Answer:
6j + 6r :)))))))))))))))))
The line integral along the given positively oriented curve is -216π. Using green's theorem, the required value is calculated.
<h3>What is green's theorem?</h3>
The theorem states that,
Where C is the curve.
<h3>Calculation:</h3>
The given line integral is
Where curve C is a circle x² + y² = 4;
Applying green's theorem,
P = 9y³; Q = -9x³
Then,
⇒ 
Since it is given that the curve is a circle i.e., x² + y² = 2², then changing the limits as
0 ≤ r ≤ 2; and 0 ≤ θ ≤ 2π
Then the integral becomes
⇒ 
⇒ 
⇒ 
⇒ 
⇒ ![-108[2\pi - 0]](https://tex.z-dn.net/?f=-108%5B2%5Cpi%20-%200%5D)
⇒ -216π
Therefore, the required value is -216π.
Learn more about green's theorem here:
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It is 24+6b im writing more because it says to put more cause my answer is to "short"
2y=29+3
2y=32
Y=32/2
Y=16
Hope that is right
Step-by-step explanation:
3a²-7a−6
Factor the expression by grouping. First, the expression needs to be rewritten as 3a
2
+pa+qa−6. To find p and q, set up a system to be solved.
p+q=−7
pq=3(−6)=−18
Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product −18.
1,−18
2,−9
3,−6
Calculate the sum for each pair.
1−18=−17
2−9=−7
3−6=−3
The solution is the pair that gives sum −7.
p=−9
q=2
Rewrite 3a
2−7a−6 as (3a
2−9a)+(2a−6).
(3a 2−9a)+(2a−6)
Factor out 3a in the first and 2 in the second group.
3a(a−3)+2(a−3)
Factor out common term a−3 by using distributive property.
(a−3)(3a+2)
