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-2i*(2+6i)=
Your answer should be a complex number in the form
a+bi where
a and
b are real numbers.

$-2 i \cdot(2+6 i)=$ $\newline$ Your answer should be a complex number in the form $a+b i$ where $a$ and $b$ are real numbers.

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Composition of linear and quadratic functions: find an equation

Full solution

Q.

-2 i \cdot(2+6 i)=

\newline

Your answer should be a complex number in the form

a+b i

where

a

and

b

are real numbers.

Distribute $-2i$ to terms inside parentheses: To multiply the complex number $-2i$ by $(2+6i)$ , we distribute $-2i$ to both terms inside the parentheses. $\newline$ Calculation: $-2i \times 2 + (-2i \times 6i)$
Multiply $-2i$ by $2$ : Multiplying $-2i$ by $2$ gives us $-4i$ . $\newline$ Calculation: $-2i \times 2 = -4i$
Multiply $-2i$ by $6i$ : Multiplying $-2i$ by $6i$ gives us $-12i^2$ . Since $i^2$ is equal to $-1$ , this simplifies to $12$ . $\newline$ Calculation: $-2i \times 6i = -12i^2 = -12(-1) = 12$
Combine the results: Now we combine the results of the two multiplications to get the final answer. $\newline$ Calculation: $-4i + 12$

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