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:
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The two intersection points are (-2.79, -0.58) and (0.79, 6.58).
<h3>
How to find the points of intersection?</h3>
Here we want to solve the system of equations:
y = 2x + 5
x² + y² = 36
To solve this, we need to replace the first equation into the second one:
x² + (2x + 5)² = 36
Now we can solve this for x:
x² + 4x² + 10x + 25 = 36
5x² + 10x - 11 = 0
This is a quadratic equation, to solve it we use the general formula:

So we have two solutions for x:
x = (-10 - 17.9)/10 = -2.79
x = (-10 + 17.9)/10 = 0.79
To get the y-values of the solutions, we evaluate the linear equation in these values of x:
y = 2*(-2.79) + 5 = -0.58
y = 2*( 0.79) + 5 = 6.58
Then the two intersection points are (-2.79, -0.58) and (0.79, 6.58).
If you want to learn more about intersection points:
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