<u>Let's name these angles</u>.
∠ A = 72°
∠ B = 54°
∠ C = 54°
Now, in this triangle, two angles, i.e., ∠ B and ∠ C are same.
According to the properties of triangle, sides opposite to two equal angles in a triangle are also equal, which means the two sides opposite to 54° in the figure will be equal.
=> <u>AB = AC</u>
Now, if there are two equal sides in a triangle then the triangle is called <u>Isosceles Triangle.</u>
So, the above triangle is an <u>Isosceles Triangle</u>.
Split up the surface

into three main components

, where

is the region in the plane

bounded by

;

is the piece of the cylinder bounded between the two planes

and

;
and

is the part of the plane

bounded by the cylinder

.
These surfaces can be parameterized respectively by

where

and

,

where

and

,

where

and

.
The surface integral of a function

along a surface

parameterized by

is given to be

Assuming we're just finding the area of the total surface

, we take

, and split up the total surface integral into integrals along each component surface. We have






Therefore
Renting you lose money even if it’s cheap you can never get your money back. When you buy a house you can resell it and get all your money back or possibly even more.
...42, -193.. thats the 4th and 5th Hope this helps
To find x, you divide both sides of the equation by 12.20. This would look like this:
12.20x/12.20 =372.10/12.20 . Then, the answer you get is 30.5, so. X=30.5