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Sati [7]
3 years ago
8

The dimensions of a conical funnel are shown below: A conical funnel is shown with the height of the cone as 8 inches and the ra

dius of the base as 3 inches. Nisha closes the nozzle of the funnel and fills it completely with a liquid. She then opens the nozzle. If the liquid drips at the rate of 15 cubic inches per minute, how long will it take for all the liquid to pass through the nozzle? (Use π = 3.14.) 3.34 minutes 5.02 minutes 13.39 minutes 15.07 minutes
Mathematics
2 answers:
ELEN [110]3 years ago
7 0

Answer:

B

Step-by-step explanation:

r=3 in

h=8 in

Volume of funnel

V=\frac{1}{3}\pi r^2h\\=\frac{1}{3} \times 3.14 \times 3^2 \times 8\\=75.36 in.^3\\time ~taken~to~drip out=\frac{75.36}{15}\\\approx 5.02~minutes.

REY [17]3 years ago
5 0

Answer:

5.02 is right

Step-by-step explanation:

i hope this helps

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Part 3 - Discussion/Explanation Question
SpyIntel [72]

Step-by-step explanation:

Vertical asymptote can be Identites if there is a factor only in the denominator. This means that the function will be infinitely discounted at that point.

For example,

\frac{1}{x - 5}

Set the expression in the denominator equal to 0, because you can't divide by 0.

x - 5 = 0

x = 5

So the vertical asymptote is x=5.

Disclaimer if you see something like this

\frac{(x - 5)(x + 3)}{(x - 5)}

x=5 won't be a vertical asymptote, it will be a hole because it in the numerator and denominator.

Horizontal:

If we have a function like this

\frac{1}{x}

We can determine what happens to the y values as x gets bigger, as x gets bigger, we will get smaller answers for y values. The y values will get closer to 0 but never reach it.

Remember a constant can be represent by

a \times  {x}^{0}

For example,

1 = 1 \times  {x}^{0}

2 =  2 \times {x}^{0}

And so on,

and

x =  {x}^{1}

So our equation is basically

\frac{1 \times  {x}^{0} }{ {x}^{1} }

Look at the degrees, since the numerator has a smaller degree than the denominator, the denominator will grow larger than the numerator as x gets larger, so since the larger number is the denominator, our y values will approach 0.

So anytime, the degree of the numerator < denominator, the horizontal asymptote is x=0.

Consider the function

\frac{3 {x}^{2} }{ {x}^{2}  + 1}

As x get larger, the only thing that will matter will be the leading coefficient of the leading degree term. So as x approach infinity and negative infinity, the horizontal asymptote will the numerator of the leading coefficient/ the leading coefficient of the denominator

So in this case,

x =  \frac{3}{1}

Finally, if the numerator has a greater degree than denominator, the value of horizontal asymptote will be larger and larger such there would be no horizontal asymptote instead of a oblique asymptote.

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2 years ago
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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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MrRissso [65]

Answer:

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