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3 years ago

A pump removes 1000 gal of water from a pool at a constant rate of 50 gal/min.

Mathematics

1 answer:

sergejj [24]3 years ago

8 0

Answer:

Since it says graph I will assume there are graphs that go along with this?

If this is an open ended question you can find out how much water is left after (t) minutes by doing this equasion:

1,000/50=20

(t) being minutes so the answer to (t) would be 20 minutes

(y) being water would be 1,000gal

Final Answer: 1,000gal in 20 min

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11-(3x+1) = 7(x-6)+2

Galina-37 [17]

Hello!

$\large\boxed{x = 5}$

11 - (3x + 1) = 7(x - 6) + 2

Distribute the coefficients outside of the parenthesis:

11 - (3x) - (1) = 7(x) + 7(-6) + 2

Simplify:

11 - 3x - 1 = 7x - 42 + 2

Combine like terms:

10 - 3x = 7x - 40

Add 3x to both sides to isolate x:

10 = 10x - 40

Add 40 to both sides:

50 = 10x

Divide both sides by 10:

x = 5

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3 years ago

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A spherical balloon currently has a radius of 19cm. If the radius is still growing at a rate of 5cm or minute, at what rate is a

miv72 [106K]

Answer:

22670.8 cm³/min

Step-by-step explanation:

Given:

Radius of the balloon at a certain time (r) = 19 cm

Rate of growth of radius is, $\frac{dr}{dt}=5\ cm/min$

The rate at which the air is pumped in the balloon can be calculated by finding the rate of increase in the volume of the balloon.

So, first we find the volume of the sphere in terms of 'r'. As the balloon is spherical in shape, the volume of the balloon is equal to the volume of a sphere. Therefore,

Volume of balloon is given as:

$V=\frac{4}{3}\pi r^3$

Now, rate of increase of volume is obtained by differentiating both sides of the equation with respect to time 't'.

Differentiating both sides with respect to time 't', we get:

$\frac{dV}{dt}=\frac{d}{dt}(\frac{4}{3}\pi r^3)\\\\\frac{dV}{dt}=\frac{4\pi}{3}(3r^2)(\frac{dr}{dt})\\\\\frac{dV}{dt}=4\pi r^2(\frac{dr}{dt})$

Now, plug in 19 cm for 'r', 5 cm per minute for $\frac{dr}{dt}$ and solve for $\frac{dV}{dt}$ . This gives,

$\frac{dV}{dt}=4\pi (19 cm)^2(5\ cm/min)\\\\\frac{dV}{dt}=4\times 3.14\times 361\times 5\ cm^3/min\\\\\frac{dV}{dt}=22670.8\ cm^3/min$

Therefore, the rate at which the air is being pumped into the balloon is 22670.8 cm³/min.

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The answer to this question is 29.542 hectares

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3 years ago

Find the gcf of 200 and 72

Aleks04 [339]

To get the Greates Common Factor (GCF) of 72 and 200 we need to factor each value first and then we choose all the copies of factors and multiply them:

72: 22233 200: 222 55GCF: 222

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Answer: 8

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Suppose n represents an even number for a simplified expression that represents the product of the next two even numbers

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N+1 x n+2 = whatever that first part means

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