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anonymous

  • 5 years ago

How long is a leg a of a right triangle whose other leg b measures 2x units, and whose hypotenuse c measures 4x units (where x 0) ?

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  1. anonymous
    • 5 years ago
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    this is sq root ((2x)^2 + (4x)^2))

  2. anonymous
    • 5 years ago
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    oopps - sorry it is sq root((4x)^2 - (2x)^2))

  3. anonymous
    • 5 years ago
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    \[(4x)^{2}-(2x)^{2}\]

  4. anonymous
    • 5 years ago
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    this comes to sq root12 x^2

  5. anonymous
    • 5 years ago
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    \[16x^{2}-4x ^{2}=12x ^{2}\]\[\sqrt{12x ^{2}}=2x \sqrt{3}\]

  6. anonymous
    • 5 years ago
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    \[\frac{\sqrt{3}}{2}x \]

  7. anonymous
    • 5 years ago
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    Thank you so much! I havent had math in a year and I'm trying to review some math

  8. anonymous
    • 5 years ago
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    If I deserve medal then give me one.

  9. anonymous
    • 5 years ago
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    How do I do that..

  10. anonymous
    • 5 years ago
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    next to my name click give medal

  11. anonymous
    • 5 years ago
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    what u see ,give medal or good answer?

  12. anonymous
    • 5 years ago
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    \[4x \text{Cos}[\text{ArcSin}[(2x)/(4x)]] =2 \sqrt{3} x \] I forgot to multiply by 4x

  13. anonymous
    • 5 years ago
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    thank

  14. myininaya
    • 5 years ago
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    \[2x \sqrt{4-x^2}\]

  15. myininaya
    • 5 years ago
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    its a right triangle so we have a^2+b^2=c^2 c=4x b=2x a=? a^2=c^2-b^2 a=sqrt{c^2-b^2} a=sqrt{(4x)^2-(2x)^2} a=sqrt{16x^2-4x^2} a=sqrt{4x^2}sqrt{4-1} a=2xsqrt{3} oh i made a mistake lol i guess it helps when u work it out

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