To determine the total thickness of the two pieces of wood, we need to add their individual thicknesses together.
Here are the steps to solve this problem:
1. Identify the given thicknesses of the two pieces of wood:
- The thickness of the first piece of wood is [tex]\( \frac{5}{16} \)[/tex] inches.
- The thickness of the second piece of wood is [tex]\( \frac{7}{8} \)[/tex] inches.
2. Convert the fractions to a common denominator if necessary:
- To add these fractions together, we need a common denominator. The denominators here are 16 and 8. The least common denominator (LCD) is 16.
3. Convert the fractions:
- The first piece [tex]\( \frac{5}{16} \)[/tex] is already expressed with the denominator of 16.
- Convert the second piece [tex]\( \frac{7}{8} \)[/tex] to have the same denominator:
[tex]\[
\frac{7}{8} = \frac{7 \times 2}{8 \times 2} = \frac{14}{16}
\][/tex]
4. Add the fractions:
- Now, add [tex]\( \frac{5}{16} \)[/tex] and [tex]\( \frac{14}{16} \)[/tex]:
[tex]\[
\frac{5}{16} + \frac{14}{16} = \frac{5 + 14}{16} = \frac{19}{16}
\][/tex]
5. Express the fraction as a mixed number (if needed):
- Divide the numerator by the denominator to get the mixed number:
[tex]\[
\frac{19}{16} = 1 + \frac{3}{16} = 1\frac{3}{16}
\][/tex]
6. Convert the mixed number to a decimal form (optional):
- [tex]\( 1\frac{3}{16} \)[/tex] can be written in decimal form as 1.1875 inches.
7. Compare this result to the given options to find the correct answer.
- A: [tex]\( \frac{11}{4} = 2.75 \)[/tex] inches
- B: [tex]\( \frac{13}{16} = 0.8125 \)[/tex] inches
- C: [tex]\( \frac{11}{8} = 1.375 \)[/tex] inches
- D: [tex]\( \frac{15}{8} = 1.875 \)[/tex] inches
The result we calculated, 1.1875 inches, matches none of the provided options as is, but can be confirmed as an exact fractional answer of [tex]\( \frac{19}{16} \)[/tex]. Notice that none of the answers misuse this form incorrectly either.
Therefore, the total thickness of the two pieces of wood glued together is [tex]\( 1\frac{3}{16} \)[/tex] inches, which simplifies accurately to 1.1875 inches, neither aligning with any of the given operators exactly in practical problem context they simulate theoretical practice.
