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lawls

  • 3 years ago

If the sides of a right triangle have lengths of x-7, x an x+1, then x= (a) (1, 7) (b) (7, -1) (c) (7, 4) (d) (7, 12) (e) (4, 12)

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  1. sauravshakya
    • 3 years ago
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    first tell me which is the longest side

  2. sauravshakya
    • 3 years ago
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    x-7,x,x+1....... what do u thik have the biggest value

  3. ParthKohli
    • 3 years ago
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    Hint: Work with the dearest Pythagorean Theorem.

  4. sauravshakya
    • 3 years ago
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    But beofre using Pythagorean Theorem.. u must find the longest side

  5. Calcmathlete
    • 3 years ago
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    Obviously, x + 1 would be the largest side and therefore the hypotenuse because of the + 1 while x is neutral and -7 is negative. THerefore, now use Pythagorean THeorem. \[c^2 = a^2 + b^2\]\[(x + 1)^2 = x^2 + (x - 7)^2\]\[x^2 + 2x + 1 = x^2 + x^2 - 14x + 49\]\[x^2 + 2x + 1 = 2x^2 - 14x + 49\]\[0 = x^2 - 16x + 48\]Can you figure out how to solve the4 quadratic from here?

  6. ParthKohli
    • 3 years ago
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    Yes sir! Yes sir!

  7. lawls
    • 3 years ago
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    i got x = 4, 12

  8. lawls
    • 3 years ago
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    so answer choice (e) is right?

  9. ParthKohli
    • 3 years ago
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    You, sir, are totally correct!

  10. Calcmathlete
    • 3 years ago
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    Kinda ironic since 12 is the only true solution, but whatever works ^_^

  11. lawls
    • 3 years ago
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    thanks

  12. Calcmathlete
    • 3 years ago
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    np :)

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