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Natasha_Volkova [10]
3 years ago
10

What is 283+t 748-235+7493-7

Mathematics
1 answer:
WARRIOR [948]3 years ago
8 0

Answer:

t^748+7534

Step-by-step explanation:


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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
A can of tomato soup contains 8 oz of soup the ounces is equal to 28.35 g is how many grams of soup are in the can?
Natalka [10]

Answer:

226.8

Step-by-step explanation:

Multiply 8 x 28.35

6 0
3 years ago
10. The Johnson family purchased a new television that had a 62 inch diagonal. The height of the television was 30 inches. Appro
lana66690 [7]

Answer:

Width = 54.3 inches

Step-by-step explanation:

We solve this question using the Pythagoras Theorem

Width² + Height² = Diagonal²

From the above question

Width= ??

Height = 30 inches

Diagonal = 62 inches

Width² + 30² = 62²

Width² = 62² - 30²

Width = √62² - 30²

Width =✓2944

Width = 54.258639865 inches

Approximately

Width = 54.3 inches

Hence, the new television was 54.3 inches wide.

7 0
3 years ago
0.6=x/5 what is the answer please
stealth61 [152]

Answer: 3

Step-by-step explanation:

reverse the problem

0.6=x/5

0.6 times 5 = 3

check your work

3/5=0.6

0.6=0.6

3 0
3 years ago
Read 2 more answers
A card has p meters of lace edging wound on it. Nerys buys n lengths of edging, each x cm long. If Q meters of edging are left o
Diano4ka-milaya [45]

Answer:

Q = P - n(x/100)

Step-by-step explanation:

P is the total number of lace edging, n is the number of times the length x is cut from P, and Q is the remnant of the reduction.

So the number of x length is subtracted from P and must be converted to meters (since it is in centimeters) to get Q. That is;

Q = P - n(x/100)

5 0
3 years ago
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