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lawls
 2 years ago
If the sides of a right triangle have lengths of x7, x an x+1, then x=
(a) (1, 7)
(b) (7, 1)
(c) (7, 4)
(d) (7, 12)
(e) (4, 12)
lawls
 2 years ago
If the sides of a right triangle have lengths of x7, 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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sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.0first tell me which is the longest side

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.0x7,x,x+1....... what do u thik have the biggest value

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0Hint: Work with the dearest Pythagorean Theorem.

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.0But beofre using Pythagorean Theorem.. u must find the longest side

Calcmathlete
 2 years ago
Best ResponseYou've already chosen the best response.2Obviously, 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?

lawls
 2 years ago
Best ResponseYou've already chosen the best response.0so answer choice (e) is right?

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0You, sir, are totally correct!

Calcmathlete
 2 years ago
Best ResponseYou've already chosen the best response.2Kinda ironic since 12 is the only true solution, but whatever works ^_^
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