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Answer 1

\[a^x \times a^y = a^{x+y}\]

Answer 2

and ur answer is incorrect

Answer 3

and also
\[\frac{ 1 }{ a^x } = a^{-x}\]

Answer 4

so its \[4^2\]

Answer 5

Close. If we follow what @rishavraj said we would still have a negative as an exponent...

\(\Huge{4^{6} \times 4^{-8} = 4^{(6 + -8)} = 4^{-2}}\)

Since the exponent is still a negative we would have it in fraction form to rid of the negative...

\(\Huge{\frac{1}{4^{2}}=\frac{1}{16}}\)

Answer 6

Do you understand?

Answer 7

No im confused with the \[\frac{ 1 }{ 4 } = \frac{ 1 }{ }\]

Answer 8

excuse me 1 over 16

Answer 9

That is the simplification of the fraction....

Like @rishavraj said...

\(\Huge{a^{-x}=\frac{1}{a^{x}}}\)

This shows that if you have a negative exponent you would then convert it to fraction form in which i did with your problem....

\(\Huge{4^{-2}=\frac{1}{4^{2}}}\)

This is done so we can rid of the negative...

Now \(\Large{4^{2}}\) can be simplified by multiplying itself `4 x 4` in which we would get 16 but \(\Large\frac{1}{16}\) is not one of the choices so we would leave it as \(\LARGE{\frac{1}{4^{2}}}\)...

Answer 10

Ok that makes more sense to me now thanks im not really good at math thanks so much

Answer 11

np :)