document.write( "Question 1203019: I have a problem with questions relating to variation as the sum of two parts.
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document.write( "Here is a worked example.
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document.write( "A quantity, p, is the sum of two terms, one of which is constant while the other varies inversely as the square of q.
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document.write( "When q = 1, p=-1;
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document.write( "when q = 2, p=2;
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document.write( "find the positive value of q when p=2.75
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document.write( "If a and b are constants, p=a+b/q2;
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document.write( "q = 1, p= -1
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document.write( "-1 = a +b/12;
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document.write( "therefore -1 = a+b (equation 1)
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document.write( "when q = 2, p=2
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document.write( "2 = a+b/22
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document.write( "2=a+b/4
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document.write( "8 = 4a+b (equation 2)
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document.write( "Subtracting equation 1 from 2 gives
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document.write( "9=3a
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document.write( "so a =3
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document.write( "Substituting a = 3 into equation 1 we get
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document.write( "-1 = 3+b
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document.write( "b=-4
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document.write( "so p=3-4/q2
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document.write( "so when p = 2.75, 2.75 = 3-4/q2
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document.write( "4/q2 = ΒΌ
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document.write( "q2 = 16
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document.write( "q=+/- 4 and we want the positive value so q = 4
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document.write( "I have tried to apply the same approach to the following problem:
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document.write( "A quantity C is the sum of two parts. The first part varies directly as the cube of t;
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document.write( "the second part varies inversely as the square of t.
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document.write( "C = 74, t = 1
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document.write( "C = 34, t = 2
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document.write( "Find C when t = 3
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document.write( "I wrote C = a3 +b/t2
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document.write( "74 = a3+b/12
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document.write( "74 = a3 +b (equation 1).\r
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document.write( "34 = a3+b/22
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document.write( "34 = a3+b/4
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document.write( "136 = 4a3+b (equation 2)\r
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document.write( "Subtract equation 1 from 2
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document.write( "136 = 4a3+b
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document.write( "74 = a3 +b
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document.write( "This gives 62 = 3a3
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document.write( "a3 = 20.67
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document.write( "a = 2.74\r
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document.write( "Substituting into the first equation
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document.write( "74 = 20.67 + b/12
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document.write( "74 = 20.67+b
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document.write( "b=53.33\r
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document.write( "To find C when t = 3
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document.write( "C = a3+b/t2
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document.write( "C = 20.67 + 53.33/9
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document.write( "9C = 186.03+53.33
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document.write( "9C = 239.36
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document.write( "C = 25.6 but the answer is 62.
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document.write( "Where have I gone wrong?\r
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document.write( "
Algebra.Com's Answer #838252 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "You are doing most of the work right; but your equation for C in the new problem is not what you intended. \n" ); document.write( "NOTE: For readability, use \"^\" (shift-6) to denote an exponent -- e.g. \"q^2\" and \"t^3\" instead of \"q2\" and \"t3\". (It might be that you were using some other special character to represent and exponent, and that special character doesn't show up at all.) \n" ); document.write( "In the new problem, you wrote \n" ); document.write( " \n" ); document.write( "But what you meant was \n" ); document.write( " \n" ); document.write( "Fix that and solve the problem using the methods you show and you will get the correct answer of 62. \n" ); document.write( "I'll go ahead and finish solving the problem; but you will get more benefit if you finish it yourself and compare your work and final answer to mine. \n" ); document.write( "t=1: \n" ); document.write( "t=2: \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "The equation is \n" ); document.write( " \n" ); document.write( "When t=3... \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " |