Answer: 10
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Explanation:
We'll start off converting each mixed number into an improper fraction.
The formula to use is [tex]A \frac{b}{c} = \frac{A*c+b}{c}[/tex]
So,
[tex]A \frac{b}{c} = \frac{A*c+b}{c}\\\\2 \frac{1}{8} = \frac{2*8+1}{8}\\\\2 \frac{1}{8} = \frac{16+1}{8}\\\\2 \frac{1}{8} = \frac{17}{8}\\\\[/tex]
And,
[tex]A \frac{b}{c} = \frac{A*c+b}{c}\\\\2 \frac{2}{3} = \frac{2*3+2}{3}\\\\2 \frac{2}{3} = \frac{6+2}{3}\\\\2 \frac{2}{3} = \frac{8}{3}\\\\[/tex]
So the task of computing [tex]2 \frac{1}{8} \times 2 \frac{2}{3}[/tex] is exactly the same as computing [tex]\frac{17}{8} \times \frac{8}{3}[/tex]
Notice how we have an 8 up top and an 8 down below. Those 8's cancel out and we're left with [tex]\frac{17}{3}[/tex]. That fraction does not reduce any further.
The last step is to convert that improper fraction result to a mixed number.
[tex]\frac{17}{3} = \frac{15+2}{3}\\\\\frac{17}{3} = \frac{15}{3}+\frac{2}{3}\\\\\frac{17}{3} = 5+\frac{2}{3}\\\\\frac{17}{3} = 5 \frac{2}{3}\\\\[/tex]
Or you could note that 17/3 leads to 5 remainder 2. The 5 is the whole part and the 2 forms the numerator of the fractional part 2/3.
The value is in [tex]A \frac{b}{c}[/tex] form where
Therefore, A+b+c = 5+2+3 = 10
