Answer:
6
Step-by-step explanation:
Answer:
No. The new height of the water is less than the height of the glass(6.33 cm<10 cm)
Step-by-step explanation:
-For the water in the glass to overflow, the volume of the inserted solid must be greater than the volume of the empty space or the ensuing height of water >height of glass.
#Volume of the golf ball:
#The volume of the water in the glass:
We then equate the two volumes to the glass' volume to determine the new height of the water:
Hence, the glass will not overflow since the new height of the water is less than the height of the glass(6.33 cm<10cm).
Answer:
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Problem One
Call the radius of the second can = r
Call the height of the second can = h
Then the radius of the first can = 1/3 r
The height of the first can = 3*h
A1 / A2 = (2*pi*(1/3r)*(3h)] / [2*pi * r * h]
Here's what will cancel. The twos on the right will cancel. The 3 and 1/3 will multiply to one. The 2 r's will cancel. The h's will cancel. Finally, the pis will cancel
Result A1 / A2 = 1/1
The labels will be shaped differently, but they will occupy the same area.
Problem Two
It seems like the writer of the problem put some lids on the new solid that were not implied by the question.
If I understand the problem correctly, looking at it from the top you are sweeping out a circle for the lid on top and bottom, plus the center core of the cylinder.
One lid would be pi r^2 = pi w^2 and so 2 of them would be 2 pi w^2
The region between the lids would be 2 pi r h for the surface area which is 2pi w h
Put the 2 regions together and you get
Area = 2 pi w^2 + 2 pi w h
Answer: Upper left corner <<<<< Answer
