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anonymous

  • 5 years ago

i am having trouble with exponets. can u help me?

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  1. anonymous
    • 5 years ago
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    sure, what is the problem?

  2. anonymous
    • 5 years ago
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    i have more than one problem if thats ok

  3. anonymous
    • 5 years ago
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    I will tell you how to do one problem and give you the tools necessary to solve the rest of hte problems. don't expect me to solve all your problems for you.

  4. anonymous
    • 5 years ago
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    ok, the problem is -2b(squaredtothe3rdpower)x7b(tothe6thpower)

  5. anonymous
    • 5 years ago
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    \[(-2b ^{2}) ^{3} \times 7b ^{6}\]

  6. anonymous
    • 5 years ago
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    is that it? use the equation button below to type in your equation

  7. anonymous
    • 5 years ago
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    yes that is it, but there are no parenthesees around the -2b

  8. anonymous
    • 5 years ago
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    I need you to write it out and take a picture and attach it here or write it out using the equation button

  9. anonymous
    • 5 years ago
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    sorry i dont use computers that much, what is the equation button? srry

  10. anonymous
    • 5 years ago
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    it is on the screen right where the arrow is pointing | | v

  11. anonymous
    • 5 years ago
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    it is on the left of the post button and attach file button

  12. anonymous
    • 5 years ago
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    sorry i dontn o what to do, the problem that u put in earlier, look up, the one i told u about, help me with that one

  13. anonymous
    • 5 years ago
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    \[-2b ^{3} \times 7b ^{6}\]

  14. anonymous
    • 5 years ago
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    right?

  15. anonymous
    • 5 years ago
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    yes

  16. anonymous
    • 5 years ago
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    the basic rules you have to follow are : \[x ^{a} \times x ^{b} = x ^{a+b}\] \[x ^{a} \div x ^{b} = x ^{a-b}\]

  17. anonymous
    • 5 years ago
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    ok then what

  18. anonymous
    • 5 years ago
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    \[-2b ^{3} \times 7b ^{6} = -14 (b ^{3} \times b ^{6})\]

  19. anonymous
    • 5 years ago
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    ok, got it then what

  20. anonymous
    • 5 years ago
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    \[= -14b ^{3+6} = -14b ^{9}\]

  21. anonymous
    • 5 years ago
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    beast, that is so cool

  22. anonymous
    • 5 years ago
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    thank you

  23. anonymous
    • 5 years ago
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    you are welcome.

  24. anonymous
    • 5 years ago
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    lol do u help people just for fun

  25. anonymous
    • 5 years ago
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    yes

  26. anonymous
    • 5 years ago
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    wow u are a really kind person, thank u again u are a big help and might help me get a better grade on my test coming up

  27. anonymous
    • 5 years ago
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    awww. you made me blush. all the best for you exam.

  28. anonymous
    • 5 years ago
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    thank u have a great night

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spraguer (Moderator)
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