First, we can see that he used the Distributive property. Then after that, he used the Additive property. After that, he combined like terms and the used the additive property again. Finally, he used the divisive property.
Answer:
Step-by-step explanation:
Answer:
The area of the resulting cross section is
Step-by-step explanation:
we know that
The resulting cross section is a circle congruent with the circle of the base of cylinder
therefore
The area is equal to
we have
-----> the radius is half the diameter
substitute the values