Which recursive formula can be used to generate the sequence below, where f(1) = 3 and n ≥ 1? 3, –6, 12, –24, 48, … f (n + 1) = –3 f(n ) f (n + 1) = 3 f(n ) f (n + 1) = –2 f(n ) f (n + 1) = 2 f(n)

Which recursive formula can be used to generate the sequence below, where f(1) = 3 and n ≥ 1?

3, –6, 12, –24, 48

The recursive formula for this sequence is

f (n + 1) = –2 f(n)

at n=1 f(n)= 3

at n = 2

f(2) = -2 (3) = -6

n = 3

f(3) = -2 (-6) = 12 and so on

5x+40

5(x + 8)

7x - 49

7(x - 7)

18x + 9

9(2x + 1)

15x - 10

5(3x - 2)

step-by-step explanation:

for number one all you have to do is factor 5 out of 5x + 40

for number 2 factor 7 out of 7x - 49

for number three factor 9 out of 18x + 9

for number four factor 5 out of 15x - 10

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.