The answer is D
Because I worked it out
Answer:
![x = 3 + 4\sqrt{3}](https://tex.z-dn.net/?f=x%20%3D%203%20%2B%204%5Csqrt%7B3%7D)
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](https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2B%20%28x%2B5%29%5E%7B2%7D%20%3D%20%28x%2B8%29%5E%7B2%7D%5C%5Cx%5E%7B2%7D%20%2B%20x%5E%7B2%7D%20%2B%2010x%20%2B25%20%3D%20x%5E%7B2%7D%20%2B%2016x%20%2B%2064%5C%5Cx%5E%7B2%7D%20%2B%20x%5E%7B2%7D%20-%20x%5E%7B2%7D%20%2B%2010x%20-%2016x%20%2B%2025%20-%2064%20%3D%200%5C%5Cx%5E%7B2%7D%20-%206x%20-39%20%3D%200)
<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}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B-b%20%5C%C2%B1%20%5Csqrt%7Bb%5E%7B2%7D-4ac%7D%7D%7B2a%7D%5C%5Cx%20%3D%20%5Cfrac%7B-%28-6%29%20%5C%C2%B1%20%5Csqrt%7B%28-6%29%5E%7B2%7D-4%281%29%28-39%29%7D%7D%7B2%281%29%7D%5C%5Cx%20%3D%20%5Cfrac%7B6%7D%7B2%7D%20%5C%C2%B1%20%5Cfrac%7B%5Csqrt%7B192%7D%7D%7B2%7D%5C%5Cx%20%3D%203%20%5C%C2%B1%20%5Cfrac%7B8%5Csqrt%7B3%7D%7D%7B2%7D%5C%5Cx%20%3D%203%20%5C%C2%B1%204%5Csqrt%7B3%7D)
![x = 3 + 4\sqrt{3} = 9.928(4s.f.)](https://tex.z-dn.net/?f=x%20%3D%203%20%2B%204%5Csqrt%7B3%7D%20%3D%209.928%284s.f.%29)
OR
(Length can't be negative)
∴ ![x = 3 + 4\sqrt{3}](https://tex.z-dn.net/?f=x%20%3D%203%20%2B%204%5Csqrt%7B3%7D)
bruh
why
what
?
osea si pero al a vez cuando pasa osea si parece pero no lo es pero lo es cuando puede ser
Answer:
B. 5/6 of a number cannot be greater than the number.
7/3 = 2 1/3 > 1 2/3