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anonymous

  • 5 years ago

i need help with the following attachment ....

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  1. anonymous
    • 5 years ago
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  2. anonymous
    • 5 years ago
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    What do you need help with specifically. Long problem, has a lot of work on it already?

  3. anonymous
    • 5 years ago
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    to check #6 if i did it correctlty?

  4. anonymous
    • 5 years ago
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    Good work, unfortunately some algebra mistakes. When working on an equation, If you do something to one side, you have to do to the other side (all terms). Likewise when you do something to a radical expression p/q; you must do top and bottom.

  5. anonymous
    • 5 years ago
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    ahh i see so then that would make it x = 400*666.7?

  6. anonymous
    • 5 years ago
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    On your first step you square, square root (400 + x^2) and 1000 but you didn't square 1500. In any event, it is not recommended to go this route. When you want to square to remove radical, you should first isolate the radical to one side and all other terms to the other side. So your first step, it is recommended to multiply through by sq rt (400+x^2) not to remove radical but to cancel it from the denominator.

  7. anonymous
    • 5 years ago
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    okay give me a sec and ill do it

  8. anonymous
    • 5 years ago
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    so x=(2/3)?

  9. anonymous
    • 5 years ago
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    That was a big jump from one step to the next, may be it's right but the next step is \[-1000(\sqrt{400+x ^{2}}) +1500x=0\]

  10. anonymous
    • 5 years ago
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    aww no im not right i did something else..okay let me see if i can get it

  11. anonymous
    • 5 years ago
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    is the next step moving the "1000" over the "=" sign making it a positive?

  12. anonymous
    • 5 years ago
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    You can, but it is just as easy to divide by -1000.

  13. anonymous
    • 5 years ago
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    so it just basically cancels out?

  14. anonymous
    • 5 years ago
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    \[\sqrt{400+x ^{2}}=(-1500x)/(-1000)\]

  15. anonymous
    • 5 years ago
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    wouldnt it be positive 1000?

  16. anonymous
    • 5 years ago
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    oh wait nevermind...

  17. anonymous
    • 5 years ago
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    would it be -1500x=\[\sqrt{400+x ^{2}}\](-1000) ?

  18. anonymous
    • 5 years ago
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    The purpose of doing this is to isolate the radical, then square it and get rid of it\[400+x ^{2}=(9x ^{2})/4\]

  19. anonymous
    • 5 years ago
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    how did you get \[(9x ^{2})/4??\]

  20. anonymous
    • 5 years ago
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    \[[(-1500x)/(-1000)]^{2}=(15x/10)^{2}=(3x/2)^{2}\]

  21. anonymous
    • 5 years ago
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    does x= -31???

  22. anonymous
    • 5 years ago
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    I am not confident in your work, you jump too many steps; it's not the way you do mathematics. Work it from there. I'll check it later. It is definitely not -31

  23. anonymous
    • 5 years ago
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    okay i went from your step to \[x^{2}=(9^{x ^{2}}-400]??

  24. anonymous
    • 5 years ago
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    \[x ^{2}=(9x ^{2}-400)\]

  25. anonymous
    • 5 years ago
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    9xsquared /4 that is sorry

  26. anonymous
    • 5 years ago
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    i got \[100\sqrt{2}\]

  27. anonymous
    • 5 years ago
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    \[x ^{2}-(9x ^{2})/4=-400\]

  28. anonymous
    • 5 years ago
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    \[x ^{2}(1-[9/4]=-400\]

  29. anonymous
    • 5 years ago
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    \[x ^{2}(-5/4)=-400\]

  30. anonymous
    • 5 years ago
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    isee how your getting this:D

  31. anonymous
    • 5 years ago
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    \[x ^{2}=400(4/5)\]

  32. anonymous
    • 5 years ago
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    now we divide -400 and (-5/4)?

  33. anonymous
    • 5 years ago
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    Yes\[x =\sqrt{320}\]\[x =6\sqrt{5}\]

  34. anonymous
    • 5 years ago
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    ahhh i had \[\sqrt{200}\] .... boy was i off ... well thank you so much for helping me i have a better understanding for calculus im in 10th grade so im new to this ... you were tons of help ..bye bye?!

  35. anonymous
    • 5 years ago
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    Good work, you are doing good. It took years of working with a tutor for me to feel this comfortable.

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