Answer:
quotion is 104 and remainder is 6
In Cartesian coordinates, the region (call it
) is the set

In the plane
, we have

which is a circle with radius
. Then we can better describe the solid by

so that the volume is

While doable, it's easier to compute the volume in cylindrical coordinates.

Then we can describe
in cylindrical coordinates by

so that the volume is

Answer:
Congruent and isometries.
Step-by-step explanation:
Congruent meaning the shape is still the same size and shape
(which is true because translations and rotations don't change size or shape)
And isometries because it is one way of mapping points to a different location.
Answer:
C=5/9(F-32)
Step-by-step explanation:
there you go
I’m pretty sure you do need to subtract 30-8.