To find out when Isaac's racquet hits the ball, we need to set the two equations equal to each other, since that will give us the point in time (x milliseconds) when both the height of the ball and the height of the racquet are the same.
The equations we have are:
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For Samantha's ball: \[ y = -3 |x - 4| + 20 \]
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For Isaac's racquet: \[ y = \frac{1}{3}x + 4 \]
Setting them equal to each other:
\[ -3 |x - 4| + 20 = \frac{1}{3}x + 4 \]
Next, we can solve this equation for \(x\). We will handle the absolute value by considering two cases:
Case 1: \(x - 4 \geq 0\) (which means \(x \geq 4\))
In this case, \(|x - 4| = x - 4\), so the equation becomes:
\[ -3(x - 4) + 20 = \frac{1}{3}x + 4 \]
Expanding and simplifying:
\[ -3x + 12 + 20 = \frac{1}{3}x + 4 \] \[ -3x + 32 = \frac{1}{3}x + 4 \]
To eliminate the fraction, multiply throughout by 3:
\[ -9x + 96 = x + 12 \]
Combine like terms:
\[ -10x = -84 \qquad \Rightarrow \qquad x = 8.4 \]
Since \(x = 8.4\) is greater than 4, this solution is valid for Case 1.
Case 2: \(x - 4 < 0\) (which means \(x < 4\))
In this case, \(|x - 4| = -(x - 4) = 4 - x\), so the equation becomes:
\[ -3(4 - x) + 20 = \frac{1}{3}x + 4 \]
Expanding and simplifying:
\[ -12 + 3x + 20 = \frac{1}{3}x + 4 \] \[ 3x + 8 = \frac{1}{3}x + 4 \]
Multiply through by 3 to eliminate the fraction:
\[ 9x + 24 = x + 12 \]
Combine like terms:
\[ 8x = -12 \qquad \Rightarrow \qquad x = -1.5 \]
Since \(x = -1.5\) is less than 4, this result is not useful for the context of the racquetball game since negative time doesn't make sense.
Thus, the only valid solution, where Isaac's racquet hits the ball, is from Case 1:
\[ x = 8.4 \text{ milliseconds} \]
So the answer is A. after 8.4 milliseconds.