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Evgen [1.6K]
3 years ago
8

A sequence is defined by the recursive function f(n+1)=f(n). If f(3)=9, what is f(1)?

Mathematics
2 answers:
vekshin13 years ago
6 0
Put n=2 ,
f(2+1) = f(2) => f(3) = f(2) = 9
now, put n=1,
f(1+1) = f(1) => f(2) = f(1) = 9
shusha [124]3 years ago
5 0

Answer:  The required value of f(1) is 9.

Step-by-step explanation:  Given that a sequence is defined by the following recursive function :

f(n+1)=f(n)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)

and f(3) = 9.

We are to find the value of f(1).

Putting n = 2 in equation (i), we have

f(2+1)=f(2)\\\\\Rightarrow f(3)=f(2).

Since f(3) = 9, so we get

f(2)=9.

Again, putting n = 1 in equation (i), we get

f(1+1)=f(1)\\\\\Rightarrow f(2)=f(1).

Since f(2) = 9, so we arrive at

f(1)=9.

Thus, the required value of f(1) is 9.

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Answer:

Quarters = q = 200

Dimes = d = 756

Step-by-step explanation:

Let

Dimes = d

Quarter = q

Value of each Dime = 0.10

Value of each Quarter = 0.25

d + q = 956 (1)

0.1d + 0.25q = $125.60 (2)

From (1)

d = 956 - q

Substitute d = 956 - q into (2)

0.1d + 0.25q = $125.60

0.1(956 - q) + 0.25q = $125.60

95.6 - 0.1q + 0.25q = $125.60

- 0.1q + 0.25q = 125.60 - 95.6

0.15q = 30

Divide both sides by 0.15

q = 30 / 0.15

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q = 200

Substitute q = 200 into (1)

d + q = 956

d + 200 = 956

d = 956 - 200

= 756

d = 756

Therefore,

Quarters = q = 200

Dimes = d = 756

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