The answer is 39, jk=18 because j multiplied by k is 18 and then i added 3 and i got 21, then k (6) multiplied by 3 is 18, so then i added 21 and 18 and i got 39
That's impossible to say. The radius of a square is meaningless.
Answer:
The system of the equations has no solution; the two lines are parallel.
Step-by-step explanation:
The equations have the same slope. In standard form, this can be seen as the same coefficients for x and y. We will multiply the second equation by -12 to reveal this.

This means the equations are parallel and will never cross. There is no solution.




<h3>The answers are :</h3><h3>x = 6</h3><h3>y = 11</h3><h3>s = 9</h3><h3>n = 8.5</h3><h3>Hope it helps!</h3><h3 />
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)
[∵ 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.