If x=2 and xy=0, what is the value of x^4y^2-(xy)^2/ x^3y^2 ?
How do you come to the conclusion?
2007-03-26 23:05:51 · 5 answers · asked by thatfella 1 in Science & Mathematics ➔ Mathematics
x = 2
xy = 0
2y = 0
y = 0/2
y = 0
x^4y^2 = x^4 * 0 = 0
(xy)^2 = 0^2 = 0
x^3y^2 = x^3 * 0 = 0
x^4y^2-(xy)^2/ x^3y^2 = 0/0 = infinity
2007-03-26 23:15:18 · answer #1 · answered by Akilesh - Internet Undertaker 7 · 0⤊ 0⤋
If x=2 and xy=0, Then y=0
what is the value of x^4y^2-(xy)^2/ x^3y^2 = 0/0 which is indeterminent/
2007-03-27 06:11:32 · answer #2 · answered by John S 6 · 0⤊ 0⤋
If x=2 and xy=0, what is the value of
x^4 y^2 -(xy)^2 / x^3y^2
x = 2
xy = 0
2y = 0
y = 0
then
x^4 y^2 -(xy)^2 / x^3y^2
= x^4 0^2 -(x0)^2 / x^30^2
= 0/0
= indeterminate. any number can't be divided by zero.
2007-03-27 06:34:15 · answer #3 · answered by rooster1981 4 · 0⤊ 0⤋
if xy = 0 and x = 2, that means
xy = 0
(-2)y = 0
y = 0
x^4y^2-(xy)^2/ x^3y^2
This equation is undefined because you can't have a zero in the denominator. (x^3y^2 will equal zero if y = 0.)
2007-03-27 06:18:36 · answer #4 · answered by Mathematica 7 · 0⤊ 0⤋
the answer is zero because xy=0 and there was the y^2 and xy automatically they will be zero
.
2007-03-27 06:11:17 · answer #5 · answered by Andrew T 1 · 0⤊ 1⤋