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lara [203]
3 years ago
13

Find the value of x.

Mathematics
2 answers:
Lisa [10]3 years ago
8 0

{x}^{2}  +  {(x  + 3)}^{2}  =   {( \sqrt{117} )}^{2}   \\ 2 {x}^{2}  + 6x + 9 = 117 \\ 2 {x}^{2}   + 6x - 108  = 0\\  {x }^{2}  + 3x - 54 = 0 \\ (x + 9)(x - 6) = 0 \\ x =  - 9 \: or \: 6 \\ but \: x > 0 \\ so \: the \: answer \: is \: x = 6
hope this would help you
Cerrena [4.2K]3 years ago
3 0
x^2 + (x + 3)^2 = (\sqrt{117})^2

x^2 + x^2 + 6x + 9 = 117

2x^2 + 6x -108 = 0

x^2 + 3x - 54 = 0

(x + 9)(x - 6) = 0

x + 9 = 0 ~~or~~ x - 6 = 0

x = -9 ~~ or~~ x = 6

Since the side of a triangle cannot have a negative length, we discard the solution x = -9.

Answer: x = 6
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Step-by-step explanation:

Since the foci are at(0,±c) = (0,±63) and vertices (0,±a) = (0,±91), the major axis is the y- axis. So, we have the equation in the form (with center at the origin) \frac{x^{2} }{b^{2} } + \frac{y^{2} }{a^{2} }.

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So the equation is

\frac{x^{2} }{(14\sqrt{22}) ^{2} } + \frac{y^{2} }{91^{2} } = \frac{x^{2} }{4312 } + \frac{y^{2} }{8281 }

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