![y'=(t+y)^2-1](https://tex.z-dn.net/?f=y%27%3D%28t%2By%29%5E2-1)
Substitute
, so that
, and
![u'=u^2-1](https://tex.z-dn.net/?f=u%27%3Du%5E2-1)
which is separable as
![\dfrac{u'}{u^2-1}=1](https://tex.z-dn.net/?f=%5Cdfrac%7Bu%27%7D%7Bu%5E2-1%7D%3D1)
Integrate both sides with respect to
. For the integral on the left, first split into partial fractions:
![\dfrac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)=1](https://tex.z-dn.net/?f=%5Cdfrac%7Bu%27%7D2%5Cleft%28%5Cfrac1%7Bu-1%7D-%5Cfrac1%7Bu%2B1%7D%5Cright%29%3D1)
![\displaystyle\int\frac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)\,\mathrm dt=\int\mathrm dt](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint%5Cfrac%7Bu%27%7D2%5Cleft%28%5Cfrac1%7Bu-1%7D-%5Cfrac1%7Bu%2B1%7D%5Cright%29%5C%2C%5Cmathrm%20dt%3D%5Cint%5Cmathrm%20dt)
![\dfrac12(\ln|u-1|-\ln|u+1|)=t+C](https://tex.z-dn.net/?f=%5Cdfrac12%28%5Cln%7Cu-1%7C-%5Cln%7Cu%2B1%7C%29%3Dt%2BC)
Solve for
:
![\dfrac12\ln\left|\dfrac{u-1}{u+1}\right|=t+C](https://tex.z-dn.net/?f=%5Cdfrac12%5Cln%5Cleft%7C%5Cdfrac%7Bu-1%7D%7Bu%2B1%7D%5Cright%7C%3Dt%2BC)
![\ln\left|1-\dfrac2{u+1}\right|=2t+C](https://tex.z-dn.net/?f=%5Cln%5Cleft%7C1-%5Cdfrac2%7Bu%2B1%7D%5Cright%7C%3D2t%2BC)
![1-\dfrac2{u+1}=e^{2t+C}=Ce^{2t}](https://tex.z-dn.net/?f=1-%5Cdfrac2%7Bu%2B1%7D%3De%5E%7B2t%2BC%7D%3DCe%5E%7B2t%7D)
![\dfrac2{u+1}=1-Ce^{2t}](https://tex.z-dn.net/?f=%5Cdfrac2%7Bu%2B1%7D%3D1-Ce%5E%7B2t%7D)
![\dfrac{u+1}2=\dfrac1{1-Ce^{2t}}](https://tex.z-dn.net/?f=%5Cdfrac%7Bu%2B1%7D2%3D%5Cdfrac1%7B1-Ce%5E%7B2t%7D%7D)
![u=\dfrac2{1-Ce^{2t}}-1](https://tex.z-dn.net/?f=u%3D%5Cdfrac2%7B1-Ce%5E%7B2t%7D%7D-1)
Replace
and solve for
:
![t+y=\dfrac2{1-Ce^{2t}}-1](https://tex.z-dn.net/?f=t%2By%3D%5Cdfrac2%7B1-Ce%5E%7B2t%7D%7D-1)
![y=\dfrac2{1-Ce^{2t}}-1-t](https://tex.z-dn.net/?f=y%3D%5Cdfrac2%7B1-Ce%5E%7B2t%7D%7D-1-t)
Now use the given initial condition to solve for
:
![y(3)=4\implies4=\dfrac2{1-Ce^6}-1-3\implies C=\dfrac3{4e^6}](https://tex.z-dn.net/?f=y%283%29%3D4%5Cimplies4%3D%5Cdfrac2%7B1-Ce%5E6%7D-1-3%5Cimplies%20C%3D%5Cdfrac3%7B4e%5E6%7D)
so that the particular solution is
![y=\dfrac2{1-\frac34e^{2t-6}}-1-t=\boxed{\dfrac8{4-3e^{2t-6}}-1-t}](https://tex.z-dn.net/?f=y%3D%5Cdfrac2%7B1-%5Cfrac34e%5E%7B2t-6%7D%7D-1-t%3D%5Cboxed%7B%5Cdfrac8%7B4-3e%5E%7B2t-6%7D%7D-1-t%7D)
Answer:
40 student 110 adult
this is using desmos
but to smash the equation you can go like basically we have to cancel out each variable to solve for the other
3x + 8y = 1000
x + y < =150
-3(x + y <= 150)
-3x -3y = -450
3x + 8y = 1000
5y = 550
y= 110
3x + 8 y = 1000
-8(x+y<=150) = -8x - 8y <= -1200
-5x = -200
![\frac{ - 5x}{ - 5 } = x](https://tex.z-dn.net/?f=%20%5Cfrac%7B%20-%205x%7D%7B%20-%205%20%7D%20%20%20%3D%20x)
![\frac{ - 200}{ - 5} = 40](https://tex.z-dn.net/?f=%20%5Cfrac%7B%20-%20200%7D%7B%20-%205%7D%20%20%3D%2040)
x= 40
40 kids 110 adults
Answer: Yes
Step-by-step explanation:
1 ton = 2000 pounds and 500 is 1/4 of 2000
Approx - 10%
72 students signed up for canoeing + the 23 students who signed up for trekking + the 13 students who signed up for both = 108 students
there are 120 students total, so you have to subtract 120 - 108 = 12
Now to find the percentage, you have to divide the number of students who didn't sign up for either, by the total
which is .1 = 10/100 = 10%
5s-100-2s=4s-20-2s
3s-100. = 2s-20
+20. +20
3s-80. = 2s
-3s. -3s
-80. = -1s
-1 -1
80. = s
s=80