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anonymous

  • one year ago

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

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    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?

  3. anonymous
    • one year ago
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    Confusing... @Nnesha

  4. anonymous
    • one year ago
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    What part is confusing you?

  5. anonymous
    • one year ago
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    every part...

  6. Nnesha
    • one year ago
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    1st write those exponents rules on a piece of paper(or notebook)!!!!!!!!PLEASE!!!

  7. anonymous
    • one year ago
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    ok

  8. anonymous
    • one year ago
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    done

  9. Nnesha
    • one year ago
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    no.

  10. anonymous
    • one year ago
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    huh i meant done as in im done writing them down

  11. Nnesha
    • one year ago
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    did you write all of them just in 2 minutes ?

  12. anonymous
    • one year ago
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    yes i did ill show u

  13. Nnesha
    • one year ago
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    nice and then answer his/her question ^^^^ he/she will help you good luck!

  14. anonymous
    • one year ago
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  15. anonymous
    • one year ago
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    PLease excuse my handwriting

  16. Nnesha
    • one year ago
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    looks good.

  17. anonymous
    • one year ago
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    umm i guess she meant u had to help me @Calcmathlete

  18. anonymous
    • one year ago
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    *thinking* -I'm to stupid for her-

  19. anonymous
    • one year ago
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    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?

  20. anonymous
    • one year ago
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    yes

  21. anonymous
    • one year ago
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    u did 2*2*2 which = 8 and then 2*2 which = 4 then u multiplied them and got 32

  22. anonymous
    • one year ago
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    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?

  23. anonymous
    • one year ago
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    yes

  24. anonymous
    • one year ago
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    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), \(\large{9^{-4} \times9^{10}=9^{?}}\)

  25. anonymous
    • one year ago
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    so would u subtract the exponents now or just multiply to get -40

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

  27. anonymous
    • one year ago
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    so 6

  28. anonymous
    • one year ago
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    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 \(\frac{a^b}{a^c}=a^{b-c}\)?

  29. anonymous
    • one year ago
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    In other words, \(\large{\frac{9^8}{9^6}=9^{?}}\)

  30. anonymous
    • one year ago
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    umm do i just use the exponents like 8-6 or something

  31. anonymous
    • one year ago
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    cuz thats 2 so that means its c

  32. anonymous
    • one year ago
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    Yes. You're dividing, so you subtract the exponents. And yes, it is C. Good job.

  33. anonymous
    • one year ago
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    Yay tysm!!!

  34. anonymous
    • one year ago
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    I have 1 more question could u help me with it in a new post?

  35. anonymous
    • one year ago
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    Sure.

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