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4 years ago

-2x+5c+6c-3x= Simplify and write the answer.

Mathematics

2 answers:

Irina18 [472]4 years ago

8 0

-2x+5c+6c-3x
-5x+11c=

Shalnov [3]4 years ago

7 0

Answer:

-5x+11c

Step-by-step explanation:

add the alike terms

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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

The vertices of a triangle are P(−3, −4), Q(3, 4), and R(−6, −3). Name the vertices of Rx = 0 (PQR).

Katyanochek1 [597]

The correct answer is A) P'(3, −4), Q'(−3, 4), R'(6, −3)

Rx = 0 indicates a reflection over the y-axis.

The rule for such a transformation is:
(x, y) --> (-x, y)
which means that the x-coordinate changes sign and the y-coordinate stays the same.

Therefore:
P(-3, -4) --> P'(3, -4)
Q(3, 4) --> Q'(-3, 4)
R(-6, -3) --> R'(6, -3)

These points are those in option A).

6 0

3 years ago

timama [110]

Answer:

25π units^2

Step-by-step explanation:

The shape is a quarter circle with radius 10 units.

area of circle = πr^2

area of a quarter circle = (1/4)π(10 units)^2

area = 25π units^2

5 0

3 years ago

Find the equation of the line passing through the points (2, 4) and (3, 2). y = [? ]x + [ ]

PtichkaEL [24]

Answer:

first find slope m

m=(4-2)/(2-3)

m= -2

so equation line is

y-y1=m(x-x1)

y-4=-2(x-2)

y-4=-2x+4

y=-2x+4+4

y=-2x+8

?=-2

7 0

3 years ago

I need the answer just put the work or answer

Serjik [45]

Answer:

65000

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers