In Exercises 45–50, express each repeating decimal as a fraction ...

Question

In Exercises 45–50, express each repeating decimal as a fraction in lowest terms.

0.47 ̅ (repeating 47) = 47/100 + 47/10,000 + 47/1,000,000 + ...

Accepted Answer

Hello. Everyone in this video. We're going to be looking at this practice problem where we want to rewrite and simplify the reoccurring decimal 0.39 as a single fraction. So you can see that this decimal is written out with the fractions associated with it. So 0.39 repeating is equal to 39 divided by 100 plus 39 divided by 10,000 plus divided by one million. And between these terms you can see that we're adding two Zeros each time which means that there's some sort of ratio between these terms. And this gives me a hint that I'm going to be working with a geometric sequence. So the first thing I want to do when I'm working with geometric sequences is to try to determine the common ratio, Which I'll denote as our so to find our I can take my second term 39 divided by 10, And I'll divide the second term by the first term which is divided by 100. I'm gonna change this division to multiplication by switching the numerator and denominator. So my second fraction is now 100 over 39. And here you can see that the 30 nine's cancel out to one and I can Get rid of two Zeros on the denominator of my first fraction and two zeroes in the numerator of my second fraction. So that leaves me with R is equal to one over 100. And what this means is that when I multiply the first term 39 divided by 100 By one over 100 I get my second term which is over 10,000 and the same. If I multiply my second term by one over 100 I get my third term, 39/1 million. So given that it's a geometric sequence and the common ratio is less than one but greater than zero, I can go ahead and find the infinite sum of this sequence to write it as a fraction. So the infinite sum for a geometric sequence Is a one divided by 1 -2. So in this case a one is the first term of the sequence. And are is that common ratio that we determined? So I can plug in values for this. So the infinite sum is equal to the first term 39 divided by 100 divided by 1 -2. Where R is equal to one over 100. And I'm actually going to rewrite this one As over 100. To make it easier to simplify. So if I simplify that I now have 39 over 100 divided by 100 -1 is 99 divided by 100. And once again I'm going to change this Division to multiplication by switching the numerator and denominator of my second fraction. So my second fraction is now 100 Over in 99. And from here you can see that these 100's cancel out to be one. So I'm left that with the infinite sum is equal to over 99 Which means that the repeating decimal 0.39 Is equal to Over 99. If I simplify this fraction by dividing by three, I'm left with 13/33, so this becomes my final answer for this problem, and you can see that this aligns with answer choice C. So hopefully this video helped and see you next time.

In Exercises 45–50, express each repeating decimal as a fraction in lowest terms. 0.47 ̅ (repeating 47) = 47/100 + 47/10,000 + 47/1,000,000 + ...

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