The solution is not the ones in the parenthesis.
your solution is, (0,16)
102
the angles are alternate so they add up to 180
180-78=102
hope this helps :)
if you need me to explain more just let me know :)
Answer:
x=2 x=-2
Step-by-step explanation:
x² + 2x - 4 = 2x
Subtract 2x from each side
x² + 2x-2x - 4 = 2x-2x
x^2 -4 = 0
Factor
We know this is the difference of squares
(x-2)(x+2) =0
Using the zero product property
x-2 =0 x+2 =0
x=2 x=-2
Answer:
This contradicts the Mean Value Theorem since there exists a c on (1, 7) such that f '(c) = f(7) − f(1) (7 − 1) , but f is not continuous at x = 3
Step-by-step explanation:
The given function is
![f(x)=(x-3)^{-2}](https://tex.z-dn.net/?f=f%28x%29%3D%28x-3%29%5E%7B-2%7D)
When we differentiate this function with respect to x, we get;
![f'(x)=-\frac{2}{(x-3)^3}](https://tex.z-dn.net/?f=f%27%28x%29%3D-%5Cfrac%7B2%7D%7B%28x-3%29%5E3%7D)
We want to find all values of c in (1,7) such that f(7) − f(1) = f '(c)(7 − 1)
This implies that;
![0.06-0.25=-\frac{2}{(c-3)^3} (6)](https://tex.z-dn.net/?f=0.06-0.25%3D-%5Cfrac%7B2%7D%7B%28c-3%29%5E3%7D%20%286%29)
![-0.19=-\frac{12}{(c-3)^3}](https://tex.z-dn.net/?f=-0.19%3D-%5Cfrac%7B12%7D%7B%28c-3%29%5E3%7D%20)
![(c-3)^3=\frac{-12}{-0.19}](https://tex.z-dn.net/?f=%28c-3%29%5E3%3D%5Cfrac%7B-12%7D%7B-0.19%7D)
![(c-3)^3=63.15789](https://tex.z-dn.net/?f=%28c-3%29%5E3%3D63.15789)
![c-3=\sqrt[3]{63.15789}](https://tex.z-dn.net/?f=c-3%3D%5Csqrt%5B3%5D%7B63.15789%7D)
![c=3+\sqrt[3]{63.15789}](https://tex.z-dn.net/?f=c%3D3%2B%5Csqrt%5B3%5D%7B63.15789%7D)
![c=6.98](https://tex.z-dn.net/?f=c%3D6.98)
If this function satisfies the Mean Value Theorem, then f must be continuous on [1,7] and differentiable on (1,7).
But f is not continuous at x=3, hence this hypothesis of the Mean Value Theorem is contradicted.
Answer:
Y= 1
Step-by-step explanation:
Given the expression f(x) = 3x - 3.
To solve for the asymptote y
Firstly we need to solve for x
3x-3=0
3x=3
x= 3/3
x=1
Hence the asymptote of the function y=1