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3 years ago

Pls help I don’t get it

Mathematics

2 answers:

faust18 [17]3 years ago

6 0

Here you go!!!! sorry for taking long to answer

andrezito [222]3 years ago

4 0

Can't really explain it but the answer would be

x=1

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If the discriminate of a quadratic equation is 28 describe the roots

katrin2010 [14]

B) two real roots is the answer because it is bigger than 0.

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3 years ago

(x^2y+e^x)dx-x^2dy=0

klio [65]

It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor µ(x, y) such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$

The modified DE,

$\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0$

is now exact:

$\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}$

So we look for a solution of the form F(x, y) = C. This solution is such that

$\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}$

Integrate both sides of the first condition with respect to x :

$F(x,y) = -e^{-x}y - \dfrac1x + g(y)$

Differentiate both sides of this with respect to y :

$\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C$

Then the general solution to the DE is

$F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}$

5 0

3 years ago

EXPERT HELP I'LL GIVE BRAINLIEST:

Kisachek [45]

Answer:

Step-by-step explanation:

Its D because they Because they wouldnt just do one school they would do multipule school to see the amount and how much

3 0

3 years ago

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Find the distance between each pair of points

dolphi86 [110]

Answer: the correct answer is 20

Step-by-step explanation:

The formula for determining the distance between two points on a straight line is expressed as

Distance = √(x2 - x1)² + (y2 - y1)²

Where

x2 represents final value of x on the horizontal axis

x1 represents initial value of x on the horizontal axis.

y2 represents final value of y on the vertical axis.

y1 represents initial value of y on the vertical axis.

From the graph given,

x2 = - 7

x1 = 5

y2 = - 7

y1 = 9

Therefore,

Distance = √(- 7 - 5)² + (- 7 - 9)²

Distance = √(- 12²) + (- 16)²

= √(144 + 256) = √400

Distance = 20

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3 years ago

The price that a company charged for a basketball hoop is given by the equation 50-5x^2 where x is the number of hoops that are

Debora [2.8K]

Given:
Profit : 15,000,000
Cost: 30 per basketball hoop
production: 1 million hoops
price: 50 - 5x²

Profit = Sales - Cost
15,000,000 = sales - 30(1,000,000)
15,000,000 + 30,000,000 = sales
45,000,000 = sales

45,000,000 / 1,000,000 = 45 sales price.

3 0

3 years ago

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