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gladu [14]
2 years ago
6

I need some help again

Mathematics
1 answer:
Marrrta [24]2 years ago
4 0

A = small box of oranges

B = large box of oranges

3A + 14B = 203

11A + 11B = 220

Take the second equation and divide it by 11 as both A and B have the same coefficient, which will equal:

\frac{11A+11B}{11} = A + B

220 will also have to be divided by 11, giving you 20

A + B = 20

You can cancel out 3A in the first equation by subtracting 3A + 3B = 60 from it.

11B = 143

143/11 = 13

B (large box of oranges) costs $13.

Substitute 13 for B back into one of our equations, in this case 3A + 3B = 60

3A + 3*13 = 60

3A + 39 = 60

3A = 21

A = 7

Thus:

A small box of oranges costs $7, and a large box of oranges costs $13.

Hope this helped! (please brainliest)

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Answer:

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Step-by-step explanation:

Given Equation:  

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Putting the given value of 'x' in the equation:

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At x=1, the value of y is y=-3

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(x^2y+e^x)dx-x^2dy=0
klio [65]

It looks like the differential equation is

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\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 <em>µ(x, y)</em> 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 <em>µ(x, y)</em> = <em>µ(x)</em> be independent of <em>y</em>, then <em>∂µ/∂y</em> = 0 and we can solve for <em>µ</em> :

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 <em>F(x, y)</em> = <em>C</em>. 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 <em>x</em> :

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

Differentiate both sides of this with respect to <em>y</em> :

\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}

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Anybody know how to do this ​
ICE Princess25 [194]

Answer:

y = 2

y = 4

y = 8

y = 2

y = 3

y = 6

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3 years ago
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