The solution for given linear equation systems is (3,-4).
Option: C
<u>Step-by-step explanation:</u>
The given equations are 5x+3y=3 and x+y= -1.
Let us isolate x value from equation 5x+3y=3.
5x = 3-3y.
.
Substitute this x value in equation x+y= -1.
.
After taking LCM for the denominator,
.
3-3y+5y=-5.
2y=-8.
y=-4.
For substitute the y value.
.
.
.
x=3.
∴ The solution is (3,-4).
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:
.3636363636363636363636363636363636363...so on so forth.
Step-by-step explanation:
repeating
