Answer:
Therefore the value of y(1)= 0.9152.
Step-by-step explanation:
According to the Euler's method
y(x+h)≈ y(x) + hy'(x) ....(1)
Given that y(0) =3 and step size (h) = 0.2.
Putting the value of y'(x) in equation (1)
Substituting x =0 and h= 0.2
![y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]](https://tex.z-dn.net/?f=y%280%2B0.2%29%5Capprox%20y%280%29%2B0.2%5B0%5Ctimes%20y%280%29-%5Cfrac12%20%28y%280%29%29%5E2%5D)
![\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]](https://tex.z-dn.net/?f=%5CRightarrow%20y%280.2%29%5Capprox%203%2B0.2%5B-%5Cfrac12%20%5Ctimes3%5D)
[∵ y(0) =3 ]
Substituting x =0.2 and h= 0.2
![y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]](https://tex.z-dn.net/?f=y%280.2%2B0.2%29%5Capprox%20y%280.2%29%2B0.2%5B%280.2%29%5E2%5Ctimes%20y%280.2%29-%5Cfrac12%20%28y%280.2%29%29%5E2%5D)
![\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]](https://tex.z-dn.net/?f=%5CRightarrow%20y%280.4%29%5Capprox%20%202.7%2B0.2%5B%280.2%29%5E2%5Ctimes%202.7-%20%5Cfrac12%282.7%29%5E2%5D)
Substituting x =0.4 and h= 0.2
![y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]](https://tex.z-dn.net/?f=y%280.4%2B0.2%29%5Capprox%20y%280.4%29%2B0.2%5B%280.4%29%5E2%5Ctimes%20y%280.4%29-%5Cfrac12%20%28y%280.4%29%29%5E2%5D)
![\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]](https://tex.z-dn.net/?f=%5CRightarrow%20y%280.6%29%5Capprox%20%201.9926%2B0.2%5B%280.4%29%5E2%5Ctimes%201.9926-%20%5Cfrac12%281.9926%29%5E2%5D)
Substituting x =0.6 and h= 0.2
![y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]](https://tex.z-dn.net/?f=y%280.6%2B0.2%29%5Capprox%20y%280.6%29%2B0.2%5B%280.6%29%5E2%5Ctimes%20y%280.6%29-%5Cfrac12%20%28y%280.6%29%29%5E2%5D)
![\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]](https://tex.z-dn.net/?f=%5CRightarrow%20y%280.8%29%5Capprox%20%201.6593%2B0.2%5B%280.6%29%5E2%5Ctimes%201.6593-%20%5Cfrac12%281.6593%29%5E2%5D)
Substituting x =0.8 and h= 0.2
![y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]](https://tex.z-dn.net/?f=y%280.8%2B0.2%29%5Capprox%20y%280.8%29%2B0.2%5B%280.8%29%5E2%5Ctimes%20y%280.8%29-%5Cfrac12%20%28y%280.8%29%29%5E2%5D)
![\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]](https://tex.z-dn.net/?f=%5CRightarrow%20y%281.0%29%5Capprox%20%200.8800%2B0.2%5B%280.8%29%5E2%5Ctimes%200.8800-%20%5Cfrac12%280.8800%29%5E2%5D)
Therefore the value of y(1)= 0.9152.
Answer: 36 degrees
Step-by-step explanation:
subtract 5 from 30=25
add 11 to 25
25+11=36
Answer: Hello mate!
We know that you have a graph of y vs t, and we know that this is an "altitude" graph.
Then we can assume that the y represents the altitude, and t represents the time, and this graph shows the altitude of something as a function of the time.
the t-intercept means that the graph passes through the y-axis, this means that, in this point, y is equal to zero:
Then, at the t-intercept, we have y = 0, which means that at this time (where is the intersection) the altitude is equal to zero.
The y-intercept means that the graph passes through the y-axis, where t = 0
this is the initial value of the altitude, where t = 0 usually denotes the time where we start to measure.
