If x > 0, then 1/[v(2x)+vx] =
A. 1/v(3x)
B. 1/[2v(2x)]
C. 1/(xv2)
D. (v2-1)/vx
E. (1+v2)/vx
pls help
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if nothing is not to the power of something else (!) (i mean if i didnt misunderstand)
then:
1/[v(2x)+vx] = 1/[vx(2+1)] = 1/[3vx]
then:
1/[v(2x)+vx] = 1/[vx(2+1)] = 1/[3vx]
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If you see the algebra, it's very quick:anjaligeorge1 wrote:If x > 0, then 1/[v(2x)+vx] =
A. 1/v(3x)
B. 1/[2v(2x)]
C. 1/(xv2)
D. (v2-1)/vx
E. (1+v2)/vx
1/[v(2x)+vx] = 1/(2vx + 1vx) = 1/3vx
If you don't see the algebra, then this is a perfect question for picking numbers.
let's let v=3 and x=7 (picking fairly random wacky numbers reduces the chance that more than 1 answer will work out).
Original: 1/(3*2*7 + 3*7) = 1/(42+21) = 1/63
Choices:
a) 1/(3*3*7) = 1/63.. yay!
if we want to be 100% sure:
b) 1/(2*3*2*7) = 1/84
c) 1/(7*3*2) = 1/42
d) way too big, don't bother
e) way too big, don't bother
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