what is the value of x if (.0268=((x-414635.67)/x) and how did you work it out
2007-06-08 10:37:49 · 11 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
426053.91
2007-06-08 10:42:41 · answer #1 · answered by Deevah 2 · 0⤊ 0⤋
It's very easy: .0268 =x - 414635.67 / x
.0268 x = x- 414635.67
414635.67 = x ( 1- .0268 )
x= 414635.67/ .9732 = 426053.91
2007-06-08 17:47:06 · answer #2 · answered by melao3 7 · 0⤊ 0⤋
well that's do this step by step....
the equation you have is .0268=(x-414635.67)/x
step 1: multiply both side by x
you get .0268x=x-414635.67
step 2: take away x both side which mean minus x on both side
you would get .0268x-x=-414635.67
which equal to -.9732x=-414635.67
step 3: divide both side by -.9732
you would get x=426053.9149
hope this help
2007-06-08 17:50:29 · answer #3 · answered by Mrs. Jenny 2 · 0⤊ 0⤋
(.0268=((x-414635.67)/x)
Multiply both sides by x
0.0268x = x - 414635.67
Subtract x from both sides
-0.9732x = -414635.67
x = 426,053.91
.
2007-06-08 17:50:20 · answer #4 · answered by Robert L 7 · 0⤊ 0⤋
.0268 = (x-414635.67)/x
.0268x = x-414635.67
-.9732x = -414635.67
.9732x = 414635.67
x = 414635.67/.9732
x = 426053.9149198520345252774352651
2007-06-08 17:45:06 · answer #5 · answered by Anonymous · 0⤊ 0⤋
X=46053.93
2007-06-08 17:45:51 · answer #6 · answered by fenrirsoul 1 · 0⤊ 0⤋
o.0268x=x-414635.67
.0268x-x=-414635.67
x(.0268-1)=-414635.67
-0.9732x=-414635.67
x=414635.67/0.9732
x=426053.9149
2007-06-08 17:46:44 · answer #7 · answered by novajr17 3 · 0⤊ 0⤋
0.0268 = (x - 414635.67) / x
Multiplying both sides by x,
0.0268x = x - 414635.67
Subtracting x,
-0.9732x = -414635.67
Dividing by -0.9732,
x = approx. 426053.914919852
2007-06-08 17:45:45 · answer #8 · answered by Louise 5 · 0⤊ 0⤋
0.0268 = (x-414635.67) / x
0.0268 x = x - 414635.67
414635.67 + 0.0268 x = x
414635.67 = x - 0.0268 x = (1-0.0268) x
414635.67/(1-0.0268) = x
So x = 414635.67/(1-0.0268) = 426053.9
2007-06-08 17:46:46 · answer #9 · answered by ? 6 · 0⤊ 0⤋
id otnt nknow
2007-06-08 17:45:17 · answer #10 · answered by Anonymous · 0⤊ 0⤋