What is the value of the expression?
A.
9−48
B.
9−2
C.
92
D.
922

- anonymous

What is the value of the expression?
A.
9−48
B.
9−2
C.
92
D.
922

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- anonymous

- anonymous

Do you remember your exponent rules?\[a^b\times a^c = a^{b + c}\]\[\frac{a^b}{a^c}=a^{b-c}\]So, try working this one step at a time.\[\frac{9^8}{9^{-4}\times9^{10}}=\frac{9^8}{9^{-4}}\times\frac{1}{9^{10}}\]Does this help?

- anonymous

Confusing...
@Nnesha

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## More answers

- anonymous

What part is confusing you?

- anonymous

every part...

- Nnesha

1st write those exponents rules on a piece of paper(or notebook)!!!!!!!!PLEASE!!!

- anonymous

ok

- anonymous

done

- Nnesha

no.

- anonymous

huh i meant done as in im done writing them down

- Nnesha

did you write all of them just in 2 minutes ?

- anonymous

yes i did ill show u

- Nnesha

nice
and then answer his/her question ^^^^
he/she will help you
good luck!

- anonymous

##### 1 Attachment

- anonymous

PLease excuse my handwriting

- Nnesha

looks good.

- anonymous

umm i guess she meant u had to help me @Calcmathlete

- anonymous

*thinking* -I'm to stupid for her-

- anonymous

Alright. So, the first one. (\(a^b \times a^c = a^{b+c}\)). Let's do a smaller example so you can see what this means. \[2^3\times 2^2 = ?\]These numbers are small enough that we can do it by hand. \(2^3 = 8\) and \(2^2 = 4\), right? If we use this method, \(2^3\times2^2 = (8)\times(4) = 32\). However, if you notice, if we add the exponents instead, \(2^3 \times 2^2 = 2^{3 + 2} = 2^5 = 32\). Do you see what happened there?

- anonymous

yes

- anonymous

u did 2*2*2 which = 8 and then 2*2 which = 4 then u multiplied them and got 32

- anonymous

And if you're wondering why this works, think about it this way. When you have \(2^3\), what you're really doing is multiplying 2 by itself 3 times. So, \(2^3 = 2\times2\times2\). So, it should also make sense that \(2^3 \times2^2 = (2\times2\times2)\times(2\times2) = 2\times2\times2\times2\times2 = 2^5\) Does this part make sense?

- anonymous

yes

- anonymous

Ok. Let's work on the original problem then. Work on it little by little. The entire thing is \[\frac{9^8}{9^{-4}\times 9^{10}}\]So, let's work with the denominator first. Using what we did above (adding exponents), \(\large{9^{-4} \times9^{10}=9^{?}}\)

- anonymous

so would u subtract the exponents now or just multiply to get -40

- anonymous

Well, adding a negative number is subtracting, technically. If we follow what we did above, you would add (-4) to 10.

- anonymous

so 6

- anonymous

Yes. So, we can say that \(9^{-4}\times9^{10} = 9^6\). From there, plug it back in and move onto the second step. \[\frac{9^8}{9^{-4}\times9^{10}} \rightarrow\frac{9^8}{9^6}\]Here, we're dividing instead of multiplying. So, what would you do with the exponents now if we follow the fact that \(\frac{a^b}{a^c}=a^{b-c}\)?

- anonymous

In other words, \(\large{\frac{9^8}{9^6}=9^{?}}\)

- anonymous

umm do i just use the exponents like 8-6 or something

- anonymous

cuz thats 2 so that means its c

- anonymous

Yes. You're dividing, so you subtract the exponents. And yes, it is C. Good job.

- anonymous

Yay tysm!!!

- anonymous

I have 1 more question could u help me with it in a new post?

- anonymous

Sure.

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