To find the relationship between \( P \) and \( Q \), given that \( P \) is partly constant and partly varies inversely as the square root of \( Q \), we can express \( P \) in the form:
\[ P = k + \frac{c}{\sqrt{Q}} \]
where \( k \) is a constant and \( c \) is a coefficient that relates to the inverse variation with respect to the square root of \( Q \).
Given the conditions:
- \( P = 6 \) when \( Q = 9 \)
- \( P = 10 \) when \( Q = 16 \)
We can substitute these values into the equation to create a system of equations.
- From \( P = 6 \) when \( Q = 9 \):
\[ 6 = k + \frac{c}{\sqrt{9}} \implies 6 = k + \frac{c}{3} \]
This can be rearranged to:
\[ k + \frac{c}{3} = 6 \quad \text{(Equation 1)} \]
- From \( P = 10 \) when \( Q = 16 \):
\[ 10 = k + \frac{c}{\sqrt{16}} \implies 10 = k + \frac{c}{4} \]
This can be rearranged to:
\[ k + \frac{c}{4} = 10 \quad \text{(Equation 2)} \]
Now we have a system of equations from Equations 1 and 2:
\[ k + \frac{c}{3} = 6 \] \[ k + \frac{c}{4} = 10 \]
Next, we can eliminate \( k \) by isolating it in both equations.
From Equation 1:
\[ k = 6 - \frac{c}{3} \]
From Equation 2:
\[ k = 10 - \frac{c}{4} \]
Setting the two expressions for \( k \) equal to each other:
\[ 6 - \frac{c}{3} = 10 - \frac{c}{4} \]
Now, we can solve for \( c \). First, multiply through by 12 to eliminate the fractions:
\[ 12 \cdot 6 - 12 \cdot \frac{c}{3} = 12 \cdot 10 - 12 \cdot \frac{c}{4} \]
This simplifies to:
\[ 72 - 4c = 120 - 3c \]
Now, rearranging gives:
\[ 72 - 120 = c - 4c \]
\[ -48 = -c \]
Thus, we find:
\[ c = 48 \]
Now substitute \( c = 48 \) back into Equation 1 to find \( k \):
\[ k + \frac{48}{3} = 6 \] \[ k + 16 = 6 \implies k = 6 - 16 = -10 \]
Now we have:
\[ k = -10 \quad \text{and} \quad c = 48 \]
Thus, the relationship between \( P \) and \( Q \) is:
\[ P = -10 + \frac{48}{\sqrt{Q}} \]
Now to find \( P \) when \( Q = 4 \):
\[ P = -10 + \frac{48}{\sqrt{4}} = -10 + \frac{48}{2} = -10 + 24 = 14 \]
Therefore, the value of \( P \) when \( Q = 4 \) is:
\[ \boxed{14} \]
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