Perform the indicated operations and write the result in standard form, \( a+b i \). \( 6 \sqrt{-64}+3 \sqrt{-81} \) \( 6 \sqrt{-64}+3 \sqrt{-81}=\square \) (Simplify your answer.)

To simplify the expression \( 6 \sqrt{-64} + 3 \sqrt{-81} \), we first evaluate each square root separately. Starting with \( \sqrt{-64} \): \[ \sqrt{-64} = \sqrt{64} \sqrt{-1} = 8i \] Thus, \( 6 \sqrt{-64} = 6 \times 8i = 48i \). Next, we evaluate \( \sqrt{-81} \): \[ \sqrt{-81} = \sqrt{81} \sqrt{-1} = 9i \] Therefore, \( 3 \sqrt{-81} = 3 \times 9i = 27i \). Now, we can combine these results: \[ 6 \sqrt{-64} + 3 \sqrt{-81} = 48i + 27i = (48 + 27)i = 75i \] Putting it all together, we find: \[ 6 \sqrt{-64} + 3 \sqrt{-81} = 0 + 75i \] So, the final result in standard form \( a + bi \) is: \[ \boxed{0 + 75i} \]