Answer: i'm pretty sure Meg would have skated 1/2 of a mile in an hour
The left side is the derivative of
:
![\left((y')^2\right)'=2y'y''](https://tex.z-dn.net/?f=%5Cleft%28%28y%27%29%5E2%5Cright%29%27%3D2y%27y%27%27)
So we can integrate both sides of
![\left((y')^2\right)'=1\implies (y')^2=x+C\implies y'=\pm\sqrt{x+C}](https://tex.z-dn.net/?f=%5Cleft%28%28y%27%29%5E2%5Cright%29%27%3D1%5Cimplies%20%28y%27%29%5E2%3Dx%2BC%5Cimplies%20y%27%3D%5Cpm%5Csqrt%7Bx%2BC%7D)
Then integrate again to solve for
:
![y=\pm\dfrac23(x+C_1)^{3/2}+C_2](https://tex.z-dn.net/?f=y%3D%5Cpm%5Cdfrac23%28x%2BC_1%29%5E%7B3%2F2%7D%2BC_2)
With the given initial conditions, we find
![y(0)=2\implies 2=\pm\dfrac23{C_1}^{3/2}+C_2](https://tex.z-dn.net/?f=y%280%29%3D2%5Cimplies%202%3D%5Cpm%5Cdfrac23%7BC_1%7D%5E%7B3%2F2%7D%2BC_2)
![y'(0)=1\implies 1=\pm\sqrt{C_1}](https://tex.z-dn.net/?f=y%27%280%29%3D1%5Cimplies%201%3D%5Cpm%5Csqrt%7BC_1%7D)
The second equation says
is either 1 or -1, but in the latter case, we would get
in the first equation, which is undefined over the real numbers, so
.
So there are two candidate solutions,
![y_1=\dfrac23(x+1)^{3/2}+\dfrac43](https://tex.z-dn.net/?f=y_1%3D%5Cdfrac23%28x%2B1%29%5E%7B3%2F2%7D%2B%5Cdfrac43)
![y_2=-\dfrac23(x+1)^{3/2}+\dfrac83](https://tex.z-dn.net/?f=y_2%3D-%5Cdfrac23%28x%2B1%29%5E%7B3%2F2%7D%2B%5Cdfrac83)
However, the second equation doesn't satisfy the initial value of the derivative, since
. So the solution is
![\boxed{y(x)=\dfrac23(x+1)^{3/2}+\dfrac43}](https://tex.z-dn.net/?f=%5Cboxed%7By%28x%29%3D%5Cdfrac23%28x%2B1%29%5E%7B3%2F2%7D%2B%5Cdfrac43%7D)
Answer:
130
Step-by-step explanation:
i think it's 130 cuz we have to multiply 10 by 13 yard and we get 130 for that.
Answer:
the image is not clear dear, please write another question with a clear image or type it