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Sergio039 [100]
3 years ago
14

The rectangle has an area….

Mathematics
1 answer:
Anon25 [30]3 years ago
4 0

Answer:

5.5193237 times 10 to the power of -10

Step-by-step explanation:

To solve you just need to divided the area by the length or width the question has provided.  Please note I am in 7th grade.  

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Write a question about a situation that has 24-x for the answer
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A bottle holds 24 ounces of water. It has x ounces of water in it.

What does 24 - x represent in this situation?

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Christine purchased a prepaid phone card for $20. Long distance calls cost 18 cents a minute using this card. Christine used her
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4 minutes and 32 seconds

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card $20.00-money left over $15.68            $20.00-$15.68  4.32

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720 pt in a hour converted to quarts in a min
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2 years ago
Nine hundred light bulbs are packaged and shipped to a retail store. Thirteen percent of the light bulbs arrive broken. Which eq
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7 0
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Read 2 more answers
use green's theorem to evaluate the line integral along the given positively oriented curve. c 9y3 dx − 9x3 dy, c is the circle
Rina8888 [55]

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,

\int_CPdx+Qdy = \int\int_D(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})dx dy

Where C is the curve.

<h3>Calculation:</h3>

The given line integral is

\int_C9y^3dx-9x^3dy

Where curve C is a circle x² + y² = 4;

Applying green's theorem,

P = 9y³; Q = -9x³

Then,

\frac{\partial P}{\partial y} = \frac{\partial 9y^3}{\partial y} = 27y^2

\frac{\partial Q}{\partial x} = \frac{\partial -9x^3}{\partial x} = 27x^2

\int_C9y^3dx-9x^3dy = \int\int_D(-27x^2 - 27y^2)dx dy

⇒ -27\int\int_D(x^2 + y^2)dx dy

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

-27\int\limits^{2\pi}_0\int\limits^2_0r^2. r dr d\theta

⇒ -27\int\limits^{2\pi}_0\int\limits^2_0 r^3dr d\theta

⇒ -27\int\limits^{2\pi}_0 (r^4/4)|_0^2 d\theta

⇒ -27\int\limits^{2\pi}_0 (16/4) d\theta

⇒ -108\int\limits^{2\pi}_0 d\theta

⇒ -108[2\pi - 0]

⇒ -216π

Therefore, the required value is -216π.

Learn more about green's theorem here:

brainly.com/question/23265902

#SPJ4

3 0
1 year ago
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