# Cups A and B are cone shaped and have heights of #24 cm# and #15 cm# and openings with radii of #5 cm# and #8 cm#, respectively. If cup B is full and its contents are poured into cup A, will cup A overflow? If not how high will cup A be filled?

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To determine if cup A will overflow when the contents of cup B are poured into it, we need to compare the volumes of the two cups. If the volume of cup B's contents is greater than the remaining volume in cup A after filling it to its opening, then cup A will overflow. Otherwise, cup A will not overflow, and its height after being filled will be the height of cup B's contents.

The volume of a cone can be calculated using the formula: ( V = \frac{1}{3} \pi r^2 h ), where ( r ) is the radius of the cone's base and ( h ) is the height of the cone.

First, let's calculate the volumes of cups A and B:

For cup A: [ V_A = \frac{1}{3} \pi (5^2)(24) ]

For cup B: [ V_B = \frac{1}{3} \pi (8^2)(15) ]

Next, let's calculate the volume of cup B's contents: [ V_{\text{contents}} = V_B ]

Now, let's compare the volume of cup B's contents to the remaining volume in cup A after filling it to its opening: [ \text{Remaining volume in cup A} = V_A - \text{Volume of cup B's contents} ]

If the remaining volume in cup A is greater than or equal to zero, then cup A will not overflow, and its height after being filled will be the height of cup B's contents. Otherwise, if the remaining volume in cup A is negative, cup A will overflow.

Finally, to find the height to which cup A will be filled, we can use the formula for the volume of a cone: [ V = \frac{1}{3} \pi r^2 h ]

We can rearrange this formula to solve for ( h ): [ h = \frac{3V}{\pi r^2} ]

We'll substitute the calculated volume of cup B's contents and the radius of cup A into this formula to find the height to which cup A will be filled.

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To determine if cup A will overflow when the contents of cup B are poured into it, we need to compare the volumes of the two cups.

The volume of a cone is given by the formula ( V = \frac{1}{3} \pi r^2 h ), where ( r ) is the radius and ( h ) is the height.

For cup A: Radius ( r = 5 ) cm Height ( h = 24 ) cm

For cup B: Radius ( r = 8 ) cm Height ( h = 15 ) cm

Let's calculate the volumes of both cups:

For cup A: [ V_A = \frac{1}{3} \pi (5^2) (24) ] [ V_A = \frac{1}{3} \pi (25) (24) ] [ V_A = \frac{1}{3} \times \frac{75 \pi}{3} ] [ V_A = 200 \pi , \text{cm}^3 ]

For cup B: [ V_B = \frac{1}{3} \pi (8^2) (15) ] [ V_B = \frac{1}{3} \pi (64) (15) ] [ V_B = \frac{1}{3} \times \frac{960 \pi}{3} ] [ V_B = 320 \pi , \text{cm}^3 ]

Since cup B is full, its volume is ( 320 \pi , \text{cm}^3 ). When poured into cup A, cup A will not overflow because its volume is ( 200 \pi , \text{cm}^3 ), which is greater than the volume of cup B's contents.

Therefore, cup A will be filled to a height where its volume matches the volume of cup B's contents. Let's denote this height as ( h' ). We can rearrange the volume formula to solve for ( h' ):

[ V_A = \frac{1}{3} \pi (5^2) h' ] [ 200 \pi = \frac{1}{3} \pi (25) h' ] [ 600 = 25h' ] [ h' = \frac{600}{25} ] [ h' = 24 , \text{cm} ]

So, cup A will be filled to a height of ( 24 ) cm when cup B's contents are poured into it.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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