Answer:
C. y₂ = (1 + (t/n))²
Step-by-step explanation:
yₙ₊₁ = yₙ + Δt F(tₙ, yₙ)
yₙ₊₁ = yₙ + Δt yₙ
yₙ₊₁ = yₙ + (t/n) yₙ
When n=0:
y₁ = y₀ + (t/n) y₀
y₁ = 1 + (t/n)
When n=1:
y₂ = y₁ + (t/n) y₁
y₂ = 1 + (t/n) + (t/n) (1 + (t/n))
y₂ = 1 + (t/n) + (t/n) + (t/n)²
y₂ = 1 + 2(t/n) + (t/n)²
y₂ = (1 + (t/n))²
Answer:
Provide question
Step-by-step explanation:
Provide question
Answer: c.+dhf+2-4= 6)
Step-by-step explanation: Hope this helps!
Answer:
D. (25π - 50) sq m
Step-by-step explanation:
Since ABCD is a square with the area of 100, we know that the dimensions are 10 by 10. With that, we can find that triangle ABC = 50 sq m because of (h*l)/2. Since A is the center of the circle that contains arc BD, we know that 10 is the radius of the quarter-circle ABC. Using (πr^2)/4 to get the area of the quarter-circle, we get that the quarter-circle ABC = 25π sq m. But the original triangle ABC in the quarter-circle is not shaded. Therefore, we subtract 50 from 25π, which leads to (25π - 50) sq m.
I hope this helped! :D
