4 years ago

PLZZZZZZZZZZZZZZZZZZZZZZZZZ HELP WILL GIVE BRAINLIEST

Mathematics

2 answers:

elixir [45]4 years ago

8 0

You're answer would be B. Vertical Angles.

Hope this helps, Kam

motikmotik4 years ago

3 0

Answer:

Vertical I believe. #Hopethishelps

Step-by-step explanation:

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Solve this for Brainliest! See the attachment [Level - GCSE]

madreJ [45]

Answer:

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Step-by-step explanation:

OPQ is a right angle triangle.

Using Pythagoras

$x^{2} + (x+5)^{2} = (x+8)^{2}\\x^{2} + x^{2} + 10x +25 = x^{2} + 16x + 64\\x^{2} + x^{2} - x^{2} + 10x - 16x + 25 - 64 = 0\\x^{2} - 6x -39 = 0$

Using quadratic formula

$x = \frac{-b \± \sqrt{b^{2}-4ac}}{2a}\\x = \frac{-(-6) \± \sqrt{(-6)^{2}-4(1)(-39)}}{2(1)}\\x = \frac{6}{2} \± \frac{\sqrt{192}}{2}\\x = 3 \± \frac{8\sqrt{3}}{2}\\x = 3 \± 4\sqrt{3}$

$x = 3 + 4\sqrt{3} = 9.928(4s.f.)$

$x = 3 - 4\sqrt{3} = -3.928(4s.f.)$ (Length can't be negative)

∴ $x = 3 + 4\sqrt{3}$

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torisob [31]

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