Is 0.666666 irracional?

1 answer:

4 0

While it may seem irrational because the decimal does not terminate. It is actually rational because repeating .666 can be written as a fraction which is 2/3. And any number written as a fraction is rational number.

You might be interested in

Question 2 of 5

AlekseyPX

Given:

The different recursive formulae.

To find:

The explicit formulae for the given recursive formulae.

Solution:

The recursive formula of an arithmetic sequence is $f(n)=f(n-1)+d, f(1)=a,n\geq 2$ and the explicit formula is $f(n)=a+(n-1)d$ , where a is the first term and d is the common difference.

The recursive formula of a geometric sequence is $f(n)=rf(n-1), f(1)=a,n\geq 2$ and the explicit formula is $f(n)=ar^{n-1}$ , where a is the first term and r is the common ratio.

The first recursive formula is:

$f(1)=5$

$f(n)=f(n-1)+5$ for $n\geq 2$ .

It is the recursive formula of an arithmetic sequence with first term 5 and common difference 5. So, the explicit formula for this recursive formula is:

$f(n)=5+(n-1)(5)$

$f(n)=5+5(n-1)$

Therefore, the correct option is A, i.e., $f(n)=5+5(n-1)$ .

The second recursive formula is:

$f(1)=5$

$f(n)=3f(n-1)$ for $n\geq 2$ .

It is the recursive formula of a geometric sequence with first term 5 and common ratio 3. So, the explicit formula for this recursive formula is:

$f(n)=5(3)^{n-1}$

Therefore, the correct option is F, i.e., $f(n)=5(3)^{n-1}$ .

The third recursive formula is:

$f(1)=5$

$f(n)=f(n-1)+3$ for $n\geq 2$ .

It is the recursive formula of an arithmetic sequence with first term 5 and common difference 3. So, the explicit formula for this recursive formula is:

$f(n)=5+(n-1)(3)$