<p>Each of the containers shown above has a radius r of 3 inches and a height of 6 inches. If the cone shaped container is filled with water and this water is then poured into the empty cylinderical container, what will be the depth, in inches, of water in the cylinderical container? (Volume of cone = Pir^2/3)</p>
<p>I dont have the image, but it is just an image of an ordinary cone and cylinder.
The question is a grid in. </p>
<p>(Answer=2) Help anyone?</p>
<p>your volume formula is not correct.</p>
<p>anyway volume = 1/3 * h * r^2*pi, this liquid volume is constant regardless of the container.
so plug in h and r, v = 1/3 * 6 * 3^2 *pi= 18pi</p>
<p>cylinder is just area of circle * h, so
pi<em>r^2</em>h = 18 pi
r^2<em>h= 18
3</em>3*h=18
h = 2</p>
<p>basically set the volume equal to each other, and plug in r, and solve for H in cylinder eqn.</p>
<p>I understand now lol, depth equal height not volume. Thanks.</p>