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miss Akunina [59]
3 years ago
9

Consider a can in the shape of a right circular cylinder. the top and bottom of the can is made of a material that costs 4 cents

per square centimeter, and the side is made of a material that costs 3 cents per square centimeter. we want to find the dimensions of the can which has volume 72 π cubic centimeters, and whose cost is as small as possible. (a) find a function f(r) which gives the cost of the can in terms of radius r. be sure to specify the domain. (b) give the radius and height of the can with least cost. (c) explain how you known you have found the can of least cost.
Mathematics
1 answer:
SVEN [57.7K]3 years ago
5 0
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Answer:

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Step-by-step explanation:

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y = 3x + 1 ( subtract 1 from both sides )

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\frac{y-1}{3} = x

Change y back into terms of x, thus

f^{-1}(x) = \frac{x-1}{3}

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? Divided by ? = 1/9
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Answer:

0.11

Step-by-step explanation:

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Emily has $110 in a savings account this year. She plans on depositing $10 into the savings account each following year. How muc
Masteriza [31]

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The boundary of a lamina consists of the semicircles y = 1 − x2 and y = 16 − x2 together with the portions of the x-axis that jo
oksano4ka [1.4K]

Answer:

Required center of mass (\bar{x},\bar{y})=(\frac{2}{\pi},0)

Step-by-step explanation:

Given semcircles are,

y=\sqrt{1-x^2}, y=\sqrt{16-x^2} whose radious are 1 and 4 respectively.

To find center of mass, (\bar{x},\bar{y}), let density at any point is \rho and distance from the origin is r be such that,

\rho=\frac{k}{r} where k is a constant.

Mass of the lamina=m=\int\int_{D}\rho dA where A is the total region and D is curves.

then,

m=\int\int_{D}\rho dA=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}rdrd\theta=k\int_{}^{}(4-1)d\theta=3\pi k

  • Now, x-coordinate of center of mass is \bar{y}=\frac{M_x}{m}. in polar coordinate y=r\sin\theta

\therefore M_x=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA

=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\sin\theta)rdrd\theta

=k\int_{0}^{\pi}\int_{1}^{4}r\sin\thetadrd\theta

=3k\int_{0}^{\pi}\sin\theta d\theta

=3k\big[-\cos\theta\big]_{0}^{\pi}

=3k\big[-\cos\pi+\cos 0\big]

=6k

Then, \bar{y}=\frac{M_x}{m}=\frac{2}{\pi}

  • y-coordinate of center of mass is \bar{x}=\frac{M_y}{m}. in polar coordinate x=r\cos\theta

\therefore M_y=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA

=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\cos\theta)rdrd\theta

=k\int_{0}^{\pi}\int_{1}^{4}r\cos\theta drd\theta

=3k\int_{0}^{\pi}\cos\theta d\theta

=3k\big[\sin\theta\big]_{0}^{\pi}

=3k\big[\sin\pi-\sin 0\big]

=0

Then, \bar{x}=\frac{M_y}{m}=0

Hence center of mass (\bar{x},\bar{y})=(\frac{2}{\pi},0)

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3 years ago
The following frequency table shows the number of fish caught by each of Igor's family members.
Maslowich

Step-by-step explanation:

The values on the left of the table represent the number of fish caught, and the number of the right of the table represents how many family members caught that amount of fish.

Therefore, the first row means that 0 family members caught 0 fish.

The second row means that 3 family members caught 1 fish.

The third row would mean 1 family member caught 2 fish.

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And the final row would mean 4 family members caught 4 fish.

The question does not ask for the total amount of fish caught; rather is ask for the maximum number of fish that a single family member caught.

Therefore, the maximum amount of fish that a single family member catches is 4. (And 4 family members did so. But individually, the maximum amount of fish one person caught is 4).

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