Answer:
Hello your question is incomplete below is the complete question
What is wrong with the equation? integral^2 _3 x^-3 dx = x^-2/-2]^2 _3 = -5/72 f(x) = x^-3 is continuous on the interval [-3, 2] so FTC2 cannot be applied. f(x) = x^-3 is not continuous on the interval [-3, 2] so FTC2 cannot be applied. f(x) = x^-3 is not continuous at x = -3, so FTC2 cannot be applied The lower limit is less than 0, so FTC2 cannot be applied. There is nothing wrong with the equation. If f(2) = 14, f' is continuous, and f'(x) dx = 15, what is the value of f(7)? F(7) =
answer : The value of f(7) = 29
Step-by-step explanation:
Attached below is the detailed solution
Hence : F(7) - 14 = 15
F(7) = 15 + 14 = 29
Answer:
-12
Step-by-step explanation:
Use synthetic division to find the remainder:
a + 2 yields the synthetic division divisor -2. The four coefficients of a^3 - 4 are {1, 0, 0, -4}. Setting up synthetic division results in:
-2 / 1 0 0 -4
-2 4 -8
-------------------------
1 -2 4 -12
The remainder is -12. This is also the value of a^3 - 4 when a = -2.
Answer:
Yo my guy, this is really easy
Step-by-step explanation: