What is wrong with the equation? 2 x−3 dx = x−2 −2 2 −3 = − 5 72 −3 f(x) = x−3 is 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. f(x) = x−3 is not continuous on the interval [−3, 2] so FTC2 cannot be applied. There is nothing wrong with the equation. The lower limit is less than 0, so FTC2 cannot be applied.

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

$4034.40

the perimeter of the field is the sum of the lengths of the four sides:

47 yd + 158 yd + 47 yd + 158 yd = 410 yd

the cost of 3 ft (1 yd) of fence is 3×$3.28 = $9.84.

so the cost of jake's wood perimeter fance is

(410 yd) × ($9.84 /yd) = $4034.40

step-by-step explanation:

a and d