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3 years ago

What choices are correct?

Mathematics

2 answers:

motikmotik3 years ago

7 0

The answer is D.The domain is all real numbers

Snezhnost [94]3 years ago

4 0

Answer:

its b

Step-by-step explanation:

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(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

At Doggy Daycare, there are 3 trainers assigned for every 12 dogs. There are 72 dogs enrolled for the daycare on Tuesday. Determ

fenix001 [56]

Answer:

Step-by-step explanation:

make a ratio relation:

3 trainers : 12 dogs

x trainers : 72 dogs

3/12 = x/72

solve for x, and you get 18

7 0

3 years ago

Can someone help me please In the diagram, VWX and XYZ are congruent equilateral triangles, and ∠VXY is 80o. What is the measure

Lelu [443]

Answer:

Step-by-step explanation:

its vwx = xyz

5 0

4 years ago

Read 2 more answers

Which of the following is the solution set of x2 + 8x + 12 = 0? (-2.-6 2.6

algol [13]

To solve this polynomial equation, we first factor the left side.

To factor x² + 8x + 12 = 0, look for factors of 12 that add to 8.

These factors are 6 and 2 so we have (x + 6)(x + 2) = 0.

So either x + 6 = 0 or x + 2 = 0 and solving for

x in each equation, our answer is {-6, -2}.

6 0

3 years ago

1 the average of 3 6 9

Brut [27]

1. 3 + 6 + 9 = 18
18/3 = 6 --> The average is 6
2. 15 + 5 + 8 + 4 = 32
32/4 = 8 --> The average is 8
3. 6 + 10 + 5 = 21
21/3 = 7 --> The average is 7
4. 14 + 9 + 7 = 30
30/3 = 10 --> The average is 10

6 0

3 years ago