So if I have: 42*r2/(r2-.26)=5.52. How do I solve for r2?
2006-09-19 15:05:19 · 8 answers · asked by eltel2910 1 in Science & Mathematics ➔ Mathematics
Thanks everyone I got it from mfrsiska's post.
2006-09-19 15:23:53 · update #1
you are right -- it is a simple problem
2006-09-19 15:08:53 · answer #1 · answered by oracle 5 · 0⤊ 0⤋
42 * r2/(r2 - .26) = 5.52 => 42 * r2 = 5.52 (r2 - .26) => 42(r2) - 5.52(r2) = 5.52 * (-.26) => 42-5.52(r2) = -1.4352 => r2 = (-1.4352/36.48) => r2 = -.03934...; What you are trying to do is isolate r2 on one side of the equation and everything else on the other side. mfrsiska almost has it, but she forgot the decimal point in the term .26.
2006-09-19 15:18:30 · answer #2 · answered by flyfisher_20750 3 · 0⤊ 0⤋
42 x r2 / (r2 - 26) = 5.5.2
42 x r2 = 5.52 x (r2 - 26)
42 x r2 = 5.52 x r2 - 143,52
42 x r2 - 5.52 x r2 = - 143.52
36.48 x r2 = -143.52
r2 = -143.52 / 38.48
r2 = - 3.729
2006-09-19 15:09:24 · answer #3 · answered by Anonymous · 1⤊ 0⤋
I dont understand your problem because you say r2, is this a ("2" times variable "r") or (variable "r" squared).
If it is 2 times variable r-
Lets substitute r2 with x.
42x/(x-2.6)=5.52
multiply (x-2.6) on both sides to get-
42x=5.52(x-2.6)
simplify-
42x= 5.52x - 14.352
add 5.52x to both sides
42x + 5.52x = -14.352
divide the common variable out on the left side-
x(42+5.52) simplify-
x(47.52)
divide both sides by 47.52
x= -0.302
which is r2 = -0.302
so now if it was 2 times variable r, the answer would be -0.302/2
and if it was variable r squared the answer would be the square root of -0.302
2006-09-19 15:18:01 · answer #4 · answered by Morph 1 · 0⤊ 0⤋
What ever you do to one side you have to the other.
(/r2)
5.52/r2=42/.74
5.52/r2=56.756756756756756756756756756757....
(*r2)
56.756756756756...r2=5.52
(/56.756756756756...)
r2=0.097257142857142857142857142857154
there you go it works
2006-09-19 15:42:32 · answer #5 · answered by Anonymous · 0⤊ 0⤋
r2= .26/(1-1/(5.52/42))
2006-09-19 15:24:22 · answer #6 · answered by trollmannen 2 · 0⤊ 0⤋
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2006-09-19 15:09:57 · answer #7 · answered by cotterall&elaineadams 2 · 1⤊ 0⤋
sorry i thought i had it but i dont
2006-09-19 15:08:30 · answer #8 · answered by JB 2 · 0⤊ 0⤋