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:
2
Step-by-step explanation:
Start by identifying any factor of 50 and then rewriting 50 to include that one factor and another:
50 = 5 * 10 5 is prime, but 10 is not. Thus, factor 10:
50 = 5 * 5 * 2 This is 50 as a product of its prime factors 5, 5 and 2.
Lets write down the information that we know.
Jessie has 7 stacks of 6 dimes, so Jessie has a total of 6*7 dimes
Dimes=7*6=42dimes
Jessie has 42 dimes, but then she found another one so lets add one more dime.
42+1=43 dimes
Jessie has 8 stacks of 5 nickels, so Jessie has 8*5 nickels
Nickels=8*5=40nickels
Jessie has 40 nickels, but then she found another one so lets add 1 to the total number of nickels.
40+1=41nickels
Now we can answer the question.
How many dimes and nickels does jessie have in all?
nickels+dimes=
41+43=84 coins
Jessie has 84 coins in total
