The answer would be A = 54raiz (3) + 18raiz (91)
Formula:
A = Ab + Al Where, Ab=base area A= lateral area
The area of the base is: Ab = (3/2) * (L ^ 2) * (root (3)) Where, L= side of the hexagon. Substitute: Ab = (3/2) * (6 ^ 2) * (root (3)) Ab = (3/2) * (36) * (root (3)) Ab = 54raiz (3)
The lateral area is: Al = (6) * (1/2) * (b) * (h) Where, b= base of the triangle h= height of the triangle Substitute: Al = (6) * (1/2) * (6) * (root ((8) ^ 2 + ((root (3) / 2) * (6)) ^ 2)) Al = 18 * (root (64 + 27)) Al = 18raiz (91)
The total area is: A = 54raiz (3) + 18raiz (91)
Answer:
129.996 cubic feet
Step-by-step explanation:
Assuming that the barrel is a perfect cylinder, we can use the formula to find the volume:
First, we have to find the radius, which is:
D = 2r
So the radius is 3 feet, since it is half of the diameter. Then, we plug in the values.
Now, we solve. Exponents are first...
3.14(9)(4.6)
Now multiply left to right.
28.26(4.6)
129.996 cubic feet
Therefore, the barrel can hold 129.996 cubic feet of oil. Hope this helps you!
Answer: 42
Step-by-step explanation:
We know that because 40% percent of 70 equals 28. So we do the following-
70 minus 28 equals 42!
Glad to help!
The answer is (1/2)xe^(2x) - (1/4)e^(2x) + C
Solution:
Since our given integrand is the product of the functions x and e^(2x), we can use the formula for integration by parts by choosing
u = x
dv/dx = e^(2x)
By differentiating u, we get
du/dx= 1
By integrating dv/dx= e^(2x), we have
v =∫e^(2x) dx = (1/2)e^(2x)
Then we substitute these values to the integration by parts formula:
∫ u(dv/dx) dx = uv −∫ v(du/dx) dx
∫ x e^(2x) dx = (x) (1/2)e^(2x) - ∫ ((1/2) e^(2x)) (1) dx
= (1/2)xe^(2x) - (1/2)∫[e^(2x)] dx
= (1/2)xe^(2x) - (1/2) (1/2)e^(2x) + C
where c is the constant of integration.
Therefore,
∫ x e^(2x) dx = (1/2)xe^(2x) - (1/4)e^(2x) + C
