To determine whether the camera will be able to capture Jada's jump before she hits the water, we need to analyze the two functions \( f(x) \) and \( g(x) \) and see if they intersect, which would indicate that the camera captures Jada at that point.
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Equations:
- Jada's dive: \[ f(x) = -0.1(x - 3)^2 + 10 \]
- Camera's view: \[ g(x) = 0.67x + 3 \]
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Intersect the two equations: To find where the camera captures Jada, we set \( f(x) \) equal to \( g(x) \): \[ -0.1(x - 3)^2 + 10 = 0.67x + 3 \] Simplifying this: \[ -0.1(x - 3)^2 + 10 - 3 = 0.67x \] \[ -0.1(x - 3)^2 + 7 = 0.67x \] Rearranging gives: \[ -0.1(x - 3)^2 - 0.67x + 7 = 0 \]
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Expand and simplify: Expanding \((x - 3)^2\): \[ (x - 3)^2 = x^2 - 6x + 9 \] Substitute it back: \[ -0.1(x^2 - 6x + 9) - 0.67x + 7 = 0 \] This expands and simplifies to: \[ -0.1x^2 + 0.6x - 0.9 - 0.67x + 7 = 0 \] \[ -0.1x^2 - 0.07x + 6.1 = 0 \] Multiply through by -1 for simplicity: \[ 0.1x^2 + 0.07x - 6.1 = 0 \]
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Use the quadratic formula: The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 0.1 \), \( b = 0.07 \), and \( c = -6.1 \). \[ b^2 - 4ac = (0.07)^2 - 4(0.1)(-6.1) = 0.0049 + 2.44 = 2.4449 \] \[ x = \frac{-0.07 \pm \sqrt{2.4449}}{0.2} \] \[ x = \frac{-0.07 \pm 1.5655}{0.2} \] Calculating the two potential solutions: \[ x_1 = \frac{1.4955}{0.2} = 7.4775 \] \[ x_2 = \frac{-1.6355}{0.2} = -8.1775 \text{ (not applicable as distance must be positive)} \]
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Find the vertical distance when \( x = 7.4775 \): Substitute \( x \) back into either original function. We will use \( g(x) \): \[ g(7.4775) = 0.67(7.4775) + 3 \] \[ g(7.4775) = 5.012525 + 3 = 8.012525 \] Rounding to the nearest integer gives \( 8 \) feet.
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Conclusion: The camera will capture Jada's jump at approximately 8 feet above the surface of the water.
Thus, the answer is: B. 8 feet