A snow cone is a tasty treat with flavored ice and a spherical bubble gum ball at the bottom. The radius of the cone is 1.25 inches, and its height is 2.75 inches. If the diameter of the bubble gum ball is 0.5 inches, what is the closest approximation of the volume of the cone that can be filled with flavored ice? (picture coming in a sec)
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4.43 0.07 13.50 4.50
solved.
how do you do it and whats the answer
The formula for the volume of a cone is \[V= \frac{ 1 }{ 3 } \pi r ^{2} h \]
give it a try
its not one of the answer choices
$$ \text{volume for a cone}\\ \cfrac{\pi r^2 h}{3}\\ \text{volume for a sphere}\\ \cfrac{4}{3}\pi r^3 $$ so substract the volume of the gum ball from the volume of the cone
Try subtracting the space the bubble gum takes up form the area and see if you get the answer.
the diameter of the gum is given so the radius is 1/2 d the volume of a sphere is \[V =\frac{ 4 }{ 3 } \pi r ^{3} \]
.......not as easy as you all put it
keep in mind the bubble gum DIAMETER is 0.5", not the radius
its 4.5 ok
so that means that the radius is .25
wait let me double check
nvm the correct answer is 4.43
Volume of Cone. \[\large V = \cfrac{1}{3} \pi r^2 h\]\[\large r = 1.25 \ in \ , \ h = 2.75 \ in\]\[\large V_{1} = \cfrac{1}{3} \pi (1.25)^2 (2.75)\]\[\large V_1=\cfrac{275}{192} \pi\]Volume of Sphere \[\large V = \cfrac{4}{3} \pi r^3\]\[\large d=0.5 \ , \ r = \cfrac{0.5}{2}= 0.25 \]\[\large V_2 = \cfrac{4}{3} \pi (0.25)^3\]\[\large V_2 = \cfrac{1}{48}\pi\]
because i didnt know what to do with the ball but thanks
Now Subtract \(\large V_1 - V_2\)
well, is a bubble gum ball, so, what to do with a bubble gum? well, just chew it :)
after you eat the ice of course
true lol
exchange medals
\[\large V_1 - V_2 = \cfrac{275}{192} \pi - \cfrac{1}{48}\pi \approx 4.43 \ in^3 \]
thanks everyone
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