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Mekhanik [1.2K]
3 years ago
9

Plz help me with this

Mathematics
2 answers:
enyata [817]3 years ago
5 0

I’m pretty sure the answer is B.

pashok25 [27]3 years ago
5 0

Answer:   \bold{b)\quad \dfrac{-2\pm \sqrt{14}}{2}}

<u>Step-by-step explanation:</u>

2x^2+4x-5=0\qquad \implies\qquad a=2,\ b=4,\ c=-5\\\\\text{The quadratic equation is not factorable so use the Quadratic Formula:}\\x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\x=\dfrac{-4\pm \sqrt{4^2-4(2)(-5)}}{2(2)}\\\\\\.\ =\dfrac{-4\pm \sqrt{16+40}}{4}\\\\\\.\ =\dfrac{-4\pm \sqrt{56}}{4}\\\\\\.\ =\dfrac{-4\pm 2\sqrt{14}}{4}\\\\\\.\ =\large\boxed{\dfrac{-2\pm \sqrt{14}}{2}}

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1. A pair of supplementary angles: ∠IJH and ∠HJG, ∠IJH and ∠HJG

2. A pair of complementary angles: ∠JGK and ∠KGC, ∠FGE and ∠EGD

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Sum of the adjacent angles in a straight line = 180°

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∠JGK + ∠KGC = 90°

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Therefore ∠FGE and ∠EGD are complementary angles.

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