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Sholpan [36]
2 years ago
5

For the last three days, it has rained 0.13 inches, 0.08 inches, and 0.18 inches. What is the best estimate of the total rainfal

l?
A. 0.2 inch.

B. 0.5 inch.

C. 0.3 inch.

D. 0.4 inch.
Mathematics
1 answer:
Natali [406]2 years ago
6 0

Answer:

(D) . The best estimate of the total rainfall is 0.4 inches.

Step-by-step explanation:

Here, as given in the data:

The rainfall on the first day = 0.13 inches

Now, the 'nearest tenth' estimation of 0.13  = 0.10 as ( 3 < 5)

So, the estimated rainfall on first day = 0.10 inches

The rainfall on the second  day = 0.08 inches

Now, the 'nearest tenth' estimation of 0.08  = 0.10 as ( 8 >  5)

So, the estimated rainfall on second day = 0.10 inches

The rainfall on the third  day = 0.18 inches

Now, the 'nearest tenth' estimation of  0.18  = 0.20 as ( 8 >  5)

So, the estimated rainfall on third day =0.20 inches

So, total estimated rainfall = Estimated rainfall on ( first + second + third day)

= 0.10 inches  + 0.10 inches + 0.20 inches  

= 0.40 inches

Hence,  the best estimate of the total rainfall is 0.4 inches.

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2 years ago
(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

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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