A spherical cantaloupe is cut in half. The visible cross-section has an area of 5252 in2. The volume of the cantaloupe is _____
1 answer:
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You can think of this question like the photo attached above.
Hi!
I think for b, the answer would be:
You can construct an angle that is one fourth the measure of angle JKL, by dividing angle either angle MKL or JKM in half.
This is because one-fourth is also equal to one quarter (1/4). If you split angle JKL into 4 equal angles, you would have an angle that is one-fourth the measure (original angle) of angle JKL.
I think for c, the answer would be:
You can construct a 15 degree angle from a given 60 degree angle, by dividing the 60 degree angle into 4 equal angles.
This would work, because each of the 4 angles would be 15 degrees.
Hope this helps! Best of luck!
Hi!
I think for b, the answer would be:
You can construct an angle that is one fourth the measure of angle JKL, by dividing angle either angle MKL or JKM in half.
This is because one-fourth is also equal to one quarter (1/4). If you split angle JKL into 4 equal angles, you would have an angle that is one-fourth the measure (original angle) of angle JKL.
I think for c, the answer would be:
You can construct a 15 degree angle from a given 60 degree angle, by dividing the 60 degree angle into 4 equal angles.
This would work, because each of the 4 angles would be 15 degrees.
Hope this helps! Best of luck!