Question

6^-2

Answer 1

whats your question?

Answer 2

That^

Answer 3

do you need to simplify

Answer 4

hint \[b^{-n}=\frac{1}{b^n}\] but \(b=6, n=2\)

Answer 5

i mean "PUT"

Answer 6

But it says that it's wrong

Answer 7

12

Answer 8

did you write \[\frac{1}{6^2}\]?

Answer 9

No

Answer 10

then it was wrong
unless you wrote \[\frac{1}{36}\]

Answer 11

Yes that was right thanks sir

Answer 12

lol i like "sir"

Answer 13

oh my bad. i confused it. I thought it was 6*-2, but it was actually 6*6

Answer 14

@helpneeded1914
Had it been 6 * (-2), the answer would have been -12, not 12.
Had it been 6 * 6, the answer would have been 36, but that's not it either.
In fact, it is neither of those.

To evaluate \(\Large 6^{-2} \), use the definition of a negative exponent:

\(\Large a^{-n} = \dfrac{1}{a^n} \)

Solving this specific problem:

\(\Large 6^{-2} = \dfrac{1}{6^2} = \dfrac{1}{6 \times 6} = \dfrac{1}{36} \)

Answer 15

@mathstudent55 yes due to the fact that it would have ended in a decimal where the 7 continously repeated, so the number would be better reffered to in a decimal.