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WITCHER [35]
2 years ago
7

The stadium is like a hemisphere with a radius of 150 feet last month the owners of the stadium paid 0. 05 per cubic foot to cov

er the cost of utilities. What was the total cost for utilities last month
Mathematics
1 answer:
GREYUIT [131]2 years ago
6 0

The total cost for utilities last month is 353,571.43.

<h3>What is the total cost for utilities last month?</h3>

The total cost for utilities last month can be determined by multiplying the volume of the stadium by the cost per cubic foot.

The volume of the stadium can be determined by using the formula for the volume of a hemisphere.

Volume of a hemisphere = Volume of a hemisphere = (2/3) x (n) x (r^3)

n = 22/7

r = radius

2/3 x 22/7 x 150³ = 7071,428.57

The total cost for utilities last month =  7071,428.57 x 0.05 = 353,571.43

To learn more about volume, please check: brainly.com/question/13705125

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Answer:

[0, 3, -6]

Step-by-step explanation:

{4x + y = 3

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The volume V of an ice cream cone is given by V = 2 3 πR3 + 1 3 πR2h where R is the common radius of the spherical cap and the c
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Answer:

The change in volume is estimated to be 17.20 \rm{in^3}

Step-by-step explanation:

The linearization or linear approximation of a function f(x) is given by:

f(x_0+dx) \approx f(x_0) + df(x)|_{x_0} where df is the total differential of the function evaluated in the given point.

For the given function, the linearization is:

V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh

Taking R_0=1.5 inches and h=3 inches and evaluating the partial derivatives we obtain:

V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh\\V(R, h) = V(R_0, h_0) + (\frac{2 h \pi r}{3}  + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh

substituting the values and taking dx=0.1 and dh=0.3 inches we have:

V(R_0+dR, h_0+dh) =V(R_0, h_0) + (\frac{2 h \pi r}{3}  + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh\\V(1.5+0.1, 3+0.3) =V(1.5, 3) + (\frac{2 \cdot 3 \pi \cdot 1.5}{3}  + 2 \pi 1.5^2)\cdot 0.1 + (\frac{\pi 1.5^2}{3} )\cdot 0.3\\V(1.5+0.1, 3+0.3) = 17.2002\\\boxed{V(1.5+0.1, 3+0.3) \approx 17.20}

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