1) you can expand the brackets, simplify, set the equation = 0 and then do a double bracket factorisation.
![(x - 4)^{2} - 28 = 8 \\ x - 8x + 16 - 28 = 8 \\ x - 8x - 12 = 8 \\ x - 8x - 20 = 0 \\ (x - 10)(x + 2) = 0 \\ x = 10 \\ x = - 2](https://tex.z-dn.net/?f=%28x%20-%204%29%5E%7B2%7D%20%20%20-%2028%20%3D%208%20%5C%5C%20x%20-%208x%20%2B%2016%20-%2028%20%3D%208%20%5C%5C%20x%20-%208x%20-%2012%20%3D%208%20%5C%5C%20x%20-%208x%20-%2020%20%3D%200%20%5C%5C%20%28x%20-%2010%29%28x%20%20%2B%202%29%20%3D%200%20%5C%5C%20%20x%20%3D%2010%20%5C%5C%20x%20%3D%20%20-%202)
2)
![x^{2} - 6x - 7 = 0 \\ (x - 7)(x + 1) = 0 \\ \\ x - 7 = 0 \\ x = 7 \\ \\ x + 1 = 0 \\ x = - 1](https://tex.z-dn.net/?f=x%5E%7B2%7D%20%20-%206x%20-%207%20%3D%200%20%5C%5C%20%28x%20-%207%29%28x%20%2B%201%29%20%3D%200%20%5C%5C%20%20%5C%5C%20x%20%20-%207%20%3D%200%20%20%5C%5C%20x%20%3D%207%20%5C%5C%20%5C%5C%20%20x%20%2B%201%20%3D%200%20%5C%5C%20x%20%3D%20%20-%201)
therefore Josh's solution is wrong asx = 7 and x= -1
Answer:
OK I'll help
Step-by-step explanation:
But I don't have any assignments to be submitted now really.
Here a tip : if the top number is less then half of the bottom number, then it is less then 1/2.
less then 1/2 : 1/4, 1/5, 2/12
to be greater then 1, the top number must be bigger then the bottom number
greater then 1 : 6/5, 14/8, 11/10
If a function f(x) has an inverse f ⁻¹(x), then by definition
![f\left(f^{-1}(x)\right) = x](https://tex.z-dn.net/?f=f%5Cleft%28f%5E%7B-1%7D%28x%29%5Cright%29%20%3D%20x)
Differentiating both sides with respect to x yields
![f'\left(f^{-1}(x)\right) \times \left(f^{-1}\right)'(x) = 1 \implies \left(f^{-1}\right)'(x) = \dfrac1{f'\left(f^{-1}(x)\right)}](https://tex.z-dn.net/?f=f%27%5Cleft%28f%5E%7B-1%7D%28x%29%5Cright%29%20%5Ctimes%20%5Cleft%28f%5E%7B-1%7D%5Cright%29%27%28x%29%20%3D%201%20%5Cimplies%20%5Cleft%28f%5E%7B-1%7D%5Cright%29%27%28x%29%20%3D%20%5Cdfrac1%7Bf%27%5Cleft%28f%5E%7B-1%7D%28x%29%5Cright%29%7D)
Now, if f(a) = b and b = f ⁻¹(a), then
![\left(f^{-1}\right)'(a) = \dfrac1{f'\left(f^{-1}(a)\right)} = \dfrac1{f'(b)}](https://tex.z-dn.net/?f=%5Cleft%28f%5E%7B-1%7D%5Cright%29%27%28a%29%20%3D%20%5Cdfrac1%7Bf%27%5Cleft%28f%5E%7B-1%7D%28a%29%5Cright%29%7D%20%3D%20%5Cdfrac1%7Bf%27%28b%29%7D)
Answer:
i dunno sorry
Step-by-step explanation:
I think what you have to do is related to PEMDAS