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3 years ago

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AngleFBC and AngleCBG are supplements, AngleDBG and AngleDBF are supplements, and AngleCBG Is-congruent-to AngleDBF. 4 lines are

shown. A horizontal line with points F, B, G intersects a vertical line with points A, B, E at point B and forms a right angle. Another diagonal line with points D, B, C intersects the other 2 lines at point B between angle F B E and angle A B G. A fourth line intersects 3 lines at points A, C, G. By the congruent supplements theorem, what can you conclude? AngleCBG Is-congruent-to AngleDBG AngleFBC Is-congruent-to AngleDBG AngleCBG is supplementary to AngleDBF. AngleFBC is supplementary to AngleDBG.

1 answer:

aniked [119]3 years ago

5 0

Answer:

b

Step-by-step explanation:

i just did it on edginity

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<h2>Question :</h2>

$\tt \dfrac{x+2}{x-2} + \dfrac{x-2}{x+2} = \dfrac{5}{6}$

<h2>Answer :</h2>

$\large \underline{\boxed{\bf{x = \dfrac{\pm 2\sqrt{119}}{7}}}}$

<h2>Explanation :</h2>

$\tt : \implies \dfrac{x+2}{x-2} + \dfrac{x-2}{x+2} = \dfrac{5}{6}$

$\tt : \implies \dfrac{(x+2)(x+2) + (x-2)(x-2)}{(x-2)(x+2)} = \dfrac{5}{6}$

$\tt : \implies \dfrac{(x+2)^{2} + (x-2)^{2}}{(x-2)(x+2)} = \dfrac{5}{6}$

<u>Now, we know that</u> :

$\large \underline{\boxed{\bf{(a+b)^{2} = a^{2} + b^{2}+ 2ab}}}$
$\large \underline{\boxed{\bf{(a-b)^{2} = a^{2} + b^{2} - 2ab}}}$
$\large \underline{\boxed{\bf{(a+b)(a-b) = a^{2} - b^{2}}}}$

$\tt : \implies \dfrac{x^{2}+2^{2}+ 2 \times x \times 2 + x^{2}+2^{2} - 2 \times x \times 2 }{x^{2}-2^{2}} = \dfrac{5}{6}$

$\tt : \implies \dfrac{x^{2}+ 4 + \cancel{4x} + x^{2}+ 4 - \cancel{4x}}{x^{2}-4} = \dfrac{5}{6}$

$\tt : \implies \dfrac{x^{2} + x^{2} + 4 + 4}{x^{2}-4} = \dfrac{5}{6}$

$\tt : \implies \dfrac{2x^{2} + 8}{x^{2}-4} = \dfrac{5}{6}$

<u>By cross multiply</u> :

$\tt : \implies (2x^{2} + 8)6= 5(x^{2}-4)$

$\tt : \implies 12x^{2} + 48 = 5x^{2}-20$

$\tt : \implies 12x^{2} + 48 - 5x^{2} + 20 = 0$

$\tt : \implies 7x^{2} + 68 = 0$

$\tt : \implies 7x^{2} + 0x + 68 = 0$

<u>Now, by comparing with ax² + bx + c = 0, we have</u> :

a = 7
b = 0
c = 68

<u>By using quadratic formula</u> :

$\large \underline{\boxed{\bf{x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}}}}$

$\tt : \implies x = \dfrac{-(0) \pm \sqrt{(0)^{2} - 4(7)(68)}}{2(7)}$

$\tt : \implies x = \dfrac{0 \pm \sqrt{0 - 1904}}{14}$

$\tt : \implies x = \dfrac{\pm \sqrt{- 1904}}{14}$

$\tt : \implies x = \dfrac{\pm \sqrt{2\times 2\times 2\times 2\times 7\times 17}}{14}$

$\tt : \implies x = \dfrac{\pm \cancel{2} \times 2\sqrt{7\times 17}}{\cancel{14}}$

$\tt : \implies x = \dfrac{\pm2\sqrt{119}}{7}$

$\large \underline{\boxed{\bf{x = \dfrac{\pm 2\sqrt{119}}{7}}}}$

Hence value of $\bf x =\dfrac{\pm 2\sqrt{119}}{7}$

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3 years ago

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