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2 years ago

Am I correct? Tell me if I’m wrong and what is the correct answer.

Mathematics

2 answers:

tia_tia [17]2 years ago

6 0

It’s the other way around

lyudmila [28]2 years ago

4 0

Answer:

3.6 is the correct answer because it is bigger,

Step-by-step explanation:

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Type the correct equation of the line. (0, 1), (2, -3), (4, -7)

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Answer:

1(2-3)(4-7)

(1)(-1)(4-7)

-1(4-7)

(-1)(-3)

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3 years ago

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3 years ago

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3 years ago

Keisha pumped 6 gallons of water into her pool each minute for 48 minutes what was the total change in the amount of water in th

Eduardwww [97]

The total change in the amount of water in the pool is 288 gallons.

<h3>What is Volume?</h3>

Volume is a scalar quantity expressing the amount of three-dimensional space enclosed by a closed surface.

Here, Rate of water pumped into her pool = 6 gallons / minute.

Total Duration = 48 minutes

Total amount of water = Rate X Time

= 6 X 48

= 288 gallons

Thus, the total change in the amount of water in the pool is 288 gallons.

Learn more about Volume from:

brainly.com/question/1578538

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3 0

2 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

3 years ago