Question

Guys, if you're factoring, and you have a problem that has \[(r^0)5\] there wouldn't be a r, would there??

Answer 1

Maybe it will help if you post the whole problem, but mathematically r^0=1, so that is a number.

Answer 2

The original problem is this: \[(-2r^0s^5)^5\]

Answer 3

anyhow, ANY VALUE raised to 0 is 1, ANY

Answer 4

So then it would be \[(-2r^1s^5)^5\]

Answer 5

anyhow, ANY VALUE raised to 0 is \(\huge \color{red}1\), ANY

Answer 6

... I understand. Is that what i asked though? No. I asked if that was correct, which I have yet to be answered. >.<

Answer 7

hehe

Answer 8

I never said it was raised to 1, I said IT WAS 1

Answer 9

I'm not solving for vaiables though. i'm factoring. >.< "factoring"

Answer 10

same applies, say r=2, 2^0=1, say r=1,000,000,000, 000, 1,000,000,000, 000^2=1

Answer 11

Makes no sense.

Answer 12

wooops, I meant 1,000,000,000, 000^0 btw :)

Answer 13

no, there would be any r
and any value (except 0) raised to 0 =1

Answer 14

so is my problem where the r is correct?

Answer 15

dunno, what do you think is \(\large r^{0*5}\)?

Answer 16

(r^0)^5 = 1^5=1