What is the product of the numerator and the denominator when $0.\overline{09}$ is expressed as a fraction in lowest terms?
I'm not sure how to explain this, but when you have a line, the fraction denominator is the number of digits and 9 amount of that.
So under this line, there are 2 digits, so the denominator is 2 nines, 99.
The numberator is the number under the line, 9.
So our fraction is 9/99 = 1/11.
That would make the product 11.
Sorry for the sad explanation.
I hope this helped. :))
=^._.^=
There is an algebraic way through manipulation:
Let $x=0.\overline{09}$
Multiply by 100 to get $100x=9.\overline{09}$
Subtract x from 100x to get $100x-x=9.\overline{09}-0.{09}$ which cancels out the infinite decimals leaving $99x=9$
Dividing both sides by 99 gives $x=\frac{9}{99}$, which simplifies to $x=\frac{1}{11}$. The product of the numerator and denominator is $1*11=\boxed{11}$
If you are familiar with geometric series there is also another way:
x=0*10^-1+9*10^-2+0*10^-3+9*10^-4+0*10^-5+...
simplifying gives x=9*10^-2+9*10^-4+9*10^-6+... (you got lucky the 0's canceled out, there is a trick to dealing with them when there is no 0's to cancel out)
We can see this is an infinite geometric series with first term 9*10^-2 and common ration 10^-2. Using the formula for the sum of an infinite geometric series gives:
$S=\frac{9*10^-2}{1-10^-2}$
$S=\frac{\frac{9}{100}}{\frac{99}{100}}$
$S=\frac{9}{99}$
$S=\frac{1}{11}$
The product is 11 :) These are the two algebraic solutions