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zepelin [54]
3 years ago
8

The students in Mr. Sanchez's class are converting distances measured in miles to kilometers. To estimate the number of kilomete

rs, Abby takes the number of miles, doubles it, then subtracts 20% of the result. Renato first divides the number of miles by 5, then multiplies the result by 8. Write an algebraic expression that shows the process that both students used. Abby's expression: Renato's expression:
Mathematics
1 answer:
BaLLatris [955]3 years ago
6 0

Step-by-step explanation:

Abby’s Expression:

Double m, giving 2m. She then takes 20% of the result, which we can write 0.2(2m). Finally she subtracts this from 2m, giving 2m−(0.2)2m

2m − (0.2)2m

Renato’s Expression:

Divide m by 5, giving m ÷ 5 = m/5, and then multiplies the result by 8, giving:

8(m/5)

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(x^2y+e^x)dx-x^2dy=0
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It looks like the differential equation is

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Check for exactness:

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

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*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

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

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\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

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

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

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Then the general solution to the DE is

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

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