First, we need to simplify the expression \((3i-5)[(i+2)+(2i-4)]\).
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Start by simplifying the term inside the brackets: \[ (i+2)+(2i-4) = i + 2 + 2i - 4 = (i + 2i) + (2 - 4) = 3i - 2 \]
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Now substitute this back into the expression: \[ (3i-5)(3i-2) \]
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Next, distribute the terms: \[ = 3i \cdot 3i + 3i \cdot (-2) + (-5) \cdot 3i + (-5) \cdot (-2) \] \[ = 9i^2 - 6i - 15i + 10 \] \[ = 9i^2 - 21i + 10 \]
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Recall that \(i^2 = -1\), so we can replace \(9i^2\) with \(9(-1)\): \[ = -9 - 21i + 10 \] \[ = 1 - 21i \]
Thus, the expression simplifies to \(1 - 21i\).
Now, we can fill in the blank: \[ 1 - 21i \implies \underline{1} + \underline{-21}i \]
So the answer is \(1 + -21i\).