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ANEK [815]
3 years ago
15

Using f (x) = 4x + 3 and g(x) = x - 2, f(g(5))=

Mathematics
1 answer:
il63 [147K]3 years ago
5 0

Answer:

I think It is 7

Step-by-step explanation:

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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 <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}

5 0
3 years ago
Find an equation of the tangent line to the curve at the given point. y = x + tan(x) at (pi, pi)
S_A_V [24]
<span>Differentiating :

y' = 1 + sec^2(x),
</span><span>
cosπ=−1</span><span>

plug in pi = -1 ,for x:

1 + sec^2(pi) = 1 +  (-1)^2 
                      = 2</span>
8 0
4 years ago
Select the correct answer.
alekssr [168]

Using scientific notation, the correct statement is given by:

C. Corporation C owns the most land. Corporation A owns 8 times more land than corporation B.

<h3>What is scientific notation?</h3>

A number in scientific notation is given by:

a \times 10^b

With the base being a \in [1, 10).

Considering the amounts in scientific notation, the standard amounts in acres of each corporation are given as follows:

  • Corporation A: 3.2 \times 10^5 = 320000 = 320,000.
  • Corporation B: 4 \times 10^4 = 40000 = 40,000.
  • Corporation C: 4.3 \times 10^5 = 430000 = 430,000.

Hence corporation C owns the most land. 320,000/40,000 = 8, hence Corporation A owns 8 times more land than corporation B, and statement C is correct.

More can be learned about scientific notation at brainly.com/question/16394306

#SPJ1

3 0
2 years ago
Pls open first to answer correctly get brainlist
KengaRu [80]

Answer:

C AND D

Step-by-step explanation:

it is a rational and real number

5 0
3 years ago
Read 2 more answers
What times 1/10 equals 0.047
Zina [86]

Answer:

.47

Step-by-step explanation:

1/10= .1

0.0.47/.1=.47

-1 x .47=0.47

5 0
3 years ago
Read 2 more answers
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