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3 years ago

11

picture attached please help I WILL MARK AS BRAINLIEST!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

!!!!!!!!!!!!!!!!!!1

2 answers:

Tomtit [17]3 years ago

5 0

Where's the pic? Lol

Vedmedyk [2.9K]3 years ago

3 0

It dosent let me load

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4 years ago

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

madreJ [45]

Answer:

$x = 3 + 4\sqrt{3}$

Step-by-step explanation:

OPQ is a right angle triangle.

<u>Using Pythagoras</u>

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

<u>Using quadratic formula</u>

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

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

OR

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

∴ $x = 3 + 4\sqrt{3}$

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2 years ago

Read 2 more answers

Nyla incorrectly added the fraction 2/3,1/2 and 5/12 she said that to add fractions with different denominators you use the comm

lord [1]

Answer:

The denominators are not equal.

Step-by-step explanation:

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3 years ago