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1)F=9/5C+32; for C
2)A=P(1+rt);for t
3)I=nE/(R+nr);for n
4)I=nE/(nR+r);for r
5)1/f=1/v -1/u;for u

2007-03-02 01:55:55 · 4 answers · asked by Nitin T F1 fan 5 in Science & Mathematics Mathematics

4 answers

I'm assuming you mean you want to solve for each of the indicated variables:

1) F-32=9/5C
5/9(F-32)=C

2) A/P=1+rt
A/P-1=rt
(A/P-1)/r=t

3)l(R+nr)=nE
lR+nlr=ne
lR=ne-nlr
lR=n(e-lr)
n=lR/(e-lr)

4)lnR+lr=nE
lr=nE-lnR
r=(nE-lnR)/l

5)1/u=1/v-1/f
1=u(1/v-1/f)
u=1(1/v-1/f)

2007-03-02 02:07:02 · answer #1 · answered by MISSYCL 2 · 0 0

1. f= 9/5C + 32
F -32= 9/5C
5(F - 32)=9C
5(F- 32)/9 =C

2.A= P(1+ rt )
A/P=1+ rt
A/P- 1= rt
THEN DIVIDE BOTH SIDES BY R
from this you should be able to do the remaining

2007-03-02 02:18:12 · answer #2 · answered by jay gal 3 · 0 0

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2016-12-18 13:40:25 · answer #3 · answered by ? 4 · 0 0

1) 9/5C = F-32
9C = 5F-160
C = 5/9F-160/9

Etc.... you do the rest

2007-03-02 02:10:44 · answer #4 · answered by Jack 2 · 0 0

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