Paula always withdraws 20 dollars more than she needs from the bank.
1 answer:
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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
