Transposing Formulas

dbeaudry · #1 Jan 27, 2009, 06:27 PM

How do I solve the the following:
A=3C+4C for C

5/9X - 13 = 27

Thanks

ROLCAM · Jan 27, 2009, 06:46 PM

A=3C+4C
A = 7C
A/7 = C

C = 1/7 A

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5/9X - 13 = 27

5/9x = 27 +13 when 13 moves sides it changes its sign.

5/9x = 40

x = 40* 9/5 when 5/9 changes side it reverses itself.

x = 40*9/5
x = 8*9
x = 72

bones252100 · Jan 27, 2009, 06:52 PM

Add the two C multipliers then divide both sides by the sum.

Add 13 to both sides then divide both sides by the fraction. Remember the inversion factor. The 9 is a divisor which becomes a multiplier on the other side. The five is a multiplier which becomes a divisor on the other side.

shegun4u · Nov 4, 2009, 04:16 AM

E=mgh+1-/2mv(squared) for h

shegun4u · Nov 4, 2009, 04:16 AM

E=mgh+1/2mv(squared) for h

Nhatkiem · Nov 4, 2009, 11:08 AM

There is no need to post twice here and post 2 different topics in the homework section. Also if this is homework Shegun4u, you are not allowed to post here for answers.



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