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3 years ago

10

Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

2 answers:

Goshia [24]3 years ago

7 0

Answer:

is in the photo

Step-by-step explanation:

hope this helps

Studentka2010 [4]3 years ago

4 0

Answer:

This is the correct graph but can it fit on your graph?

Step-by-step explanation:

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How do you write the logarithmic equation in exponential form when you use ln?

goldfiish [28.3K]

E^0.693 = 1/2 ln = log e

5 0

3 years ago

Complete the equation that involves multiplication contains the variable x, and has a solution of 7.

Wittaler [7]

Answer:

56 divided by 8

Step-by-step explanation:

7 times 8 equals 56

therefore<u> 56 divided by 8 equals 7 </u><u>ANSWER</u>

and 56 divided by 7 is 8

Fact families

5 0

2 years ago

Write is the equation of the line with a slope of m=-4/3 that passes through the point (12, - 4)

Anika [276]

Answer:

4x +3y=36

Step-by-step explanation:

Equation of line is (y-y1) = m(x-x1)

(y+4) = (-4/3)(x-12)

3(y+4) = (-4)(x-12)

3y +12 = -4x + 48

4x+3y = 36

4 0

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 congruents

MakcuM [25]

The same shape and size. Two shapes are congruent when you can Turn, Flip and/or Slide one so it fits exactly on the other. ... Angles are congruentwhen they are the same size (in degrees or radians). Sides arecongruent when they are the same length. Congruent.

5 0

3 years ago

Read 2 more answers