When I leave home to walk to school2And what I think I've seen3Now, what can I say4And I've kept careful track5Just a broken down wagon6When Isay that i saw it on Mulberry St7The story would really be better to hear8A reindeer is better; he's fast and he's fleet9To pull a thing that runs on wheels10Jack or Fred of Joe or NAT-11Say! That makes a story that no one can beat12But he'd look simply grand13If a man sits and listens while hitched on behind14But now what worries me is this...15To guide them through where's traffic's thick-16And he raises his hat as they dash by the stand.17With a roar of its motor an airplane appears18A big magician doing tricks..19For I had a story that NO ONE could beat!20There was so much to tell, I JUST COULDN'T BEGIN!21"Nothing," I said, growing red as a beet
Answer:
22
Step-by-step explanation:
11 x 2 = 22
22 teams played
Answer:
The equation of the line would be y = -3/2x + 9
Step-by-step explanation:
In order to solve this, start by finding the slope of the original line. You can do this by solving for y.
2x - 3y = 12
-3y = -2x + 12
y = 2/3x - 4
Now that we have a slope of 2/3, we know that the perpendicular slope is -3/2 (since perpendicular lines have opposite and reciprocal slopes). We can use this and the new point in point-slope form to find the equation.
y - y1 = m(x - x1)
y - 6 = -3/2(x - 2)
y - 6 = -3/2x + 3
y = -3/2x + 9
Step-by-step explanation:
Total songs = 193
Probability of pop song = 39/193 = 0.202
Given a solution

, we can attempt to find a solution of the form

. We have derivatives



Substituting into the ODE, we get


Setting

, we end up with the linear ODE

Multiplying both sides by

, we have

and noting that
![\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cleft%5Bx%28%5Cln%20x%29%5E2%5Cright%5D%3D%28%5Cln%20x%29%5E2%2B%5Cdfrac%7B2x%5Cln%20x%7Dx%3D%28%5Cln%20x%29%5E2%2B2%5Cln%20x)
we can write the ODE as
![\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cleft%5Bwx%28%5Cln%20x%29%5E2%5Cright%5D%3D0)
Integrating both sides with respect to

, we get


Now solve for

:


So you have

and given that

, the second term in

is already taken into account in the solution set, which means that

, i.e. any constant solution is in the solution set.