The shaded areas add to 2/5 of the area of the circle, which is given by
A = πr²
Then the shaded area is ...
(2/5)A = (2/5)π(10 in)²
shaded area = 40π in²
A = πr²
Then the shaded area is ...
(2/5)A = (2/5)π(10 in)²
shaded area = 40π in²
Find the lower bounds for the following weights?
1 answer:
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