Hello its dedpool!
the answer here is the third one!
hope this helps
Answer:
f(0) = 1
Step-by-step explanation:
f(x) indicates the "output" value present at the "input" (Output referring to the y value and input referring to the x value). So f(0) would mean the "y" value present at x = 0; however, there are two circles at x = 0. One of the dots is open(hollow) meaning that the "x" and "y" values present there are excluded. This leaves the closed(filled) dot which means that value is included and is the answer.
I walked past the crab, which was right BY the ocean
Answer:
$1.22 per pair
The error is the contractor didnt take into account that there are 5 dozen gloves not 5 pairs of gloves. He calculated $73.20/5, which gave him $14.64.
Step-by-step explanation:
1 dozen = 12
5 dozen = 12 * 5 = 60
Therefore, you have 60 pairs of gloves.
He purchased all 60 pairs of gloves for $73.20.
73.20/60 = $1.22
This is how much one pair of gloves cost.
For the ODE
![ty'+2y=\sin t](https://tex.z-dn.net/?f=ty%27%2B2y%3D%5Csin%20t)
multiply both sides by <em>t</em> so that the left side can be condensed into the derivative of a product:
![t^2y'+2ty=t\sin t](https://tex.z-dn.net/?f=t%5E2y%27%2B2ty%3Dt%5Csin%20t)
![\implies(t^2y)'=t\sin t](https://tex.z-dn.net/?f=%5Cimplies%28t%5E2y%29%27%3Dt%5Csin%20t)
Integrate both sides with respect to <em>t</em> :
![t^2y=\displaystyle\int t\sin t\,\mathrm dt=\sin t-t\cos t+C](https://tex.z-dn.net/?f=t%5E2y%3D%5Cdisplaystyle%5Cint%20t%5Csin%20t%5C%2C%5Cmathrm%20dt%3D%5Csin%20t-t%5Ccos%20t%2BC)
Divide both sides by
to solve for <em>y</em> :
![y(t)=\dfrac{\sin t}{t^2}-\dfrac{\cos t}t+\dfrac C{t^2}](https://tex.z-dn.net/?f=y%28t%29%3D%5Cdfrac%7B%5Csin%20t%7D%7Bt%5E2%7D-%5Cdfrac%7B%5Ccos%20t%7Dt%2B%5Cdfrac%20C%7Bt%5E2%7D)
Now use the initial condition to solve for <em>C</em> :
![y\left(\dfrac\pi2\right)=9\implies9=\dfrac{\sin\frac\pi2}{\frac{\pi^2}4}-\dfrac{\cos\frac\pi2}{\frac\pi2}+\dfrac C{\frac{\pi^2}4}](https://tex.z-dn.net/?f=y%5Cleft%28%5Cdfrac%5Cpi2%5Cright%29%3D9%5Cimplies9%3D%5Cdfrac%7B%5Csin%5Cfrac%5Cpi2%7D%7B%5Cfrac%7B%5Cpi%5E2%7D4%7D-%5Cdfrac%7B%5Ccos%5Cfrac%5Cpi2%7D%7B%5Cfrac%5Cpi2%7D%2B%5Cdfrac%20C%7B%5Cfrac%7B%5Cpi%5E2%7D4%7D)
![\implies9=\dfrac4{\pi^2}(1+C)](https://tex.z-dn.net/?f=%5Cimplies9%3D%5Cdfrac4%7B%5Cpi%5E2%7D%281%2BC%29)
![\implies C=\dfrac{9\pi^2}4-1](https://tex.z-dn.net/?f=%5Cimplies%20C%3D%5Cdfrac%7B9%5Cpi%5E2%7D4-1)
So the particular solution to the IVP is
![y(t)=\dfrac{\sin t}{t^2}-\dfrac{\cos t}t+\dfrac{\frac{9\pi^2}4-1}{t^2}](https://tex.z-dn.net/?f=y%28t%29%3D%5Cdfrac%7B%5Csin%20t%7D%7Bt%5E2%7D-%5Cdfrac%7B%5Ccos%20t%7Dt%2B%5Cdfrac%7B%5Cfrac%7B9%5Cpi%5E2%7D4-1%7D%7Bt%5E2%7D)
or
![y(t)=\dfrac{4\sin t-4t\cos t+9\pi^2-4}{4t^2}](https://tex.z-dn.net/?f=y%28t%29%3D%5Cdfrac%7B4%5Csin%20t-4t%5Ccos%20t%2B9%5Cpi%5E2-4%7D%7B4t%5E2%7D)