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
6.16441 which rounds to 6.2 in the tenths place
Answer:
1.No, 143 is not a prime number. The list of all positive divisors the list of all integers that divide 143 is as follows: 1, 11, 13, 143. To be 143 a prime number, it would have been required that 143 has only two divisors, itself and 1.
2.Since the polynomial can be factored, it is not prime.
Answer:
#1 i think is 2x7 i dont know #2 sry
Step-by-step explanation:
