Question

Guys please help me with this question

Answer 1

Answer: Can I get brainliest and 5 stars? thank you.

a) $\frac{65}{99}$

b) $\frac{59}{90}$

c) $\frac{13}{20}$

Step-by-step explanation:

a) $0.65$ I can't put line above it, but the line expresses that 6 and 5 and repeating infinitely. Let X equal the decimal number.

$x=0.65$

First multiply by 1 followed by as many zeros as repeating numbers there are. In this case since 6 and 5 are the repeating numbers, we have to add 2 zeros which is basically multiplying by 100

0.65*100=65.65 (another 65 remains on the right because they're infinite.)

Now substract from the original number. Now we have a new equation that says: $100x=65.65$.

Substract equation 1 from equation 2.

$100x=65.65\\-x=0.65$

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$99x=65$

Solve just like any other equation. Divide by 99 to isolate x.

$\frac{99x}{99}=\frac{65}{99}$

$x=\frac{65}{99}$ This is the fraction.

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b) $0.65^-$This one is a little bit different because we only have 1 repeating number.

In this case we have to multiply by 10.

$0.65^-*10=6.55^-$

Equation 1: $x=0.65^-$

Equation 2: $10x=6.55^-$

Substract.

$10x=6.55^-\\-x=0.65^-$

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$9x=5.9$

divide by 9 to isolate x

$\frac{9x}{9}=\frac{5.9}{9}$

$x=\frac{5.9}{9}$

To get rid of the decimal in the numerator, multiply both numerator and denominator by 10 so that we don't change the fraction.

$x=(\frac{5.9*10}{9*10} )$

$x=\frac{59}{90}$ and this is your fraction.

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c) 0.65 this one has no repeating number. It's a finite decimal number.

I assume you already know that; $a=\frac{a}{1}$, so, we can do the same here to have a fraction.

$0.65=\frac{0.65}{1}$

Now, in order to get rid of the decimal, multiply by 1 followed by as many zeros as decimal numbers are. In this case 2 zeros because there are 2 decimal numbers, so it's 100. Remember that if you do it in the numerator, you have to do it in the denominator as well to not change the fraction.

$\frac{0.65*100}{1*100} =\frac{65}{100}$

Simplify if possible.

$\frac{65/5}{100/5} =\frac{13}{20}$ This is your fraction.