Consider the sequence defined recursively by an+1 = ( an − 1 if an ≥ 10 2an if an < 10 )
(a) Let a0 be equal to the last digit in your student number, and compute a1, a2, a3, a4.
(b) Suppose an = 1, and find an+4.
(c) If a0 = 3, does limn→[infinity] an exist?

a) a₁ = 14, a₂ = 13, a₃ = 12, a₄ = 11

b) an+4 = 16

c) Does not exist

Step-by-step explanation:

The last digit in student number is given as 7.

a) a₀ = 7

Since an < 10 we use 2an

Therefore a₁ = a₀₊₁ = 2 × a₀ = 2 × 7 = 14

                 a₂ = a₁₊₁ = a₁ - 1 = 14 -1 = 13

                a₃ = a₂₊₁ = a₂ - 1 = 13 -1 = 12

                a₄ = a₃₊₁ = a₃ - 1 = 12 -1 = 11

b) an = 1, we have an+1 = 2an

Therefore  an+2 = an+1+1 = 2 × 2 = 4

                 an+3 = an+2+1 = 2 × 4 = 8

                 an+4 = an+3+1 = 2 × 8 = 16

Therefore, an+4 = 16

c) If a₀ = 3, therefore a₁ = a₀₊₁ = 2×3 = 6

                                  a₂ = a₁₊₁ = 2×6 = 12

                                 a₃ = a₂₊₁ = a₂₋₁ = 12- 1 =11

                                 a₄ = a₃₊₁ = a₃₋₁ = 11- 1 =10

                                a₅ = a₄₊₁ = a₄₋₁ = 10- 1 = 9

                                a₆ = 2×9 = 18

We can therefore see that limn→[infinity] does not exist.

9 sandwiches per hour

step-by-step explanation:

when we have a rate, we want a 1 in the denominator

27/3 = 9

27 sandwiches/ 3 hours = 9 sandwiches/hour

i suppose it should be (y=2x-5)

step-by-step explanation: