No, the cone and the cylinder can't have congruent heights and bases.
<h3>
is it possible that the two cones have congruent bases and congruent heights?</h3>
The volume of a cylinder of radius R and height H is:
V = pi*R^2*H
And for a cone of radius R and height H is:
V = pi*R^2*H/3
So, for the same dimensions R and H, the cone has 1/3 of the volume of the cylinder.
Here, the cylinder has a volume of 120cm³ and the cone a volume of 360cm³, so the cone has 3 times the volume of the cylinder.
This means that the measures must be different, so the cone and the cylinder can't have congruent heights and bases.
If you want to learn more about volumes:
brainly.com/question/1972490
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Wrong those are considered complex numbers. pure imaginary numbers have ,0 for a and are thus in the form. 3i, -2i, 17i. etc
Answer:
IDK try searching it up
Step-by-step explanation:
I'm assuming the limit is supposed to be
Multiply the numerator by its conjugate, and do the same with the denominator:
so that in the limit, we have
Factorize the first term in the denominator as
The 
terms cancel, leaving you with
and the limand is continuous at 
, so we can substitute it to find the limit has a value of -1/18.
