anonymous
  • anonymous
What is the value of the expression? A. 9−48 B. 9−2 C. 92 D. 922
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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
  • anonymous
Confusing... @Nnesha

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anonymous
  • anonymous
What part is confusing you?
anonymous
  • anonymous
every part...
Nnesha
  • Nnesha
1st write those exponents rules on a piece of paper(or notebook)!!!!!!!!PLEASE!!!
anonymous
  • anonymous
ok
anonymous
  • anonymous
done
Nnesha
  • Nnesha
no.
anonymous
  • anonymous
huh i meant done as in im done writing them down
Nnesha
  • Nnesha
did you write all of them just in 2 minutes ?
anonymous
  • anonymous
yes i did ill show u
Nnesha
  • Nnesha
nice and then answer his/her question ^^^^ he/she will help you good luck!
anonymous
  • anonymous
anonymous
  • anonymous
PLease excuse my handwriting
Nnesha
  • Nnesha
looks good.
anonymous
  • anonymous
umm i guess she meant u had to help me @Calcmathlete
anonymous
  • anonymous
*thinking* -I'm to stupid for her-
anonymous
  • 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
  • anonymous
yes
anonymous
  • anonymous
u did 2*2*2 which = 8 and then 2*2 which = 4 then u multiplied them and got 32
anonymous
  • 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
  • anonymous
yes
anonymous
  • 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
  • anonymous
so would u subtract the exponents now or just multiply to get -40
anonymous
  • anonymous
Well, adding a negative number is subtracting, technically. If we follow what we did above, you would add (-4) to 10.
anonymous
  • anonymous
so 6
anonymous
  • 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
  • anonymous
In other words, \(\large{\frac{9^8}{9^6}=9^{?}}\)
anonymous
  • anonymous
umm do i just use the exponents like 8-6 or something
anonymous
  • anonymous
cuz thats 2 so that means its c
anonymous
  • anonymous
Yes. You're dividing, so you subtract the exponents. And yes, it is C. Good job.
anonymous
  • anonymous
Yay tysm!!!
anonymous
  • anonymous
I have 1 more question could u help me with it in a new post?
anonymous
  • anonymous
Sure.

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