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3 years ago

Help plssssssssssssssss

Mathematics

1 answer:

kherson [118]3 years ago

5 0

Answer:

x= 28

Step-by-step explanation:

10/70=4/x

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A rectangular prism is 12 inches long, 9 inches wide, and 6 inches high. What is the volume of the prism?

zmey [24]

Answer:

Volume = 12×9×6 = 648 inches^3

Step-by-step explanation:

Hope this helps! Have a nice day!

5 0

3 years ago

Which of the following expressions is equivalent to -9?

dlinn [17]

Answer:

A and C

Step-by-step explanation:

-3*3 = -9

-1*-9 = 9

-27/3 = -9

-9/-1 = 9

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3 years ago

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Kareem buys candy that costs $4 per pound. He will spend at most $48 on candy. What are the possible numbers of pounds he will b

lesantik [10]

Answer:

12 lbs

Step-by-step explanation:

48/4 =12

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3 years ago

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What is the volume of the prism

Natali [406]

The answer to this question is two hundred and eighty

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3 years ago

Water is added to a cylindrical tank of radius 5 m and height of 10 m at a rate of 100 L/min. Find the rate of change of the wat

nirvana33 [79]

Answer:

$V = \pi r^2 h$

For this case we know that $r=5m$ represent the radius, $h = 10m$ the height and the rate given is:

$\frac{dV}{dt}= \frac{100 L}{min}$

$Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}$

And replacing we got:

$\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}$

And that represent $0.127 \frac{cm}{min}$

Step-by-step explanation:

For a tank similar to a cylinder the volume is given by:

$V = \pi r^2 h$

For this case we know that $r=5m$ represent the radius, $h = 10m$ the height and the rate given is:

$\frac{dV}{dt}= \frac{100 L}{min}$

For this case we want to find the rate of change of the water level when h =6m so then we can derivate the formula for the volume and we got:

$\frac{dV}{dt}= \pi r^2 \frac{dh}{dt}$

And solving for $\frac{dh}{dt}$ we got:

$\frac{dh}{dt}= \frac{\frac{dV}{dt}}{\pi r^2}$

We need to convert the rate given into m^3/min and we got:

$Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}$

And replacing we got:

$\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}$

And that represent $0.127 \frac{cm}{min}$

5 0

4 years ago