Answer:
He gets 42 visitors 4 weeks after starting to build his website.
He gets 10 new visitors per week.
Step-by-step explanation:
Equation for the number of visitors:
The equation for the number of visitors Timmy's new website receives after t weeks is:
![v(t) = 10t + b](https://tex.z-dn.net/?f=v%28t%29%20%3D%2010t%20%2B%20b)
In which b is the number of visitors rightly after he starts.
Timmy is building a new website. Right after he starts, he has 2 visitors.
This means that
, so:
![v(t) = 10t + 2](https://tex.z-dn.net/?f=v%28t%29%20%3D%2010t%20%2B%202)
How many visitors does he get 4 weeks after starting to build his website?
This is v(4). So
![v(4) = 10(4) + 2 = 40 + 2 = 42](https://tex.z-dn.net/?f=v%284%29%20%3D%2010%284%29%20%2B%202%20%3D%2040%20%2B%202%20%3D%2042)
He gets 42 visitors 4 weeks after starting to build his website.
How any new visitors does he get per week?
After 0 weeks:
![v(0) = 10(0) + 2 = 0 + 2 = 2](https://tex.z-dn.net/?f=v%280%29%20%3D%2010%280%29%20%2B%202%20%3D%200%20%2B%202%20%3D%202)
After 1 week:
![v(1) = 10(1) + 2 = 10 + 2 = 12](https://tex.z-dn.net/?f=v%281%29%20%3D%2010%281%29%20%2B%202%20%3D%2010%20%2B%202%20%3D%2012)
2 weeks:
After 2 week:
![v(2) = 10(2) + 2 = 20 + 2 = 22](https://tex.z-dn.net/?f=v%282%29%20%3D%2010%282%29%20%2B%202%20%3D%2020%20%2B%202%20%3D%2022)
22 - 12 = 12 - 2 = 10
He gets 10 new visitors per week.
Answer:
6.25 Miles
6 1/4
Step-by-step explanation:
3 + 2 = 5 MILES
1/2 you want the denominator to equal 4 so you times it by two
= 2/4
2/4+3/4= 5/4
this gives you one full mile and 1/4
add all together 5+1+1/4
I am picking this is right :)
It would be: 37 × 0.8108 = 30 hours (approx.)
Now, we know, 1 hour = 60 min.
so, 30 hours = 60 × 30 = 1800 min.
In short, Your Answer would be 1800 min
Hope this helps!
Substitute
, so that
![\dfrac{\mathrm dv}{\mathrm dx}=1-\dfrac{\mathrm dy}{\mathrm dx}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20dv%7D%7B%5Cmathrm%20dx%7D%3D1-%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D)
Then the resulting ODE in
is separable, with
![1-\dfrac{\mathrm dv}{\mathrm dx}=v^2\implies\dfrac{\mathrm dv}{1-v^2}=\mathrm dx](https://tex.z-dn.net/?f=1-%5Cdfrac%7B%5Cmathrm%20dv%7D%7B%5Cmathrm%20dx%7D%3Dv%5E2%5Cimplies%5Cdfrac%7B%5Cmathrm%20dv%7D%7B1-v%5E2%7D%3D%5Cmathrm%20dx)
On the left, we can split into partial fractions:
![\dfrac12\left(\dfrac1{1-v}+\dfrac1{1+v}\right)\mathrm dv=\mathrm dx](https://tex.z-dn.net/?f=%5Cdfrac12%5Cleft%28%5Cdfrac1%7B1-v%7D%2B%5Cdfrac1%7B1%2Bv%7D%5Cright%29%5Cmathrm%20dv%3D%5Cmathrm%20dx)
Integrating both sides gives
![\dfrac{\ln|1-v|+\ln|1+v|}2=x+C](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cln%7C1-v%7C%2B%5Cln%7C1%2Bv%7C%7D2%3Dx%2BC)
![\dfrac12\ln|1-v^2|=x+C](https://tex.z-dn.net/?f=%5Cdfrac12%5Cln%7C1-v%5E2%7C%3Dx%2BC)
![1-v^2=e^{2x+C}](https://tex.z-dn.net/?f=1-v%5E2%3De%5E%7B2x%2BC%7D)
![v=\pm\sqrt{1-Ce^{2x}}](https://tex.z-dn.net/?f=v%3D%5Cpm%5Csqrt%7B1-Ce%5E%7B2x%7D%7D)
Now solve for
:
![x-y+1=\pm\sqrt{1-Ce^{2x}}](https://tex.z-dn.net/?f=x-y%2B1%3D%5Cpm%5Csqrt%7B1-Ce%5E%7B2x%7D%7D)
![\boxed{y=x+1\pm\sqrt{1-Ce^{2x}}}](https://tex.z-dn.net/?f=%5Cboxed%7By%3Dx%2B1%5Cpm%5Csqrt%7B1-Ce%5E%7B2x%7D%7D%7D)