15 x 20 and 10 x 30 hope this works
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.
Answer:
Step-by-step explanation:
I'm guessing we are missing a piece of the question such as the average speed at which Nadeem can ride. Let's call it v
distance is a product of velocity and time
d = vt
t = d/v
so if v is in miles/hr, an inequality could be
12/v ≤ t ≤ 15/v
I can figure out this is a frefall motion.
Starting from rest => Vo = 0
Then, use the equation: d = [1/2]gt^2 => t = √(2d/g)
d = width of a black/clear stripe pair = 5cm = 0.05m
g ≈ 10 m/s^2 (the real value is about 9.81 m/s^2)
t =√(2*0.05m/10m/s^2) = 0.1 s
Answer: approximately 0.1 s