In the third line of the solution (10.0 n/c)(x 0 -0.050) 2 =(15.0N/c)(x 0 -.100)

Question

In the third line of the solution (10.0 n/c)(x0-0.050)2=(15.0N/c)(x0-.100)2 In the fllowing line (x0-0.050)2 =(1.5 N/C)(x0-0.100)2 I was wondering what math goes on between theses two lines tomake 10 N/C and 15.0 N/C become the 1.5 N/c In the third line of the solution (10.0 n/c)(x0-0.050)2=(15.0N/c)(x0-.100)2 In the fllowing line (x0-0.050)2 =(1.5 N/C)(x0-0.100)2 I was wondering what math goes on between theses two lines tomake 10 N/C and 15.0 N/C become the 1.5 N/c I was wondering what math goes on between theses two lines tomake 10 N/C and 15.0 N/C become the 1.5 N/c

Explanation / Answer

(10.0 N / C )(x0-0.050)2=(15.0N/C)(x0-.100)2                   (x0-0.050)2= [ (15.0N/C)(x0-.100)2 ] / ( 10 N / C)                                     = 1.5 ( x0-.100)2                   (x0-0.050)2= [ (15.0N/C)(x0-.100)2 ] / ( 10 N / C)                                     = 1.5 ( x0-.100)2

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