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Sagot :
Let's find the fractional form of the repeating decimal [tex]\(0 . \overline{8}\)[/tex].
1. First, let's denote the repeating decimal by [tex]\(x\)[/tex]:
[tex]\[ x = 0.8888 \ldots \][/tex]
2. Now, multiply both sides of this equation by 10 to shift the decimal point one place to the right:
[tex]\[ 10x = 8.8888 \ldots \][/tex]
3. We now have two equations:
[tex]\[ \begin{aligned} x &= 0.8888 \ldots \quad \text{(Equation 1)} \\ 10x &= 8.8888 \ldots \quad \text{(Equation 2)} \end{aligned} \][/tex]
4. Next, subtract Equation 1 from Equation 2 to eliminate the repeating part:
[tex]\[ \begin{aligned} 10x - x &= 8.8888 \ldots - 0.8888 \ldots \\ 9x &= 8 \end{aligned} \][/tex]
5. Solve for [tex]\(x\)[/tex] by dividing both sides of the equation by 9:
[tex]\[ x = \frac{8}{9} \][/tex]
Therefore, the fractional form of [tex]\(0 . \overline{8}\)[/tex] is [tex]\(\frac{8}{9}\)[/tex].
So, the correct answer is:
C. [tex]\(\frac{8}{9}\)[/tex]
1. First, let's denote the repeating decimal by [tex]\(x\)[/tex]:
[tex]\[ x = 0.8888 \ldots \][/tex]
2. Now, multiply both sides of this equation by 10 to shift the decimal point one place to the right:
[tex]\[ 10x = 8.8888 \ldots \][/tex]
3. We now have two equations:
[tex]\[ \begin{aligned} x &= 0.8888 \ldots \quad \text{(Equation 1)} \\ 10x &= 8.8888 \ldots \quad \text{(Equation 2)} \end{aligned} \][/tex]
4. Next, subtract Equation 1 from Equation 2 to eliminate the repeating part:
[tex]\[ \begin{aligned} 10x - x &= 8.8888 \ldots - 0.8888 \ldots \\ 9x &= 8 \end{aligned} \][/tex]
5. Solve for [tex]\(x\)[/tex] by dividing both sides of the equation by 9:
[tex]\[ x = \frac{8}{9} \][/tex]
Therefore, the fractional form of [tex]\(0 . \overline{8}\)[/tex] is [tex]\(\frac{8}{9}\)[/tex].
So, the correct answer is:
C. [tex]\(\frac{8}{9}\)[/tex]
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