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anonymous

  • one year ago

What is the value of x? Round to the nearest tenth.

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  1. anonymous
    • one year ago
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  2. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle a^2+b^2=c^2 }\) where c is the hypotenuse of a right triangle, and a & b are the legs.

  3. SolomonZelman
    • one year ago
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    Would you then agree with the following set up (to find x)? \(\large\color{black}{ \displaystyle 12^2+x^2=22^2 }\)

  4. anonymous
    • one year ago
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    Yes

  5. SolomonZelman
    • one year ago
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    ok, then tell me what x is going to equal (and disregard the negative x solution)

  6. anonymous
    • one year ago
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    10

  7. SolomonZelman
    • one year ago
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    no

  8. SolomonZelman
    • one year ago
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    it is not 12+x=22 then it would have been correct, BUT///

  9. SolomonZelman
    • one year ago
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    it is 12²+x²=22²

  10. anonymous
    • one year ago
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    oh ok

  11. SolomonZelman
    • one year ago
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    first, you have to simplify the powers.

  12. SolomonZelman
    • one year ago
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    12²=? 22²=?

  13. anonymous
    • one year ago
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    144

  14. anonymous
    • one year ago
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    484

  15. SolomonZelman
    • one year ago
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    yes

  16. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle 12^2+x^2=22^2 }\) \(\large\color{black}{ \displaystyle 144+x^2=484 }\) when you subtract 144 from both sides, you get: \(\large\color{black}{ \displaystyle x^2=340 }\)

  17. SolomonZelman
    • one year ago
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    So, x² is 340

  18. anonymous
    • one year ago
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    okay

  19. SolomonZelman
    • one year ago
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    then take the square root of both sides.

  20. anonymous
    • one year ago
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    okay

  21. anonymous
    • one year ago
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    then?

  22. anonymous
    • one year ago
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    so what's the final answer? @SolomonZelman

  23. anonymous
    • one year ago
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    @SolomonZelman i running out of time please

  24. anonymous
    • one year ago
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    i'm*

  25. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle x^2=340 }\) \(\large\color{black}{ \displaystyle \sqrt{x^2}=\sqrt{340} }\) \(\large\color{black}{ \displaystyle x=\pm \sqrt{340} }\) exclude the negative result, as a side can't be negagive. \(\large\color{black}{ \displaystyle x=\sqrt{340} }\) \(\large\color{black}{ \displaystyle x=\sqrt{4\times 85} }\) \(\large\color{black}{ \displaystyle x=2\sqrt{85} }\)

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