Answer:
If all the ice melted in the cup, it would overflow because the volume of the cone-shaped paper cup (184.3 cm³) is less than the volume of the spherical snowcone (523.6 cm³).
Step-by-step explanation:
To determine whether the spherical snowcone, sold in a cone-shaped paper cup, would overflow the cup when melted, we need to calculate and compare the volumes of the snowcone and the cup.
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To find the volume of the snowcone, we can use the formula for the volume of a sphere:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Volume of a Sphere}}\\\\V=\dfrac{4}{3}\pi r^3\\\\\textsf{where:}\\ \phantom{ww}\bullet\;\textsf{$V$ is the volume.}\\\phantom{ww}\bullet\;\textsf{$r$ is the radius.}\end{array}}[/tex]
In this case, the radius is 5 cm, so we substitute r = 5 into the volume formula:
[tex]V=\dfrac{4}{3}\pi \cdot 5^3\\\\\\V=\dfrac{4}{3}\pi \cdot 125\\\\\\V=\dfrac{500}{3}\pi \\\\\\V=523.598775...\\\\\\V\approx 523.6\; \rm cm^3[/tex]
Therefore, the volume of the snowcone is 523.6 cm³.
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To find the volume of the cone-shaped paper cup, we can use the formula for the volume of a cone:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Volume of a Cone}}\\\\V=\dfrac{1}{3}\pi r^2h\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$V$ is the volume.}\\\phantom{ww}\bullet\;\textsf{$r$ is the radius of the circular base.}\\\phantom{ww}\bullet\;\textsf{$h$ is the height.}\end{array}}[/tex]
If the opening of the cup of 8 cm wide, then the radius is half of this, so:
Substitute the values into the volume formula:
[tex]V=\dfrac{1}{3}\pi \cdot 4^2 \cdot 11\\\\\\V=\dfrac{1}{3}\pi \cdot 16 \cdot 11\\\\\\V=\dfrac{1}{3}\pi \cdot 176\\\\\\V=\dfrac{176}{3}\pi \\\\\\V=184.3067690...\\\\\\V\approx 184.3\; \rm cm^3[/tex]
Therefore, the volume of the cone-shaped paper cup is 184.3 cm³.
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Since the volume of the cone-shaped paper cup (184.3 cm³) is less than the volume of the spherical snowcone (523.6 cm³), if all the ice melted, it would overflow the cup.