Certainly! Let's break down the expression [tex]\(6(d - f) + f\)[/tex] step by step with the given values of [tex]\( d \)[/tex] and [tex]\( f \)[/tex].
First, we need to interpret the given values:
- [tex]\( d = 4 \frac{7}{8} \)[/tex]
- [tex]\( f = 3 \frac{1}{2} \)[/tex]
### Step 1: Convert Mixed Numbers to Improper Fractions
Convert [tex]\( d \)[/tex] and [tex]\( f \)[/tex] to improper fractions for easier manipulation.
For [tex]\( d = 4 \frac{7}{8} \)[/tex]:
[tex]\[
4 \frac{7}{8} = \frac{4 \times 8 + 7}{8} = \frac{32 + 7}{8} = \frac{39}{8}
\][/tex]
For [tex]\( f = 3 \frac{1}{2} \)[/tex]:
[tex]\[
3 \frac{1}{2} = \frac{3 \times 2 + 1}{2} = \frac{6 + 1}{2} = \frac{7}{2}
\][/tex]
So we now have:
[tex]\[
d = \frac{39}{8}
\][/tex]
[tex]\[
f = \frac{7}{2}
\][/tex]
### Step 2: Subtract [tex]\( f \)[/tex] from [tex]\( d \)[/tex]
We need to find [tex]\( d - f \)[/tex]:
[tex]\[
d - f = \frac{39}{8} - \frac{7}{2}
\][/tex]
To subtract these fractions, we need a common denominator. The least common multiple of 8 and 2 is 8. Convert [tex]\( \frac{7}{2} \)[/tex] to something over 8:
[tex]\[
\frac{7}{2} = \frac{7 \times 4}{2 \times 4} = \frac{28}{8}
\][/tex]
Now subtract:
[tex]\[
d - f = \frac{39}{8} - \frac{28}{8} = \frac{39 - 28}{8} = \frac{11}{8}
\][/tex]
### Step 3: Multiply the Result by 6
Next, we need to multiply [tex]\( \frac{11}{8} \)[/tex] by 6:
[tex]\[
6 \times \frac{11}{8} = \frac{66}{8} = \frac{33}{4}
\][/tex]
### Step 4: Add [tex]\( f \)[/tex] Back to the Result
Finally, add [tex]\( f \)[/tex] to the product:
[tex]\[
6(d - f) + f = \frac{33}{4} + \frac{7}{2}
\][/tex]
To add these fractions, we need a common denominator. The least common multiple of 4 and 2 is 4. Convert [tex]\( \frac{7}{2} \)[/tex] to something over 4:
[tex]\[
\frac{7}{2} = \frac{7 \times 2}{2 \times 2} = \frac{14}{4}
\][/tex]
Now add:
[tex]\[
\frac{33}{4} + \frac{14}{4} = \frac{33 + 14}{4} = \frac{47}{4}
\][/tex]
So, the final value of the expression [tex]\( 6(d - f) + f \)[/tex] is:
[tex]\[
\boxed{\frac{47}{4}}
\][/tex]
In decimal form, this is:
[tex]\[
\boxed{11.75}
\][/tex]
