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3 years ago

11

What is the slope-intercept form of the following equation?

1 answer:

spin [16.1K]3 years ago

5 0

Y=-.5x+3
Explanation: Move the x to the other side by subtracting it from both sides, you end up with 10y=30-5x then you divide 10 from both sides and you end up with y=-.5x+3

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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

In Brooke’s class of 25 students, 15 students report that math is their favorite class and 40% of the students report that scien

Natasha_Volkova [10]

Answer:

The students prefer Math 60%, to Science 40%

Step-by-step explanation:

15/25= 3/5= 5 is then divided into 3 to get the per cent= 60%

3 0

3 years ago

If f of x equals 3x plus 2, find f of 4

Juli2301 [7.4K]

Answer:

f(4) = 14

Step-by-step explanation:

f(x) = 3x+2

Let x=4

f(4) = 3(4)+2

= 12+2

= 14

4 0

3 years ago

The rules of a contest say there is a 1 out of 20 chance of winning a prize. 420 people enter the contest. Predict how many peop

lapo4ka [179]

Answer:

21

Step-by-step explanation:

1/20= x/420

Multiply both by 420

420/20=x

x=21

6 0

3 years ago

There is a rectangular terrain with dimensions "x" and "y". If it is known that its area of is 22m ^ 2 and that x ^ 2 + y ^ 2

Margaret [11]

9514 1404 393

Answer:

24 m

Step-by-step explanation:

The area formula tells us ...

22 = xy . . . . . square meters

Adding twice this to the quadratic constraint gives ...

(x^2 +y^2) +2(xy) = (100) +2(22)

(x +y)^2 = 144

x +y = √144 = 12

The perimeter is double this amount, so is ...

P = 2(x +y) = 2(12) = 24 . . . . meters

5 0

3 years ago