To find the equation that connects \( x \) and \( y \), given that \( x \) is partly constant and partly varies as the square of \( y \), we can express this relationship in the form:
\[ x = k + m y^2 \]
where \( k \) is a constant and \( m \) is a coefficient that dictates how \( x \) varies with \( y^2 \).
We are given the following data points:
- When \( y = 1 \), \( x = 4.25 \) (which is \( 4 \frac{1}{4} = \frac{17}{4} \)).
- When \( y = 7 \), \( x = 196.25 \) (which is \( 196 \frac{1}{4} = \frac{785}{4} \)).
Now we can set up two equations using these points.
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For \( y = 1 \): \[ \frac{17}{4} = k + m(1)^2 \] which simplifies to: \[ \frac{17}{4} = k + m \quad \text{(Equation 1)} \]
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For \( y = 7 \): \[ \frac{785}{4} = k + m(7)^2 \] which simplifies to: \[ \frac{785}{4} = k + 49m \quad \text{(Equation 2)} \]
Now we have a system of equations:
- \( k + m = \frac{17}{4} \)
- \( k + 49m = \frac{785}{4} \)
We can solve this system step by step.
Step 1: Solve for \( k \) in terms of \( m \) using Equation 1: \[ k = \frac{17}{4} - m \]
Step 2: Substitute \( k \) into Equation 2: \[ \frac{17}{4} - m + 49m = \frac{785}{4} \] \[ \frac{17}{4} + 48m = \frac{785}{4} \]
Step 3: Isolate \( m \): \[ 48m = \frac{785}{4} - \frac{17}{4} \] \[ 48m = \frac{785 - 17}{4} = \frac{768}{4} = 192 \] \[ m = \frac{192}{48} = 4 \]
Step 4: Substitute \( m \) back into Equation 1 to find \( k \): \[ k + 4 = \frac{17}{4} \] \[ k = \frac{17}{4} - 4 = \frac{17}{4} - \frac{16}{4} = \frac{1}{4} \]
Putting the values of \( k \) and \( m \) back into the equation:
\[ x = k + m y^2 \] \[ x = \frac{1}{4} + 4y^2 \]
In conclusion, the equation connecting \( x \) and \( y \) is: \[ x = 4y^2 + \frac{1}{4} \]
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