2006-12-17 16:26:48 · 8 answers · asked by jaguarz 1 in Science & Mathematics ➔ Mathematics
40x to the 2
2006-12-17 16:28:53 · answer #1 · answered by Ilya 4 · 2⤊ 0⤋
(5x^2)(8x^2)(x^[-2])
I assume that you want to simplify this.
Since multiplication is commutative (the order in which you multiply numbers doesn't matter; 5 x 6 is the same as 6 x 5), you can rearrange it how you like. I'm going to move the numbers to the front, giving us
(5)(8)(x^2)(x^2)(x^[-2])
We don't have to move them to the front to do this, but I wanted to let you know that we can multiply them together, giving us
(40)(x^2)(x^2)(x^[-2])
Now, whenever we multiply a number with the same base but different exponent, we can ADD the exponents. In this case, since we have x^2 and x^2, this is equal to x^(2+2) = x^4. Let's start by doing that.
40(x^4) (x^[-2])
And now, like we said, we can add those exponents, giving us
40(x^[4 + (-2)])
or, quite simply
40x^2
2006-12-17 16:31:06 · answer #2 · answered by Puggy 7 · 2⤊ 0⤋
40x^4 divided by x^2= 40x^2
2006-12-17 16:33:19 · answer #3 · answered by Tony T 4 · 2⤊ 0⤋
40x^2
2006-12-17 16:29:11 · answer #4 · answered by Keith 1 · 2⤊ 0⤋
5*8*x^(2+2-2)
=40x^2
2006-12-17 16:29:07 · answer #5 · answered by raj 7 · 2⤊ 1⤋
5x^2 * 8x^2 * x^(-2)
40x^4 * x^(-2)
40x^2
Solved!!!!
2006-12-17 16:40:53 · answer #6 · answered by fortman 3 · 2⤊ 0⤋
multiply the coefficients and add the exponents
40 x^2
2006-12-17 16:39:20 · answer #7 · answered by Renaud 3 · 2⤊ 0⤋
40x^4*x^-2
40x^2
2006-12-17 16:43:18 · answer #8 · answered by Anonymous · 1⤊ 0⤋