62is the third number (58,60,62,64,66)
Answer:
The volume increases by a factor or 64. When the cube is dilated, each side length is increased by 4, so the volume is increased by 64.
Step-by-step explanation:
Let L represents the Length of the cube before it's dilated.
Volume, V1 = Length * Length * Length
Volume, V1 = L * L * L
Volume, V1 = L³
When the cube is dilated by a factor of 4.
The new Length becomes 4L.
The new volume is calculated as thus.
New Volume, V2 = 4L * 4L * 4L
New Volume, V2 = 64L³
Dividing the new volume by the old volume gives the increment factor.
Factor = New Volume ÷ Old Volume
Factor = V2/V1
Factor = 64L³/L³
Factor = 64.
Hence, when the sides of the cube is dilated by 4, the volume increases by a factor of 64.
Filling the gap of the given sentence;
"The volume increases by a factor or 64. When the cube is dilated, each side length is increased by 4, so the volume is increased by 64"
I would say 50/50 chance because it can go ether ways one can go heads the others can go tails
Answer:
13.695 m
Step-by-step explanation:
The assumption made here is that the boat/water interface is essentially frictionless, so that the center of mass of the system remains in the same place as the occupant of the boat moves around.
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We can find the sum of the moments of boat and child about the pier end:
(46 kg)(7.6 m) + (80 kg)((7.6 +9.6/2) m) = 1341.6 kg·m
After the child moves, the center of mass of boat and child is presumed to remain in the same place. If x is the new distance from the pier to the child, the sum of moments is now ...
46x +80(x-4.8)* = 1341.6
126x -384 = 1341.6
x = (1341.6 +384)/126 = 13 73/105 ≈ 13.695 . . . meters
The child is about 13.695 meters from the pier when she reaches the far end of the boat.
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* The center of mass of the boat alone is half its length closer to the pier than is the child, so is located at x-4.8 meters.
