Since
(density = mass/volume), we can get the mass/weight of the liquid by integrating the density
over the interior of the tank. This is done with the integral
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which is more readily computed in cylindrical coordinates as

Answer:
8.20
Step-by-step explanation:
6.15*4=24.6
since they only bought 3 tickets because of the deal
you divide 24.6 by 3 and get 8.20
Answer:
8) answer=5 9)answer=15
Step-by-step explanation:
all you have to do is replace in the following formula

Where
X1: is the first number of the first point
Y1: is the second number of the first point
X2: is the first number of the second point
Y2: is the second number of the second point
you replace in the formula and solve
I attach solution
Answer:
b and c
Step-by-step explanation:
Answer:
x = 2 cm
y = 2 cm
A(max) = 4 cm²
Step-by-step explanation: See Annex
The right isosceles triangle has two 45° angles and the right angle.
tan 45° = 1 = x / 4 - y or x = 4 - y y = 4 - x
A(r) = x* y
Area of the rectangle as a function of x
A(x) = x * ( 4 - x ) A(x) = 4*x - x²
Tacking derivatives on both sides of the equation:
A´(x) = 4 - 2*x A´(x) = 0 4 - 2*x = 0
2*x = 4
x = 2 cm
And y = 4 - 2 = 2 cm
The rectangle of maximum area result to be a square of side 2 cm
A(max) = 2*2 = 4 cm²
To find out if A(x) has a maximum in the point x = 2
We get the second derivative
A´´(x) = -2 A´´(x) < 0 then A(x) has a maximum at x = 2