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3 years ago

5

Which of the following statements best describes the effect of replacing the graph of f(x) with the graph of f(x) − 3?

1 answer:

exis [7]3 years ago

5 0

Answer:

The graph shifts 3 units down.

Step-by-step explanation:

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Plz help me asap worth 20 points

Ainat [17]

Answer:

OK so the equation has a exponent so if there is a exponent then you need to multiply. since 227 is the Area and the exponent is 2.

Step-by-step explanation:

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3 years ago

When Elyse moved last weekend, the circular glass top for an end table broke. To replace it, she bought a circular glass top wit

spin [16.1K]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Rounding whole number 99.59

postnew [5]

Answer:

100

Step-by-step explanation:

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3 years ago

Find a second solution y2(x) of<br> x^2y"-3xy'+5y=0; y1=x^2cos(lnx)

rosijanka [135]

We can try reduction order and look for a solution $y_2(x)=y_1(x)v(x)$ . Then

$y_2=y_1v\implies{y_2}'=y_1v'+{y_1}'v\implies{y_2}''=y_1v''+2{y_1}'v+{y_1}''v$

Substituting these into the ODE gives

$x^2(y_1v''+2{y_1}'v+{y_1}''v)-3x(y_1v'+{y_1}'v)+5y_1v=0$

$x^2y_1v''+(2x^2{y_1}'-3xy_1)v'+(x^2{y_1}''-3x{y_1}'+5y_1)v=0$

$x^4\cos(\ln x)v''+x^3(\cos(\ln x)-2\sin(\ln x))v'=0$

which leaves us with an ODE linear in $w(x)=v'(x)$ :

$x^4\cos(\ln x)w'+x^3(\cos(\ln x)-2\sin(\ln x))w=0$

This ODE is separable; divide both sides by the coefficient of $w'(x)$ and separate the variables to get

$w'+\dfrac{\cos(\ln x)-2\sin(\ln x)}{x\cos(\ln x)}w=0$

$\dfrac{w'}w=\dfrac{2\sin(\ln x)-\cos(\ln x)}{x\cos(\ln x)}$

$\dfrac{\mathrm dw}w=\dfrac{2\sin(\ln x)-\cos(\ln x)}{x\cos(\ln x)}\,\mathrm dx$

Integrate both sides; on the right, substitute $u=\ln x$ so that $\mathrm du=\dfrac{\mathrm dx}x$ .

$\ln|w|=\displaystyle\int\frac{2\sin u-\cos u}{\cos u}\,\mathrm du=\int(2\tan u-1)\,\mathrm du$

Now solve for $w(u)$ ,

$\ln|w|=-2\ln(\cos u)-u+C$

$w=e^{-2\ln(\cos u)-u+C}$

$w=Ce^{-u}\sec^2u$

then for $w(x)$ ,

$w=Ce^{-\ln x}\sec^2(-\ln x)$

$w=C\dfrac{\sec^2(\ln x)}x$

Solve for $v(x)$ by integrating both sides.

$v=\displaystyle C_1\int\frac{\sec^2(\ln x)}x\,\mathrm dx$

Substitute $u=\ln x$ again and solve for $v(u)$ :

$v=\displaystyle C_1\int\sec^2u\,\mathrm du$

$v=C_1\tan u+C_2$

then for $v(x)$ ,

$v=C_1\tan(\ln x)+C_2$

So the second solution would be

$y_2=x^2\cos(\ln x)(C_1\tan(\ln x)+C_2)$

$y_2=C_1x^2\sin(\ln x)+C_2x^2\cos(\ln x)$

$y_1(x)$ already accounts for the second term of the solution above, so we end up with

$\boxed{y_2=x^2\sin(\ln x)}$

as the second independent solution.

6 0

4 years ago

What are the solutions of this quadratic equation? x2-6x=-58

weeeeeb [17]

Answer:

$x^{2}$ -6x=-58

Step-by-step explanation:

3 0

3 years ago