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4 years ago

PLEASE HELP ILL GIVE BRAINLIEST

Mathematics

1 answer:

Hatshy [7]4 years ago

8 0

Answer:

Step-by-step explanation:

1. Find the function's zeros and vertical asymptotes, and plot them on a number line.

2. Choose test numbers to the left and right of each of these places, and find the value of the function at each test number.

3.Use test numbers to find where the function is positive and where it is negative.

4. Sketchh the function's graph, plotting additional points as guides as needed.

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Give me brilliant please

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Let C be the boundary of the region in the first quadrant bounded by the x-axis, a quarter-circle with radius 9, and the y-axis,

rewona [7]

Solution :

Along the edge $C_1$

The parametric equation for $C_1$ is given :

$x_1(t) = 9t , y_2(t) = 0 \ \ for \ \ 0 \leq t \leq 1$

Along edge $C_2$

The curve here is a quarter circle with the radius 9. Therefore, the parametric equation with the domain $0 \leq t \leq 1$ is then given by :

$x_2(t) = 9 \cos \left(\frac{\pi }{2}t\right)$

$y_2(t) = 9 \sin \left(\frac{\pi }{2}t\right)$

Along edge $C_3$

The parametric equation for $C_3$ is :

$x_1(t) = 0, \ \ \ y_2(t) = 9t \ \ \ for \ 0 \leq t \leq 1$

Now,

x = 9t, ⇒ dx = 9 dt

y = 0, ⇒ dy = 0

$\int_{C_{1}}y^2 x dx + x^2 y dy = \int_0^1 (0)(0)+(0)(0) = 0$

And

$x(t) = 9 \cos \left(\frac{\pi}{2}t\right) \Rightarrow dx = -\frac{7 \pi}{2} \sin \left(\frac{\pi}{2}t\right)$

$y(t) = 9 \sin \left(\frac{\pi}{2}t\right) \Rightarrow dy = -\frac{7 \pi}{2} \cos \left(\frac{\pi}{2}t\right)$

Then :

$\int_{C_1} y^2 x dx + x^2 y dy$

$=\int_0^1 \left[\left( 9 \sin \frac{\pi}{2}t\right)^2\left(9 \cos \frac{\pi}{2}t\right)\left(-\frac{7 \pi}{2} \sin \frac{\pi}{2}t dt\right) + \left( 9 \cos \frac{\pi}{2}t\right)^2\left(9 \sin \frac{\pi}{2}t\right)\left(\frac{7 \pi}{2} \cos \frac{\pi}{2}t dt\right) \right]$

$=\left[-9^4\ \frac{\cos^4\left(\frac{\pi}{2}t\right)}{\frac{\pi}{2}} -9^4\ \frac{\sin^4\left(\frac{\pi}{2}t\right)}{\frac{\pi}{2}} \right]_0^1$

= 0

And

x = 0, ⇒ dx = 0

y = 9 t, ⇒ dy = 9 dt

$\int_{C_3} y^2 x dx + x^2 y dy = \int_0^1 (0)(0)+(0)(0) = 0$

Therefore,

$\oint y^2xdx +x^2ydy = \int_{C_1} y^2 x dx + x^2 x dx+ \int_{C_2} y^2 x dx + x^2 x dx+ \int_{C_3} y^2 x dx + x^2 x dx$

= 0 + 0 + 0

Applying the Green's theorem

$x^2 +y^2 = 81 \Rightarrow x \pm \sqrt{81-y^2}$

$\int_C P dx + Q dy = \int \int_R\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx dy$

Here,

$P(x,y) = y^2x \Rightarrow \frac{\partial P}{\partial y} = 2xy$

$Q(x,y) = x^2y \Rightarrow \frac{\partial Q}{\partial x} = 2xy$

$\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right) = 2xy - 2xy = 0$

Therefore,

$\oint_Cy^2xdx+x^2ydy = \int_0^9 \int_0^{\sqrt{81-y^2}}0 \ dx dy$

$= \int_0^9 0\ dy = 0$

The vector field F is = $y^2 x \hat i+x^2 y \hat j$ is conservative.

5 0

3 years ago

What’s 50 times 1,300??

solmaris [256]

50 * 1300 =

5* 13 000

65000

6 0

3 years ago

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You roll a dye pick a marble and pick a card how many outcomes are possible

Alja [10]

SOLUTION

The die has 6 faces, that is 6 outcomes

They are 2 marbles that is 2 outcomes

They are 2 cards, that is 2 outcomes

Number of possible outcomes become

$6\times2\times2=24$

Hence the answer is 24 possible outcomes

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1 year ago

Help me i'm crying help me

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Answer:

Get a tissue blow your nose and wipe your tears

Step-by-step explanation:

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4 years ago

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The radius of a circular oil slick expands at a rate of 2 m/min.

zhuklara [117]

As we knw that the dr/dt = 2
And The area of the circular slick is given by formul
A = (pi)r^2.
Now lets take derivative: dA/dt=2πrdr/dt
As we knw that radius is 25
sodA/dt=2π(25)(2)=100π
which is equal to
=314.16 per minute this is how area is increasing
now for part b
 t = 4, the radius will have increased to 8 m.
now we will put the values in formula dr/dt
so it will be.dA/dt=2π(8)(2)
=32π
=100.53 is the answer
hope it helps

6 0

3 years ago