To find when Isaac's racquet hits the ball, we need to set the two equations equal to each other since they represent the same height at the same time.
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The height of the ball is given by: \[ y = -3|x - 4| + 20 \]
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The height of Isaac's racquet is given by: \[ y = \frac{3}{1}x + 4 = 3x + 4 \]
Setting the two equations equal: \[ -3|x - 4| + 20 = 3x + 4 \]
We will solve this equation by considering the two cases of the absolute value.
Case 1: \(x - 4 \geq 0\) (i.e., \(x \geq 4\))
In this case, \(|x - 4| = x - 4\). Thus, the equation becomes: \[ -3(x - 4) + 20 = 3x + 4 \] Simplifying: \[ -3x + 12 + 20 = 3x + 4 \] \[ 32 - 3x = 3x + 4 \] \[ 32 - 4 = 6x \] \[ 28 = 6x \] \[ x = \frac{28}{6} = \frac{14}{3} \approx 4.67 \]
Case 2: \(x - 4 < 0\) (i.e., \(x < 4\))
In this case, \(|x - 4| = -(x - 4) = 4 - x\). Now we solve: \[ -3(4 - x) + 20 = 3x + 4 \] Simplifying: \[ -12 + 3x + 20 = 3x + 4 \] \[ 8 = 3x + 4 \] \[ 8 - 4 = 3x \] \[ 4 = 3x \] \[ x = \frac{4}{3} \approx 1.33 \]
Summary of Solutions
From the two cases, we have:
- From Case 1: \(x \approx 4.67\)
- From Case 2: \(x \approx 1.33\)
Conclusion
Isaac's racquet hits the ball at approximately \(1.33\) milliseconds, which matches option B:
B. after 1.5 milliseconds.
Note that the closest option given for \(1.33\) in the choices is \(1.5\).