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3 years ago

PLS HELP ME PLS PLS !

Mathematics

1 answer:

BaLLatris [955]3 years ago

4 0

Answer:

ill help

Step-by-step explanation:

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What is the graph of the function rule Y= 1/3x - 1 ?

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Yes. A is the right graph because the line must travel through two points: (0 ,1)and(-3, 0)

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2 years ago

Solve for x: -6 <x-1 <9

WINSTONCH [101]

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3 years ago

3y-x=3 and x is four times the value of y what is the value of x-y?

Aleonysh [2.5K]

Answer:

x - y = -9

Step-by-step explanation:

from the equation

3y - x = 3

and x is four times the value of y

i.e x = 4y

put x = 4y into the first equation

we have;

3y - 4Y = 3

-Y = 3

Divide both sides by -1

we have

y = -3

so put y = -3 in the 2nd equation

i.e x = 4y

x = 4 × -3

x = -12

therefore x = -12 and y = -3

so tho find x - y

we have

-12 - (-3)

= -9

therefore x - y = -9

7 0

3 years ago

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

2 years ago

Please helpppppoo I’m so lost

Elena L [17]

Answer:

Step-by-step explanation:

Please helpppppoo I’m so lost

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3 years ago