2006-09-14 11:36:18 · 9 answers · asked by Cutecici 1 in Science & Mathematics ➔ Mathematics
y = sqrt(x)
y = x/40
sqrt(x) = x/40
x = (x/40)^2
x = x^2/1600
x^2/1600 - x = 0
x^2 - 1600x = 0
x(x-1600) = 0
x = 0 or x - 1600 = 0
x = 1600
2006-09-14 11:41:29 · answer #1 · answered by MsMath 7 · 0⤊ 0⤋
a effective integer Q = (a^2+b^2)/(5ab+a million) can in no way equivalent 3 on the grounds that means that 3|a^2+b^2 which incredibly means that the two 3|a and 3|b yet 5ab+a million is key to a and b so this is best to 3 forcing a 9 to divide Q = 3, a contradiction. qed thank you! basically call me flipper. Now assume that 5ab+a million = 3(a^2+b^2). Divide with the help of ab: 5+a million/ab = 3(b/a + a/b). Min (a/b+b/a) = 2, Max 5+a million/ab = 6 those the two happen at a=b=a million, and the equation will fail for the different pairs. qed
2016-12-18 10:23:51 · answer #2 · answered by ? 4 · 0⤊ 0⤋
OK... we know sqrt(x) = x/40.
Multiply by 40:
40*sqrt(x) = x
Square both sides:
1600x = x^2
Rearrange:
x^2 - 1600x = 0
Factor out an x:
x*(x-1600) = 0
The product can only be 0 if one of the terms is zero.
Hence, either x=0 or x=1600.
Since 0 is not a positive number, the answer is:
x=1600.
Hopefully that helps!
2006-09-14 11:48:08 · answer #3 · answered by Bramblyspam 7 · 0⤊ 0⤋
1600
sq root of 1600=40
1600/40=40
2006-09-14 11:41:55 · answer #4 · answered by Glenn N 5 · 0⤊ 0⤋
1600
2006-09-14 11:43:43 · answer #5 · answered by btsmith_y 3 · 0⤊ 0⤋
12
2006-09-14 11:39:44 · answer #6 · answered by Alicat 6 · 0⤊ 3⤋
x^(1/2)=x/40
Which means (by squaring each side)
x=x^2/1600
Which means (by multiplying by 1600)
1600*x=x^2
Which means (by dividing by x)
x=1600
2006-09-14 11:43:29 · answer #7 · answered by cd4017 4 · 0⤊ 0⤋
1600!!!
2006-09-14 11:42:38 · answer #8 · answered by pistonfan1111 1 · 0⤊ 0⤋
4!!!!!!!!!!
2006-09-14 11:37:51 · answer #9 · answered by Anonymous · 0⤊ 3⤋