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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)
 one year ago
 one year 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)
 one year ago
 one year ago

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sauravshakyaBest ResponseYou've already chosen the best response.0
first tell me which is the longest side
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
x7,x,x+1....... what do u thik have the biggest value
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Hint: Work with the dearest Pythagorean Theorem.
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
But beofre using Pythagorean Theorem.. u must find the longest side
 one year ago

CalcmathleteBest 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?
 one year ago

lawlsBest ResponseYou've already chosen the best response.0
so answer choice (e) is right?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
You, sir, are totally correct!
 one year ago

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