It looks like you want to compute the double integral

over the region <em>D</em> with the unit circle <em>x</em> ² + <em>y</em> ² = 1 as its boundary.
Convert to polar coordinates, in which <em>D</em> is given by the set
<em>D</em> = {(<em>r</em>, <em>θ</em>) : 0 ≤ <em>r</em> ≤ 1 and 0 ≤ <em>θ</em> ≤ 2<em>π</em>}
and
<em>x</em> = <em>r</em> cos(<em>θ</em>)
<em>y</em> = <em>r</em> sin(<em>θ</em>)
d<em>x</em> d<em>y</em> = <em>r</em> d<em>r</em> d<em>θ</em>
Then the integral is

Answer:
4
Step-by-step explanation:
6(x²-4x+4-4)+1=0, 6(x-2)²-24+1=0, 6(x-2)²=23, x-2=±√(23/6), x=2±√(23/6)=2±1.95789, so x=3.95789 or 0.04211 approx. These are the zeroes.
Answer:
y=5/4-10
Step-by-step explanation:
Answer:
x=10
y=-2
(10,-2)
Step-by-step explanation:
If X=10, we can put 10 in for x in the equation.
30 + 5y = 20
Next, we need to solve to find y.
We will get y on its own so we subtract 30 from 20 to get
5y=-10
We then know that 5 goes into -10 -2 times.
y=-2