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anonymous

  • one year ago

FIND THE VALUE OF X

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @Astrophysics

  3. Astrophysics
    • one year ago
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    Hmm, I'm not very sure, as I'm not so good at geometry, and I don't want to mislead you, sorry.

  4. anonymous
    • one year ago
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    okay thanks could you try another problem? is there anyone else who could help me?

  5. Astrophysics
    • one year ago
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    @phi is good at these kinds of problems

  6. anonymous
    • one year ago
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    ok thanks

  7. Astrophysics
    • one year ago
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    Np, good luck!

  8. anonymous
    • one year ago
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    thanks!

  9. misty1212
    • one year ago
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    HI!!

  10. misty1212
    • one year ago
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    you could try \[\frac{2x}{15}=\frac{x+2}{12.5}\]

  11. misty1212
    • one year ago
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    start with \[12.5\times 2x=15(x+2)\] and solve for \(x\)

  12. Astrophysics
    • one year ago
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    Ah, so this one requires a ratio to, like your last question, thanks @misty1212

  13. misty1212
    • one year ago
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    \[\color\magenta\heartsuit\]

  14. anonymous
    • one year ago
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    hold on im sloving it

  15. anonymous
    • one year ago
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    is the answer 6?

  16. anonymous
    • one year ago
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    @misty1212

  17. Astrophysics
    • one year ago
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    Nope, you may have messed up somewhere, \[(12.5)(2x)=15x+30\]

  18. anonymous
    • one year ago
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    25x=15x+30

  19. Astrophysics
    • one year ago
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    \[25x=15x+30\]

  20. Astrophysics
    • one year ago
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    Yes

  21. Astrophysics
    • one year ago
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    Keep going

  22. anonymous
    • one year ago
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    ok

  23. anonymous
    • one year ago
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    10x=30?

  24. Astrophysics
    • one year ago
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    Yup, one more step

  25. anonymous
    • one year ago
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    x=3?

  26. Astrophysics
    • one year ago
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    Yes!

  27. anonymous
    • one year ago
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    ok thanks! i need help with a few more!

  28. anonymous
    • one year ago
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  29. anonymous
    • one year ago
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    find the value of x

  30. anonymous
    • one year ago
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    @misty1212

  31. anonymous
    • one year ago
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    @Astrophysics

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