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A conical vessel with base radius five centimeters and height 24 centimeters is full of water.
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The water is emptied into a cylindrical vessel of base radius 10 centimeters.
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Find the height of the water in the cylindrical vessel.
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Use π is equal to 22 over seven.
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Letβs start by drawing the two vessels.
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Here is the conical vessel which has a base radius of five centimeters and a height of 24 centimeters.
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This is the cylindrical vessel which has a base radius of 10 centimeters and a height, which we do not know.
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Now, we are told the water is emptied into the cylindrical vessel.
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So what this means is that all of the water that was in the conical vessel has been poured into the cylindrical vessel.
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So letβs let this line represent the water level in our cylinder.
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And we can call the length from the base of the cylinder to the top of the water level β.
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Now, we can see that the water in the cylindrical vessel actually creates a cylinder of its own.
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It has a base radius which is the same as the cylinder of 10 centimeters and a height of β.
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And now, since all of the water was emptied into the cylindrical vessel from the conical vessel, we know that the volume of water in the conical vessel is equal to the volume of water in the cylindrical vessel.
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Now, letβs recall the equations for the volume of a cone and the volume of a cylinder.
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We have that the volume of a cone is one-third ππ squared β, where π is the base radius of the cone and β is the height of the cone.
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And the volume of a cylinder is equal to ππ squared β, where π is the base radius of the cylinder and β is the height.
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Now, to help us distinguish between the base radius and height of the cone and the base radius and height of the cylinder, letβs call the base radius and height of the cone π one and β one and the base radius and height of the cylinder π two and β two.
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Now that we have these two equations, we can use them in our equation for the volume of water.
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For the volume of water in the conical vessel, we have one-third π π one squared β one.
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And we have that this is equal to the volume of water in the cylindrical vessel which is π π two squared β two.
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Now, weβre ready to substitute in π one β one, π two and β two.
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Our value of π one is the base radius of the cone, which is five centimeters.
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Our value of β one is the height of the cone, which is 24 centimeters.
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Our π two is the base radius of the water in the cylinder, which is 10 centimeters.
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And our β two is the height of the water in the cylindrical vessel or β.
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And so we can simply substitute these values into our equation, which gives us one-third π times five squared times 24 is equal to π times 10 squared times β.
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Now, we notice that we have a π on both sides of the equation.
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And so we can cancel them out.
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And now, in order to get our equation in terms of β, we can divide both sides by 10 squared.
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And this gives us that five squared times 24 over three times 10 squared is equal to β.
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This can also be written as β is equal to five times five times 24 over 10 times 10 times three.
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And since 10 is equal to five times two, we can cancel the two fives from the 10s with the two fives in the numerator.
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This leaves us with 24 over two times two times three.
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Now, 24 is also equal to eight times three.
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And so the three from the 24 can cancel with a three in the denominator.
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And this leaves us with eight over two times two.
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Now since eight is two cubed or two times two times two, we can cancel two of the twos from the eight with the two twos in the denominator.
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And so this leaves us with just two.
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And since β represents the height of water in the cylindrical vessel, this height of β is equal to two centimeters is our final answer.