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lawls
Group Title
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)
 2 years ago
 2 years ago
lawls Group Title
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)
 2 years ago
 2 years ago

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sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
first tell me which is the longest side
 2 years ago

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

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

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

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.2
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?
 2 years ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Yes sir! Yes sir!
 2 years ago

lawls Group TitleBest ResponseYou've already chosen the best response.0
i got x = 4, 12
 2 years ago

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

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

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