Y=-2/3x+3 please show the work thank you ;)

Mathematics

1 answer:

Ivahew [28]3 years ago

3 0

Answer:

Step-by-step explanation:

Looking at y=-%282%2F3%29x%2B3 we can see that the equation is in slope-intercept form y=mx%2Bb where the slope is m=-2%2F3 and the y-intercept is b=3

Since b=3 this tells us that the y-intercept is .Remember the y-intercept is the point where the graph intersects with the y-axis

So we have one point

Now since the slope is comprised of the "rise" over the "run" this means

slope=rise%2Frun

Also, because the slope is -2%2F3, this means:

rise%2Frun=-2%2F3

which shows us that the rise is -2 and the run is 3. This means that to go from point to point, we can go down 2 and over 3

So starting at , go down 2 units

and to the right 3 units to get to the next point

Now draw a line through these points to graph y=-%282%2F3%29x%2B3

So this is the graph of y=-%282%2F3%29x%2B3 through the points and

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1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves

erastova [34]

(a) The differential equation is separable, so we separate the variables and integrate:

$(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx$

$\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx$

$\ln|y| = x - \ln|x+1| + C$

When x = 0, we have y = 2, so we solve for the constant C :

$\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)$

Then the particular solution to the DE is

$\ln|y| = x - \ln|x+1| + \ln(2)$

We can go on to solve explicitly for y in terms of x :

$e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}$

(b) The curves y = x² and y = 2x - x² intersect for

$x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1$

and the bounded region is the set

$\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}$

The area of this region is

$\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}$