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
150 since rectangle has 4 sides, so 600 is divided to 4(sides)=150
The expression [ x² - 4 ] doesn't depend on 'a', 'b', or 'y'.
It only depends on 'x'.
You have said that 3x = 5, so x = 5/3.
Therefore, I may now write '5/3' in place of 'x' in the expression:
x² - 4 =
(5/3)² - 4 =
(25/9) - 4 =
(25/9) - (36/9) =
(25 - 36) / 9 = -11/9 = - (1 and 2/9)
Answer:
8125 mm
Step-by-step explanation:
Not that sure
Solution :
From the figure,
The coordinates of --
A = (-5,-3)
B = (-6,-1)
C = (-3,-1)
D = (-2,-3)
When it undergoes reflection about x-axis, we have the coordinates of the quadrilateral as
A' = (-5,+3)
B' = (-6,+1)
C' = (-3,+1)
D' = (-2,+3)
Now A'B'C'D' is translated 3 units to right, so the coordinates will be --
A" = (-2,3)
B" = (-3,1)
C" = (0,1)
D" = (1,3)
When the quadrilateral moves third and two units up, we get
A"' = (3,3)
B"' = (2,1)
C"' = (5,1)
D"' = (6,3)
