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3 years ago

HELP ASAP 10 POINTS AND BRAINLIST

Mathematics

1 answer:

ollegr [7]3 years ago

4 0

Step-by-step explanation:

hope it helps you...........

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Praedup uses the inequality r+2≥25 to represent his rock and shell collection, where r is the number of rocks in his collection.

Ne4ueva [31]

Yesss was a great time

3 0

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

What is the exact volume of the sphere?

Firlakuza [10]

The correct answer is C.

5 0

3 years ago

I REALLY NEED THIS PLEASE HURRY!

elena-s [515]

PROPERTY:

Distributive property:— It tells us how to solve expressions in the form of a(b + c).

ANSWER:

3xy + 6yz can be written as,

3y(x + 2z).

4 0

3 years ago

Read 2 more answers

Need to pass this test to graduate

Paraphin [41]

The 1st graph has vertex in (-3, -3) which can be translated into
Horizontal shift left 3
Vertical shift down 3
Other than that, the graph shows y=x^2 so it wasn't compressed or stretched

The 2nd graph has vertex in (0, 0) which mean there is no vertical and horizontal shift. But the graph is facing down and it was slimmer than y=x^2 graph. When x=1, the result is y=3 which was 3 times more than it supposed to be.
The graph must be:
Reflection across x-axis
Vertical stretch of 3

The 3rd graph has vertex in (3, -3) which can be translated into
Horizontal shift right 3
Vertical shift down 3
Same as the 1st graph, the 3rd graph shows y=x^2 so it wasn't compressed or stretched.

5 0

3 years ago