Hmmm...I don't know what to tell you I can't type out your answer For you.
sorry
Note that √(4 - t²) is defined only as long as 4 - t² ≥ 0, or -2 ≤ t ≤ 2. Then the real integral exists only if -2 ≤ x ≤ 2. (Otherwise we deal with complex numbers.)
If x = 2, then the integral corresponds to the area of a quarter-circle with radius 2. This means that the integral has a maximum value of 1/4 • π • 2² = π.
On the opposite end, if x = -2, then the integral has the same value, but the integral from 0 to -2 is equal to the negative integral from -2 to 0. So the minimum value is -π.
For all x in between, we observe that the integrand is continuous over the rest of its domain, so F(x) is continuous.
Then the range of F(x) is the interval [-π, π].
(f-g)(x) = f(x) - g(x) = 3^x + 10x - (4x - 2) = 3^x + 6x + 2 Answer
this question is incorrect because the decimal quantity functional persistent shouldnt be in a sequential farfictional order so the answer would be that satisfactory angle of theslop devided by pi and then multiplied by the persistent. hope this helped!
Answer:
c and d
Step-by-step explanation:
