To determine whether the camera can capture Jada's jump, we need to find the intersection of the two equations \( f(x) = -0.1(x - 3)^2 + 10 \) for her dive and \( g(x) = 0.67x + 3 \) for the camera's view.
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Find the equations:
- Jada's dive: \[ f(x) = -0.1(x - 3)^2 + 10 \]
- Camera view: \[ g(x) = 0.67x + 3 \]
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Set the equations equal to each other to find the intersection points: \[ -0.1(x - 3)^2 + 10 = 0.67x + 3 \]
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Rearranging the equation: \[ -0.1(x - 3)^2 - 0.67x + 7 = 0 \]
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Clear the equation: Let's first expand \( -0.1(x - 3)^2 \): \[ -0.1(x^2 - 6x + 9) = -0.1x^2 + 0.6x - 0.9 \] Now substitute it into the equation: \[ -0.1x^2 + 0.6x - 0.9 - 0.67x + 7 = 0 \]
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Combine like terms: \[ -0.1x^2 - 0.07x + 6.1 = 0 \]
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Multiply by -10 to eliminate decimals: \[ x^2 + 0.7x - 61 = 0 \]
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Now we can use the quadratic formula to find x: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1, b = 0.7, c = -61 \).
\[ b^2 - 4ac = 0.7^2 - 4(1)(-61) = 0.49 + 244 = 244.49 \] \[ x = \frac{-0.7 \pm \sqrt{244.49}}{2} \] \[ x = \frac{-0.7 \pm 15.634}{2} \]
This gives us two potential solutions for \( x \): \[ x_1 = \frac{14.934}{2} \approx 7.467 \] \[ x_2 = \frac{-16.334}{2} \approx -8.167 \quad(\text{not relevant since it is negative}) \]
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Find \( f(x) \) and \( g(x) \) at \( x \approx 7.467 \):
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To find the height of Jada at this x value: \[ f(7.467) = -0.1(7.467 - 3)^2 + 10 \] \[ f(7.467) = -0.1(4.467)^2 + 10 \approx -0.1(19.94) + 10 \approx -1.994 + 10 = 8.006 \]
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To find the camera’s altitude: \[ g(7.467) = 0.67(7.467) + 3 \approx 5.007 + 3 = 8.007 \]
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Both values are approximately 8 feet. Therefore, the camera captures Jada just as she is about to hit the water.
Conclusion: The camera will capture Jada's jump, and she will be approximately 8 feet above the water when captured. The answer to the query is 8 feet.