The diagram is y because it is y to that is because ahhh
If you have to compare two shapes you basically count the number of shapes there are of each type then you say for example 6:4. If they ask you to give the number of shapes to the number of shapes there are you count all the shapes and then you count the shape they asked you to find and then you will get something like 4:15
First you'll need to find the area of each side and then add those areas together. 2(6x10)+2(4x6)+2(4x10)= 120+48+80=248
then you multiply the total area by .25 and get 62. It will cost $62.
Stokes' theorem says the integral of the curl of
over a surface
with boundary
is equal to the integral of
along the boundary. In other words, the flux of the curl of the vector field is equal to the circulation of the field, such that

We have


Parameterize the ellipse
by

with
and
.
Take the normal vector to
to be

Then the flux of the curl is

The answer is :X^2 + 7x + 12
The mistake made was when expanding they added 3 and 4 together instead of multiplying it