Step-by-step explanation:
F(x) = ∫ₐˣ t⁷ dt
F(x) is the area under f(t) between t=a and t=x. When x=a, the width of the interval is 0, so the area is zero.
F(6) = 0, so a = 6.
F(x) = ∫₆ˣ t⁷ dt
F(6) = ∫₆⁶ t⁷ dt
F(6) = 0
Answer:
f(2) = 3
Step-by-step explanation:
We are given:
f(0) = 3
and
f(n+1) = -f(n) + 5
We have to find the value of f(2). In order to find f(2) we first have to find f(1)
f(n + 1) = - f(n) + 5
Using n = 0, we get:
f(0 + 1) = - f(0) + 5
f(1) = -f(0) + 5 Using the value of f(0), we get
f(1) = -3 + 5 = 2
Now using n = 1 in the function, we get:
f(1 + 1) = - f(1) + 5 Using the value of f(1), we get
f(2) = -2+ 5
f(2) = 3
Thus the value of f(2) will be 3
Answer:
See attached pictures.
Step-by-step explanation:
The sine and cosine functions have the forms: 
and 
. A is the amplitude for each function. The period is found by dividing 2π the absolute value of B or 
. C shifts the function up and down.
The sine function always starts and ends on the x-axis.
The cosine function always starts and ends at the y=A.
6.) The sine function starts at (0,0) then peaks at 5. Comes down to 0 and down to -5 before returning to 0.
The amplitude is 5.
The period is 
7.) Here A=3 so the amplitude is 3, B is 1/2 so the period is 4π. Start at (3,0) and descend down to (2π, 0). Go back up to (4π, 3).
8.) Here A = 2 so the amplitude is A. B is 2π so the period is 1. C is 1 so the graph is shifted up a unit.
Start the graph at (0,1) and go up to (0.25,3) and down to (0.5,1) and continue downward to (0.75, -3) then back up to (1,1).
Answer:
I had to look up the problem B4 because your image doesn't show the rest of the information
B4 = 12
B5 = 1,738
Step-by-step explanation:
if you need help on any other questions on this, i can't see them in the image
Answer:
To find the original dimensions, write a proportion with the scale as the first ratio and the scale dimension compared to the actual dimension as the second ratio. Use cross products to solve. To find new dimensions, write a proportion with the new scale as the first ratio and the scale dimension compared to the actual dimension as the second ratio. Use cross products to solve.
