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Q: How do you solve for p?

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2x+a=p 2x=p-a x=.5p-.5a

how to solve 7 + p = p +7

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10p + 5p - p = 15p - p = 14p

P(A)=.35 P(B given A)=0.6 P(A and B)= ?

16.52 + p = 495.50 p = 495,50 - 16,52 p= 478,98

-1507

3 + P = 8 Therefore: P = 8 - 3 P = 5

10.2

x = (P - 2W) / 2

Momentum, p, is solved by using the momentum equation: p = m*v.

8s=ps=p/8

Add 6 to each side: p = 22

Do you mean: r = (4q-5p)/9? If so then: p = (9r-4q)/-5

No. You have to solve it by logic. p:

p>9

no. you would solve and get -80=-2p or 80=2p. thus resulting in p=40

L = P/2 - W.

It is simply two terms of an algebriac expression in the form of: 12.95-p

A = P + P Y TSubtract 'P' from each side:A - P = P Y TDivide each side by PT :Y = (A - P) / PT

-4(3 - p) > 5(p + 1) => -12 + 4p > 5p + 5-17 > p, that is p < -17.

You can solve this for p:2m + 2p = 162p = 16 - 2mp = 8 - mIf you supply a value for "m", you can then calculate p.You can solve this for p:2m + 2p = 162p = 16 - 2mp = 8 - mIf you supply a value for "m", you can then calculate p.You can solve this for p:2m + 2p = 162p = 16 - 2mp = 8 - mIf you supply a value for "m", you can then calculate p.You can solve this for p:2m + 2p = 162p = 16 - 2mp = 8 - mIf you supply a value for "m", you can then calculate p.

C. p ‹ 12

V=nRT/P

16=24-p. subtract 16 both sides 0=8-p. add p both sides P=8