. \\(\\phantom{1}\\)\r\n\\(I=\\dfrac{k}{R} \\\\ \\)\r\n\\(\\begin{array}{ll}\r\n\\begin{array}{rrl}\r\n&&\\textbf{1st Data} \\\\\r\nI&=&\\text{12 A} \\\\\r\nk&=&\\text{find} \\\\\r\nR&=&240\\Omega \\\\ \\\\\r\nI&=&\\dfrac{k}{R} \\\\ \\\\\r\n\\text{12 A}&=&\\dfrac{k}{240\\Omega} \\\\ \\\\\r\nk&=&(\\text{12 A})(240\\Omega) \\\\\r\nk&=&2880\\text{ A}\\Omega\r\n\\end{array}\r\n& \\hspace{0.5in}\r\n\\begin{array}{rrl}\r\n&&\\textbf{2nd Data} \\\\\r\nI&=&\\text{find} \\\\\r\nk&=&2880 \\\\\r\nR&=&540\\Omega \\\\ \\\\\r\nI&=&\\dfrac{k}{R} \\\\ \\\\\r\nI&=&\\dfrac{2880\\text{ A}\\Omega}{540\\Omega} \\\\ \\\\\r\nI&=&5.\\bar{3}\\text{ A or }5\\dfrac{1}{3}\r\n\\end{array}\r\n\\end{array}\\)<\/li>\r\n \t
  25. \\(\\phantom{1}\\)\r\n\\(d_{\\text{s}}=km \\\\ \\)\r\n\\(\\begin{array}{ll}\r\n\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\\r\n&&\\textbf{1st Data} \\\\\r\nd_{\\text{s}}&=&18\\text{ cm} \\\\\r\nk&=&\\text{find} \\\\\r\nm&=&3\\text{ kg} \\\\ \\\\\r\n18\\text{ cm}&=&k(3\\text{ kg}) \\\\\u00a0 \\\\\r\nk&=&\\dfrac{\\text{18 cm}}{\\text{3 kg}} \\\\ \\\\\r\nk&=&\\text{6 cm\/kg}\r\n\\end{array}\r\n& \\hspace{0.5in}\r\n\\begin{array}{rrl}\r\n&&\\textbf{2nd Data} \\\\\r\nd_{\\text{s}}&=&\\text{find} \\\\\r\nk&=&\\text{6 cm\/kg} \\\\\r\nm&=&\\text{5 kg} \\\\ \\\\\r\nd_{\\text{s}}&=&(\\text{6 cm\/kg})(\\text{5 kg}) \\\\\r\nd_{\\text{s}}&=&\\text{30 cm}\r\n\\end{array}\r\n\\end{array}\\)<\/li>\r\n \t
  26. \\(\\phantom{1}\\)\r\n\\(V=\\dfrac{k}{P} \\\\ \\)\r\n\\(\\begin{array}{ll}\r\n\\begin{array}{rrl}\r\n\\\\\r\n&&\\textbf{1st Data} \\\\\r\nP&=&32\\text{ kg\/cm}^2 \\\\\r\nV&=&200\\text{ cm}^3 \\\\\r\nk&=&\\text{find} \\\\ \\\\\r\n200\\text{ cm}^3&=&\\dfrac{k}{32\\text{ kg\/cm}^2} \\\\ \\\\\r\nk&=&(200\\text{ cm}^3)(32\\text{ kg\/cm}^2) \\\\\r\nk&=&6400\\text{ kg cm}\r\n\\end{array}\r\n& \\hspace{0.5in}\r\n\\begin{array}{rrl}\r\n&&\\textbf{2nd Data} \\\\\r\nP&=&40 \\\\\r\nV&=&\\text{find} \\\\\r\nk&=&6400 \\\\ \\\\\r\nV&=&\\dfrac{6400}{40} \\\\ \\\\\r\nV&=&160\\text{ cm}^3\r\n\\end{array}\r\n\\end{array}\\)<\/li>\r\n \t
  27. \\(\\phantom{1}\\)\r\n\\(c=kP \\\\ \\)\r\n\\(\\begin{array}{ll}\r\n\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\\r\n&&\\textbf{1st Data} \\\\\r\nc&=&60,000 \\\\\r\nk&=&\\text{find} \\\\\r\nP&=&250 \\\\ \\\\\r\n60,000&=&k(250) \\\\ \\\\\r\nk&=&\\dfrac{60,000}{250} \\\\ \\\\\r\nk&=&240\r\n\\end{array}\r\n& \\hspace{0.5in}\r\n\\begin{array}{rrl}\r\n&& \\textbf{2nd Data} \\\\\r\nc&=&\\text{find} \\\\\r\nk&=&240 \\\\\r\nP&=&1,000,000 \\\\ \\\\\r\nc&=&(240)(1,000,000) \\\\\r\nc&=&240,000,000\\text{ or 240 million}\r\n\\end{array}\r\n\\end{array}\\)<\/li>\r\n \t
  28. \\(\\phantom{1}\\)\r\n\\(t=\\dfrac{k}{b} \\\\ \\)\r\n\\(\\begin{array}{ll}\r\n\\begin{array}{rrl}\r\n\\\\\r\n&&\\textbf{1st Data} \\\\\r\nt&=&5\\text{ h} \\\\\r\nk&=&\\text{find} \\\\\r\nb&=&7 \\\\ \\\\\r\n5\\text{ h}&=&\\dfrac{k}{7} \\\\ \\\\\r\nk&=&\\text{(5 h)}(7) \\\\\r\nk&=&35\r\n\\end{array}\r\n& \\hspace{0.5in}\r\n\\begin{array}{rrl}\r\n&&\\textbf{2nd Data} \\\\\r\nt&=&\\text{find} \\\\\r\nk&=&35 \\\\\r\nb&=&10 \\\\ \\\\\r\nt&=&\\dfrac{35}{10} \\\\ \\\\\r\nt&=&3.5\\text{ h}\r\n\\end{array}\r\n\\end{array}\\)<\/li>\r\n \t
  29. \\(\\phantom{1}\\)\r\n\\(\\lambda=\\dfrac{k}{f} \\\\ \\)\r\n\\(\\begin{array}{ll}\r\n\\begin{array}{rrl}\r\n\\\\\r\n&&\\textbf{1st Data} \\\\\r\n\\lambda&=&250\\text{ m} \\\\\r\nk&=&\\text{find} \\\\\r\nf&=&1200\\text{ kHz} \\\\ \\\\\r\n250&=&\\dfrac{k}{1200} \\\\ \\\\\r\nk&=&(250)(1200) \\\\\r\nk&=&300,000\r\n\\end{array}\r\n& \\hspace{0.5in}\r\n\\begin{array}{rrl}\r\n&&\\textbf{2nd Data} \\\\\r\n\\lambda&=&\\text{find} \\\\\r\nk&=&300,000 \\\\\r\nf&=&60\\text{ kHz} \\\\ \\\\\r\n\\lambda&=&\\dfrac{300,000}{60} \\\\ \\\\\r\n\\lambda&=&5000\\text{ m}\r\n\\end{array}\r\n\\end{array}\\)<\/li>\r\n \t
  30. \\(\\phantom{1}\\)\r\n\\(w=km \\\\ \\)\r\n\\(\\begin{array}{ll}\r\n\\begin{array}{rrl}\r\n\\\\\r\n&& \\textbf{1st Data} \\\\\r\nw&=&64\\text{ kg} \\\\\r\nk&=&\\text{find} \\\\\r\nm&=&96\\text{ kg} \\\\ \\\\\r\n64&=&k(96) \\\\ \\\\\r\nk&=&\\dfrac{64}{96} \\\\ \\\\\r\nk&=&\\dfrac{2}{3}\r\n\\end{array}\r\n& \\hspace{0.5in}\r\n\\begin{array}{rrl}\r\n&&\\textbf{2nd Data} \\\\\r\nw&=&\\text{find} \\\\\r\nk&=&\\dfrac{2}{3} \\\\\r\nm&=&60\\text{ kg} \\\\ \\\\\r\nw&=&\\left(\\dfrac{2}{3}\\right)(60\\text{ kg}) \\\\ \\\\\r\nw&=&40\\text{ kg}\r\n\\end{array}\r\n\\end{array}\\)<\/li>\r\n \t
  31. \\(\\phantom{1}\\)\r\n\\(t=\\dfrac{d}{v} \\\\ \\)\r\n\\(\\begin{array}{ll}\r\n\\begin{array}{rrl}\r\n&&\\textbf{1st Data} \\\\\r\nt&=&\\text{5 h} \\\\\r\nd&=&\\text{find} \\\\\r\nv&=&\\text{80 km\/h} \\\\ \\\\\r\n\\text{5 h}&=&\\dfrac{d}{\\text{80 km\/h}} \\\\ \\\\\r\nd&=&5(80) \\\\\r\nd&=&\\text{400 km}\r\n\\end{array}\r\n& \\hspace{0.5in}\r\n\\begin{array}{rrl}\r\n\\\\ \\\\\r\n&&\\textbf{2nd Data} \\\\\r\nt&=&\\text{4.2 h} \\\\\r\nd&=&\\text{400 km} \\\\\r\nv&=&\\text{find} \\\\ \\\\\r\n4.2&=&\\dfrac{400}{v} \\\\ \\\\\r\nv&=&\\dfrac{400}{4.2} \\\\ \\\\\r\nv&=&95.24\\text{ km\/h}\r\n\\end{array}\r\n\\end{array}\\)<\/li>\r\n \t
  32. \\(\\phantom{1}\\)\r\n\\(V=khr^2 \\\\ \\)\r\n\\(\\begin{array}{ll}\r\n\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\\r\n&&\\textbf{1st Data} \\\\\r\nV&=&33.5\\text{ cm}^3 \\\\\r\nk&=&\\text{find} \\\\\r\nh&=&\\text{8 cm} \\\\\r\nr&=&\\text{2 cm} \\\\ \\\\\r\n33.5&=&k(8)(2)^2 \\\\ \\\\\r\nk&=&\\dfrac{33.5}{(8)(2)^2} \\\\ \\\\\r\nk&=&1.046875\r\n\\end{array}\r\n& \\hspace{0.5in}\r\n\\begin{array}{rrl}\r\n&&\\textbf{2nd Data} \\\\\r\nV&=&\\text{find} \\\\\r\nk&=&1.046875 \\\\\r\nh&=&\\text{6 cm} \\\\\r\nr&=&\\text{4 cm} \\\\ \\\\\r\nV&=&khr^2 \\\\\r\nV&=&(1.046875)(6)(4)^2 \\\\\r\nV&=&100.5\\text{ cm}^3\r\n\\end{array}\r\n\\end{array}\\)<\/li>\r\n \t
  33. \\(\\phantom{1}\\)\r\n\\(F_{\\text{e}}=\\dfrac{kv^2}{r} \\\\ \\)\r\n\\(\\begin{array}{ll}\r\n\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\\r\n&&\\textbf{1st Data} \\\\\r\nF_{\\text{e}}&=&100\\text{ N} \\\\\r\nk&=&\\text{find} \\\\\r\nv&=&10\\text{ m\/s} \\\\\r\nr&=&\\text{0.5 m} \\\\ \\\\\r\n100\\text{ N}&=&\\dfrac{k(10 \\text{ m\/s})^2}{\\text{0.5 m}} \\\\ \\\\\r\nk&=&\\dfrac{(0.5)(100)}{(10)^2} \\\\ \\\\\r\nk&=&0.5\r\n\\end{array}\r\n& \\hspace{0.5in}\r\n\\begin{array}{rrl}\r\n&&\\textbf{2nd Data} \\\\\r\nF_{\\text{e}}&=&\\text{find} \\\\\r\nk&=&0.5 \\\\\r\nv&=&25\\text{ m\/s} \\\\\r\nr&=&1.0\\text{ m} \\\\ \\\\\r\nF_{\\text{e}}&=&\\dfrac{0.5(25)^2}{1.0} \\\\ \\\\\r\nF_{\\text{e}}&=&312.5\\text{ N}\r\n\\end{array}\r\n\\end{array}\\)<\/li>\r\n \t
  34. \\(\\phantom{1}\\)\r\n\\(L_{\\text{max}}=\\dfrac{kd^4}{h^2} \\\\ \\)\r\n\\(\\begin{array}{ll}\r\n\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\\r\n&&\\textbf{1st Data} \\\\\r\nL_{\\text{max}}&=&64\\text{ tonnes} \\\\\r\nk&=&\\text{find} \\\\\r\nd&=&2.0\\text{ m} \\\\\r\nh&=&8.0\\text{ m} \\\\ \\\\\r\n64&=&\\dfrac{k(2)^4}{(8)^2} \\\\ \\\\\r\nk&=&\\dfrac{64(8)^2}{(2)^4} \\\\ \\\\\r\nk&=&256\r\n\\end{array}\r\n& \\hspace{0.5in}\r\n\\begin{array}{rrl}\r\n&&\\textbf{2nd Data} \\\\\r\nL_{\\text{max}}&=&\\text{find} \\\\\r\nk&=&256 \\\\\r\nd&=&3.0\\text{ m} \\\\\r\nh&=&12.0\\text{ m} \\\\ \\\\\r\nL_{\\text{max}}&=&\\dfrac{(256)(3.0)^4}{(12.0)^2} \\\\ \\\\\r\nL_{\\text{max}}&=&144\\text{ tonnes}\r\n\\end{array}\r\n\\end{array}\\)<\/li>\r\n \t
  35. \\(\\phantom{1}\\)\r\n\\(V=\\dfrac{kT}{P} \\\\ \\)\r\n\\(\\begin{array}{ll}\r\n\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\\r\n&&\\textbf{1st Data} \\\\\r\nV&=&225\\text{ cc} \\\\\r\nk&=&\\text{find} \\\\\r\nT&=&300\\text{ K} \\\\\r\nP&=&100\\text{ N\/cm}^2 \\\\ \\\\\r\nV&=&\\dfrac{kT}{P} \\\\ \\\\\r\n225&=&\\dfrac{k(300)}{100} \\\\ \\\\\r\nk&=&\\dfrac{225(100)}{300} \\\\ \\\\\r\nk&=&75\r\n\\end{array}\r\n& \\hspace{0.5in}\r\n\\begin{array}{rrl}\r\n&&\\textbf{2nd Data} \\\\\r\nV&=&\\text{find} \\\\\r\nk&=&75 \\\\\r\nT&=&270 \\\\\r\nP&=&150 \\\\ \\\\\r\nV&=&\\dfrac{75(270)}{150} \\\\ \\\\\r\nV&=&135\\text{ cc}\r\n\\end{array}\r\n\\end{array}\\)<\/li>\r\n \t
  36. \\(\\phantom{1}\\)\r\n\\(R=\\dfrac{kl}{d^2} \\\\ \\)\r\n\\(\\begin{array}{ll}\r\n\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\\r\n&&\\textbf{1st Data} \\\\\r\nR&=&20\\Omega \\\\\r\nk&=&\\text{find} \\\\\r\nl&=&5.0\\text{ m} \\\\\r\nd&=&0.25\\text{ cm} \\\\ \\\\\r\nR&=&\\dfrac{kl}{d^2} \\\\ \\\\\r\n20\\Omega&=&\\dfrac{k(5.0\\text{ m})}{\\text{(0.25 cm)}^2} \\\\ \\\\\r\nk&=&\\dfrac{(20 \\Omega)\\text{(0.25 cm)}^2}{\\text{5.0 m}} \\\\ \\\\\r\nk&=&0.25\r\n\\end{array}\r\n& \\hspace{0.5in}\r\n\\begin{array}{rrl}\r\n&&\\textbf{2nd Data} \\\\\r\nR&=&\\text{find} \\\\\r\nk&=&0.25 \\\\\r\nl&=&10.0\\text{ m} \\\\\r\nd&=&0.50\\text{ cm} \\\\ \\\\\r\nR&=&\\dfrac{(0.25)\\text{(10.0 m)}}{\\text{(0.50 cm)}^2} \\\\ \\\\\r\nR&=&10\\Omega\r\n\\end{array}\r\n\\end{array}\\)<\/li>\r\n \t
  37. \\(\\phantom{1}\\)\r\n\\(V=khd^2 \\\\ \\)\r\n\\(\\begin{array}{ll}\r\n\\begin{array}{rrl}\r\n&&\\textbf{1st Data} \\\\\r\nV&=&377\\text{ m}^3 \\\\\r\nk&=&\\text{find} \\\\\r\nh&=&30\\text{ m} \\\\\r\nd&=&2.0\\text{ m} \\\\ \\\\\r\n377\\text{ m}^3&=&k(30)(2.0)^2 \\\\ \\\\\r\nk&=&\\dfrac{377}{(30)(2.0)^2} \\\\ \\\\\r\nk&=&3.1416\r\n\\end{array}\r\n& \\hspace{0.5in}\r\n\\begin{array}{rrl}\r\n&&\\textbf{2nd Data} \\\\\r\nV&=&225\\text{ m}^3 \\\\\r\nk&=&3.1416 \\\\\r\nh&=&\\text{find} \\\\\r\nd&=&1.75\\text{ m} \\\\ \\\\\r\n225&=&\\pi h(1.75)^2 \\\\ \\\\\r\nh&=&\\dfrac{225}{\\pi (1.75)^2} \\\\ \\\\\r\nh&=&23.4\\text{ m}\r\n\\end{array}\r\n\\end{array}\\)<\/li>\r\n<\/ol>","rendered":"
      \n
    1. \"x=ky\"<\/li>\n
    2. \"x=kyz\"<\/li>\n
    3. \"x=\dfrac{k}{y}\"<\/li>\n
    4. \"x=ky^2\"<\/li>\n
    5. \"x=kzy\"<\/li>\n
    6. \"x=\dfrac{k}{y^3}\"<\/li>\n
    7. \"x=ky^2\sqrt{z}\"<\/li>\n
    8. \"x=\dfrac{k}{y^6}\"<\/li>\n
    9. \"x=\dfrac{ky^3}{\sqrt{z}}\"<\/li>\n
    10. \"x=\dfrac{k}{y^2\sqrt{z}}\"<\/li>\n
    11. \"x=\dfrac{kzy}{p^3}\"<\/li>\n
    12. \"x=\dfrac{k}{y^3z^2}\"<\/li>\n
    13. \"\begin{array}{rrl}<\/li>\n
    14. \"\begin{array}{rrl}<\/li>\n
    15. \"\begin{array}{rrl}<\/li>\n
    16. \"\begin{array}{rrl}<\/li>\n
    17. \"\begin{array}{rrl}<\/li>\n
    18. \"\begin{array}{rrl}<\/li>\n
    19. \"\begin{array}{rrl}<\/li>\n
    20. \"\begin{array}{rrl}<\/li>\n
    21. \"\begin{array}{rrl}<\/li>\n
    22. \"\begin{array}{rrl}<\/li>\n
    23. \"\phantom{1}\"
      \n\"I=kV \\\"
      \n\"\begin{array}{ll}<\/li>\n
    24. \"\phantom{1}\"
      \n\"I=\dfrac{k}{R} \\\"
      \n\"\begin{array}{ll}<\/li>\n
    25. \"\phantom{1}\"
      \n\"d_{\text{s}}=km \\\"
      \n\"\begin{array}{ll}<\/li>\n
    26. \"\phantom{1}\"
      \n\"V=\dfrac{k}{P} \\\"
      \n\"\begin{array}{ll}<\/li>\n
    27. \"\phantom{1}\"
      \n\"c=kP \\\"
      \n\"\begin{array}{ll}<\/li>\n
    28. \"\phantom{1}\"
      \n\"t=\dfrac{k}{b} \\\"
      \n\"\begin{array}{ll}<\/li>\n
    29. \"\phantom{1}\"
      \n\"\lambda=\dfrac{k}{f} \\\"
      \n\"\begin{array}{ll}<\/li>\n
    30. \"\phantom{1}\"
      \n\"w=km \\\"
      \n\"\begin{array}{ll}<\/li>\n
    31. \"\phantom{1}\"
      \n\"t=\dfrac{d}{v} \\\"
      \n\"\begin{array}{ll}<\/li>\n
    32. \"\phantom{1}\"
      \n\"V=khr^2 \\\"
      \n\"\begin{array}{ll}<\/li>\n
    33. \"\phantom{1}\"
      \n\"F_{\text{e}}=\dfrac{kv^2}{r} \\\"
      \n\"\begin{array}{ll}<\/li>\n
    34. \"\phantom{1}\"
      \n\"L_{\text{max}}=\dfrac{kd^4}{h^2} \\\"
      \n\"\begin{array}{ll}<\/li>\n
    35. \"\phantom{1}\"
      \n\"V=\dfrac{kT}{P} \\\"
      \n\"\begin{array}{ll}<\/li>\n
    36. \"\phantom{1}\"
      \n\"R=\dfrac{kl}{d^2} \\\"
      \n\"\begin{array}{ll}<\/li>\n
    37. \"\phantom{1}\"
      \n\"V=khd^2 \\\"
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