Question

(3x)^0 = 1 Am I right?

Answer 1

So anything raised to the 0 power, always, will be 1. So yes, you are right!

Answer 2

Thanks!

Answer 3

Of course =)

Answer 4

The correct answer is "maybe". It depends on the value of 'x'. If x = 0, it is most assuredly NOT 1.

Answer 5

your right, assuming \(x\neq0\),

Answer 6

Ooooo that's a good point. Anything (Not 0) raised to the 0 power is = 1. But 0 raised to the 0 power is what... undefined @tkhunny ?

Answer 7

0

Answer 8

no

Answer 9

Indeed.

Answer 10

Am I right or wrong? I'm confused.

Answer 11

No your right

Answer 12

zero to any power is zero

Answer 13

I've confused myself

Answer 14

Anything raised to the 0 power, not including 0, is = 1. Where (3x)^0 is 1, where x does not equal 0.

Answer 15

\(0^0 \) is undefined

Answer 16

\(\large (3x)^0 = 1 \text{ for } x \ne 0\)

Answer 17

I thought 0^0 was undefined.

Answer 18

0^0 is not always clear. It is referred to as "indeterminate". It is a little better than undefined. It could be undefined or it could be defined. It depends on EXACTLY the nature of how you arrived at the state of 0^0.

Answer 19

In imprecise terms: anything raised to zero is one. zero raised to anything is zero. Then what is zero raised to zero? Is it zero or is it one?

Answer 20

How about 10(mn)^0

Answer 21

I don't understand this one

Answer 22

As long as m or n doesn't = 0, then it is 1

Answer 23

Not quite. The requirement is that m AND n must not be zero. More plainly, m*n must not be zero.

Answer 24

(9x)^0 - 9x^0 - (-9x)^0 I kind of understand this one, but I'm not sure...

Answer 25

Where x does not = 0, then it's 1 -1 -1 right?

Answer 26

I think the answer is -7

Answer 27

For x not equal to 0, (9x)^0 - 9x^0 - (-9x)^0 = 1 - 9*1 - 1 = -9