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Roman55 [17]
2 years ago
7

3/4 (x - 3) - 5 = 1/8 ( x - 1 )

Mathematics
1 answer:
KonstantinChe [14]2 years ago
3 0

Answer:

x = - \frac{57}{5}

Step-by-step explanation:

\frac{3}{4} (x - 3) - 5 = \frac{1}{8} (x - 1)

Multiply through by 8 to clear the fractions

6(x - 3) - 40 = x - 1 ← distribute parenthesis and simplify left side

6x - 18 - 40 = x - 1

6x - 58 = x - 1 ( subtract x from both sides )

5x - 58 = - 1 ( add 58 to both sides )

5x = - 57 ( divide both sides by 5 )

x = - \frac{57}{5}

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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

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