Question

1. \\(sqrt{5x} cdot sqrt{-5x}\\) 10) \\(sqrt{-81} cdot sqrt{-9}\\) 11) \\(\frac{1 - 4i}{3 + 2i}\\) 12) \\(i^{460}\\)

Explanation:

Step1: Simplify problem 9: Rewrite with $i$

$\sqrt{5x} \cdot \sqrt{-5x} = \sqrt{5x} \cdot i\sqrt{5x} = i \cdot (\sqrt{5x})^2 = 5xi$

Step2: Simplify problem 10: Rewrite with $i$

$\sqrt{-81} \cdot \sqrt{-9} = 9i \cdot 3i = 27i^2 = 27(-1) = -27$

Step3: Simplify problem 11: Rationalize denominator

Multiply numerator/denominator by $3-2i$:
$\frac{(1-4i)(3-2i)}{(3+2i)(3-2i)} = \frac{3-2i-12i+8i^2}{9-4i^2} = \frac{3-14i-8}{9+4} = \frac{-5-14i}{13} = -\frac{5}{13} - \frac{14}{13}i$

Step4: Simplify problem 12: Use $i^4=1$

$i^{460} = (i^4)^{115} = 1^{115} = 1$

Answer:

1. $5xi$
2. $-27$
3. $-\frac{5}{13} - \frac{14}{13}i$
4. $1$