1540.5 .
V=PI*R^2*H/3=3.14*9.5^2*5.433=1540;5
According to law of cosines the length of RQ can be written as .
Given the length PR is 6 , the length of RQ is p, the length of PQ is 8 and the angle RPQ is 39 degrees.
A length of the triangle can be written as according to law of cosines if sides are given and one angle is
We have to just put the values in the above equation.
as .
p is the side opposite to angle given , the length of other sides are 6 and 8 and angle is 39 degrees.
Hence the side can be written as according to law of cosines if the angle is 39 degrees is as .
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Let x represent the side length of the square end, and let d represent the dimension that is the sum of length and girth. Then the volume V is given by
V = x²(d -4x)
Volume will be maximized when the derivative of V is zero.
dV/dx = 0 = -12x² +2dx
0 = -2x(6x -d)
This has solutions
x = 0, x = d/6
a) The largest possible volume is
(d/6)²(d -4d/6) = 2(d/6)³
= 2(108 in/6)³ = 11,664 in³
b) The dimensions of the package with largest volume are
d/6 = 18 inches square by
d -4d/6 = d/3 = 36 inches long
Answer:
Total Rate= 220$ and length = 13 ft
13/220$
= 1/16.92$
Unit = 16.92$
Answer:
3-8I am right or wrong and the answer is 5