Since the problem gives \( ba = 1/4 \) and \( b = 8 \) when \( h = 5 \), we can start by determining the variable \( a \).
Given: \[ b a = \frac{1}{4} \] Substituting \( b = 8 \): \[ 8a = \frac{1}{4} \] To find \( a \): \[ a = \frac{1/4}{8} = \frac{1}{4} \times \frac{1}{8} = \frac{1}{32} \]
Now we have:
- \( a = \frac{1}{32} \)
- \( b = 8 \)
Now we need to establish a relationship between \( b \), \( a \), and \( h \). It's possible that \( b \times h \) is constant, or there is some other derived relationship. However, at this point, we don't have enough information about how \( b \) or \( a \) changes with \( h \).
If we assume that \( b \) is inversely proportional to \( h \) (i.e., \( b = \frac{k}{h} \) for some constant \( k \)), we can try to find \( k \) using the information we have at \( h = 5 \).
Using \( b = 8 \) and \( h = 5 \): \[ 8 = \frac{k}{5} \] So: \[ k = 8 \cdot 5 = 40 \]
Now, we can express \( b \) in terms of \( h \): \[ b = \frac{40}{h} \]
Finding \( b \) when \( h = 4 \):
Substituting \( h = 4 \): \[ b = \frac{40}{4} = 10 \]
Finding \( h \) when \( b = 4 \):
Now we rearrange the equation \( b = \frac{40}{h} \) to find \( h \) when \( b = 4 \): \[ 4 = \frac{40}{h} \] This leads us to: \[ h = \frac{40}{4} = 10 \]
Summary of Results:
- \( b \) when \( h = 4 \) is 10.
- \( h \) when \( b = 4 \) is 10.