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Lady_Fox [76]
1 year ago
13

8 : sand pouring from a chute forms a conical pile whose height is always equal to the diameter. if the height increases at a co

nstant rate of 5 feet per minute, at what rate is the sand pouring from the chute when the pile is 10 feet high?
Mathematics
1 answer:
sergeinik [125]1 year ago
3 0

When the pile is 10 feet high, the rate at which sand is released from the chute is 125\pi ft^3/min

Let x denote the cone's height.

We've been told that when sand is poured down a chute, it always creates a conical pile whose height is equal to its diameter.

Since radius is known to be equal to half of diameter, the radius of a cone would be x/2.

To resolve the issue at hand, we shall use the volume of cone formula.

v=1/3\pi r^{2} h

v= 1/3\pi(x/2)^2(x/2)

v= 1/3\pi(x^2/4)(x/2)

v= \pix^3/12

When we translate the height and radius values into terms of x, we obtain:

dv/dt =1/3 \pi(3x^2)dx/dt

By solving the equation,

dv/dt =125 \pi

Consequently, 125\pift^3/min of sand per minute are being released from the chute.

To learn more about volume of cone click here:

brainly.com/question/1578538

#SPJ4

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Guys please help me with this question
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Answer: Can I get brainliest and 5 stars? thank you.

a) \frac{65}{99}

b) \frac{59}{90}

c) \frac{13}{20}

Step-by-step explanation:

a) 0.65 I can't put line above it, but the line expresses that 6 and 5 and repeating infinitely. Let X equal the decimal number.

x=0.65

First multiply by 1 followed by as many zeros as repeating numbers there are. In this case since 6 and 5 are the repeating numbers, we have to add 2 zeros which is basically multiplying by 100

0.65*100=65.65 (another 65 remains on the right because they're infinite.)

Now substract from the original number. Now we have a new equation that says: 100x=65.65.

Substract equation 1 from equation 2.

100x=65.65\\-x=0.65

---------------------------------

99x=65

Solve just like any other equation. Divide by 99 to isolate x.

\frac{99x}{99}=\frac{65}{99}

x=\frac{65}{99} This is the fraction.

------------------------------------------------------------------------------------------------

b) 0.65^-This one is a little bit different because we only have 1 repeating number.

In this case we have to multiply by 10.

0.65^-*10=6.55^-

Equation 1: x=0.65^-

Equation 2: 10x=6.55^-

Substract.

10x=6.55^-\\-x=0.65^-

-----------------------

9x=5.9

divide by 9 to isolate x

\frac{9x}{9}=\frac{5.9}{9}

x=\frac{5.9}{9}

To get rid of the decimal in the numerator, multiply both numerator and denominator by 10 so that we don't change the fraction.

x=(\frac{5.9*10}{9*10} )

x=\frac{59}{90} and this is your fraction.

------------------------------------------------------------------------------------------------------

c) 0.65 this one has no repeating number. It's a finite decimal number.

I assume you already know that; a=\frac{a}{1}, so, we can do the same here to have a fraction.

0.65=\frac{0.65}{1}

Now, in order to get rid of the decimal, multiply by 1 followed by as many zeros as decimal numbers are. In this case 2 zeros because there are 2 decimal numbers, so it's 100. Remember that if you do it in the numerator, you have to do it in the denominator as well to not change the fraction.

\frac{0.65*100}{1*100} =\frac{65}{100}

Simplify if possible.

\frac{65/5}{100/5} =\frac{13}{20} This is your fraction.

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