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Rearrange the following equation so that n is the subject C equals 30n plus 295 over n-5?
KindaLost
Do you mean C = (30n+295) over (n-5)? If so then the solution is quite straightforward: C = (30n+295) over n-5 Multiply both sides of the equation by n-5 to eliminate the fraction: C(n-5) = 30n+295 Multiply out the brackets and bring over 295 to the LHS which now will be -295: Cn-5C-295 = 30n Collect like terms by bringing over Cn to the RHS which now will be -Cn: -5C-295 =30n-Cn Factorising 30n-Cn = n(30-C): -5-295 = n(30-C) Divide both sides of the equation by 30-C to make n the subject of the equation: n = (-5C-295) over (30-C) Check that your answer is correct by giving a value to C. For instance if C = 100 then n would = 159/14
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8
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-2x plus 3y equals 1
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The log of what number equals 2.771?
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