Sure, let's determine the height at which the cable will be anchored on the first vertical tower using the given function and [tex]\(x\)[/tex] value.
Given:
[tex]\[ f(x) = \frac{1}{8}x^2 - 5x + 52 \][/tex]
We need to find the height when [tex]\( x = 10 \)[/tex] meters. Here's the detailed solution:
1. Substitute [tex]\( x = 10 \)[/tex] into the function:
[tex]\[ f(10) = \frac{1}{8}(10)^2 - 5(10) + 52 \][/tex]
2. Calculate [tex]\( (10)^2 \)[/tex]:
[tex]\[ (10)^2 = 100 \][/tex]
3. Multiply by [tex]\( \frac{1}{8} \)[/tex]:
[tex]\[ \frac{1}{8} \times 100 = 12.5 \][/tex]
4. Multiply [tex]\( -5 \)[/tex] by [tex]\( 10 \)[/tex]:
[tex]\[ -5 \times 10 = -50 \][/tex]
5. Add the results to the constant term [tex]\( 52 \)[/tex]:
[tex]\[ 12.5 - 50 + 52 \][/tex]
6. Simplify step-by-step:
- First, add [tex]\( 12.5 \)[/tex] and [tex]\( 52 \)[/tex]:
[tex]\[ 12.5 + 52 = 64.5 \][/tex]
- Then subtract [tex]\( 50 \)[/tex]:
[tex]\[ 64.5 - 50 = 14.5 \][/tex]
Thus, when [tex]\( x = 10 \)[/tex] meters, the height at which the cable will be anchored on the first vertical tower is [tex]\(\boxed{14.5}\)[/tex] meters.
