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:
x=5
Step-by-step explanation:
2 2/7 : (0.6x) :: 4/21 : 0.25
Hence, as it is in a perfect proportion, we may use the proportion method..
We multiply the extreme terms and the moderate(near) ones and equalise them.
Hence,
(2 2/7)×(0.25) = (0.6x × 4/21)
16/7 × 25/100 = 6x/10 × 4/21
16/7× 1/4= 3x/5 × 4/21
(16/7 × 1/4)/(4/21) =3x/5
16/7×1/4×21/4×5/3=x
1/7×21×5/3=x
x=5
Yes, all fractions could be written in a repeating decimal.
If you have spent $6.50 on a pair of gloves, but now have $7.00 left and are looking for the amount of money you had originally, all you have to do is add both numbers then you'll find your answer. But here, I'll just do it for you, your answer is: $13.50
Hope that was understandable and helped you out!