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

When we differentiate this function with respect to x, we get;

We want to find all values of c in (1,7) such that f(7) − f(1) = f '(c)(7 − 1)
This implies that;




![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)

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.
Since you know that PQ=RQ, you have an equilateral triangle. This makes things very simple.
Angle R should be the same as angle P.
A triangle is equal to 180 degrees.
Add angles P and R. Subtract 180 from the answer you got. That will give you 2a. a divided by 2 will give you a.
Or, since there are two right triangles, you can add 47 and 90. Subtract 180 from that and you will get a.
Answer:
x= 0/-8
y= 0/5
Step-by-step explanation:
Brain-List?
Answer:
the answer is =-3±
/2
Step-by-step explanation: