Certainly! Let's break down the problem step-by-step:
1. Identify the twine given to the neighbor:
The friend gives the neighbor [tex]\( 13 \frac{2}{3} \)[/tex] feet of twine.
2. Identify the twine remaining on the spool:
After giving the twine to the neighbor, [tex]\( 38 \frac{2}{5} \)[/tex] feet of twine remain on the spool.
3. Convert the mixed numbers to improper fractions for easier calculation:
- [tex]\( 13 \frac{2}{3} \)[/tex] can be converted to an improper fraction:
[tex]\[
13 \frac{2}{3} = 13 + \frac{2}{3} = \frac{39}{3} + \frac{2}{3} = \frac{39 + 2}{3} = \frac{41}{3}
\][/tex]
- [tex]\( 38 \frac{2}{5} \)[/tex] can be converted to an improper fraction:
[tex]\[
38 \frac{2}{5} = 38 + \frac{2}{5} = \frac{190}{5} + \frac{2}{5} = \frac{190 + 2}{5} = \frac{192}{5}
\][/tex]
4. Add the improper fractions to find the total original length of the twine:
We need to add [tex]\( \frac{41}{3} \)[/tex] and [tex]\( \frac{192}{5} \)[/tex].
To do this, we need a common denominator, which is 15 (the least common multiple of 3 and 5).
- Convert [tex]\( \frac{41}{3} \)[/tex] to a fraction with a denominator of 15:
[tex]\[
\frac{41}{3} = \frac{41 \times 5}{3 \times 5} = \frac{205}{15}
\][/tex]
- Convert [tex]\( \frac{192}{5} \)[/tex] to a fraction with a denominator of 15:
[tex]\[
\frac{192}{5} = \frac{192 \times 3}{5 \times 3} = \frac{576}{15}
\][/tex]
5. Add the converted fractions:
[tex]\[
\frac{205}{15} + \frac{576}{15} = \frac{205 + 576}{15} = \frac{781}{15}
\][/tex]
6. Convert the improper fraction back to a mixed number:
Divide 781 by 15 to get:
[tex]\[
781 \div 15 = 52 \text{ remainder } 11
\][/tex]
So, [tex]\( \frac{781}{15} = 52 \frac{11}{15} \)[/tex].
Therefore, the total length of the twine originally on the spool is [tex]\( 52 \frac{11}{15} \)[/tex] feet. In decimal form, this result is approximately [tex]\( 52.0667 \)[/tex] feet.
