Four marbles each of radius r are placed in a conical container.  What is minimum radius and height of cone?
970422
The solution is too convoluted to put it here without diagrams and a good equation editor.  If you wish, I can email a Word document detailing the solution.  I figure that the cone is (1 + 2*sqrt(2/3) + sqrt(3))*r (about 4.365 r) high and has a base radius of ((sqrt(1/2) + sqrt(4/3) + sqrt(3/2))*r or about 3.087.
783604
volume=pir^2h=1000\nsince there are two variables we need a minimum of two equations
676849
Cylinder = PI r^2 x 6 r = 6PI r^3\nBalls = 4/3 x PI r^3 x 3 = 4 PI r^3\nratio = 4 PI r^3 / 6 PI r^3 = 4/6\n\nRatio = 2/3 of the space is used by the balls.
515192
If it's a square, (60 x 60) then probably 8.\nIf it's a circle (r=60) then probably 10.\nIf it's a cirlce (d=60) then probably 6
237156
Pi * r^2\n\n(Pi R Squared)\n\nwhere R is the radius (which is also equivalent to half the diameter)
22565
Step 1 is to find the volume of each of the individual spheres using formula (4/3)*pi*r^3\nSphere A diameter 2 cm\nSphere B diameter 3 cm\nsphere c diameter 4 cm\n\nSphere A Volume =(4/3)*pi*(1 cm)^3= approx 4.189 cm^3\nSphere B Volume =(4/3)*pi*(1.5 cm)^3= approx 14.137 cm^3\nSphere C Volume =(4/3)*pi*(2 cm)^3= approx 33.510 cm^3\n\nadd them together to get a volume of 51.836\n\nnow to find radius of new sphere.\n\n51.836=(4/3)*pi*r^3\n(51.836)*(3/4pi)=r^3\n12.375=r^3\nr=approx 2.313\n\ntherefore the diameter of the new sphere is about 4.626 cm
147975
First, find the volume of the can: [pi]r-squared x the height of the can. If the 3 balls in it each have a radius of 5 cm, then they are each 10 cm tall, and the can is 30 cm tall. So the volume of the can = [pi]x25x30 = 750[pi] cubic cm.\n\nNext, the volume of each ball is 4/3 x [pi]r-cubed, or 166.67[pi] cubic cm. Three of them would be 500[pi] cubic cm.\n\nLast, subtract the volume of the balls from the volume of the cylinder to get 250[pi] cubic cm, or about 785.4 cubic cm.
22742
4 inches. the radius is half of the diameter.
576892
You probably need more information...there can be multiple cylinders inscribed in a sphere with a radius of 5.
1106432
Diameter == 10.\nRadius == 5.\nCenter == (1,7)\n\nMethod: use Pythagorean Theorem.
405508
5 marbles will make a row 3 inches long. (3/5*5=3), so double that. 10 marbles should make a row 6 inches long.
307086
Cones Volume = (1/3) x (volume of cylinder) = (1/3) x (base x height) = (1/3) x (pi x r^2 x h)
543695
volume = (1/3) b h = 1/3 pi r 2 h \nSurface area = pr2 + prs\ns= the height of the outside of the cone, while h = the height of the cone at its center
778715
increments of radius is 0.001 in/min\nincrement of area is 3.14(1.801^2-1.8^2)=0.0113 sq in\nhence increment of volume is 6*0.0113=0.0678 cu in/min\n\nMore scientific solution\nassume radius increments = dx\nthen volume increment V will be Height*area increment\nV = 6*3.14{(R+dx)^2-R^2} = 18.84(R^2+2Rdx+dx^2-R^2)\nV=18.84 (2Rdx)  when dx close to 0 then dx^2 can be omitted\nthen dV/dt = 37.68 R dx/dt = 37.68 * 1.8 * 0.001 =0.0678 cu in/min
827739
Not quite sure what you are asking for....\n\nThe surface area of a right cone (i.e. a cone with its vertex above the center of its base) is:\npi*R*sqrt(R^2+H^2)\n\nwhere R is the radius of the base and H is the height of the cone.
29575
well, the formula for the volume of a cone is 1/3 * base * height.\n\nthe trick to doing this is to find the volume of the large cone, then subtract the volume of the small cone.  what's leftover will be the tumbler-shaped volume.  plug in numbers for the height and base for both the large and small cones, and then subtract. voila!
866787
let the radius of the cylinder be x,  drawing a cylinder in a cone on the paper, we see 2 triangles - the cone & the triangle with it's base as the top flat surface of the cylinder. follow me?\n\nif u do, then let's assume the height of the second triangle to be y. y can be found in terms of x as:\n(since both triangles are similar)\n\nx:4 :: y:12  \ni.e. x/4 = y/12\n=> y=3x\nthus ht of cylinder = ht of cone - y\n=12-3x\n\nvolume of cylinder = pi*r2*h  (r-> radius, h-> height of cylinder)\nv= pi * x2 * (12-3x) = pi * (12x2 - 3x3)\n  (the second term being 12 *x*x - 3*x*x*x ....sorry, i don't have a math equation display software here)\n\npartialy differentiate above equation wth respect to  x, we then get\ndv/dx = pi * (24*x - 9*x*x)\n\nequating dv/dx to zero, (to find a value of x that makes the volume value either maxium or minimum. u can verify it's max by again differentiating dv/dx with respect to x, equating to zero & finding if the value solved for is positive or negative)\n\nanyways, we get\n\neliminating  pi,\n24*x - 9*x*x=0\n\n=> x=24/9 = 8/3= 2.67\n\nnow, simply substitute this value of x in the volume equatiion above to get\n v= pi * (12*2.67*2.67 - 3*2.67*2.67*2.67)\n= 89.372 or so.\n\n\nuh...i guess i only confused u up, hope it helped u atleast a bit.