$i^9 = $
$i^2 = $
$-sqrt{-100}$
$-7i = $

Question

Accepted Answer

### 1. Solve $i^9$
# Explanation:
## Step1: Recall $i$ powers cycle
The imaginary unit $i$ has a cycle: $i^1 = i$, $i^2=-1$, $i^3 = -i$, $i^4 = 1$, and then repeats every 4. So we can express $i^9$ as $i^{4\times2 + 1}$.
## Step2: Simplify using exponent rules
Using $a^{m+n}=a^m\times a^n$, $i^{4\times2+1}=(i^4)^2\times i^1$. Since $i^4 = 1$, then $(1)^2\times i=i$.
# Answer:
$i^9 = i$

### 2. Solve $i^2$
# Explanation:
## Step1: Recall definition of $i$
By definition, the imaginary unit $i=\sqrt{-1}$, so $i^2 = (\sqrt{-1})^2=-1$.
# Answer:
$i^2=-1$

### 3. Solve $-\sqrt{-100}$
# Explanation:
## Step1: Simplify square root of negative number
We know that $\sqrt{-a}=i\sqrt{a}$ for $a>0$. So $\sqrt{-100}=i\sqrt{100}$.
## Step2: Calculate $\sqrt{100}$ and simplify
$\sqrt{100} = 10$, so $\sqrt{-100}=10i$. Then $-\sqrt{-100}=- 10i$.
# Answer:
$-\sqrt{-100}=-10i$

### 4. Solve $-7i$ (This is already in simplest form, but if we consider it as is)
# Explanation:
## Step1: Recognize the form
$-7i$ is an imaginary number in the form $bi$ where $b = - 7$, and it cannot be simplified further in terms of basic imaginary unit operations (unless combined with other complex numbers, but as a standalone, it's $-7i$).
# Answer:
$-7i=-7i$ (or just state it's in simplest form as $-7i$)

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QUESTION IMAGE

Question

Explanation:

1. Solve \(i^9\)

Step1: Recall \(i\) powers cycle

Step2: Simplify using exponent rules

Step1: Recall definition of \(i\)

Step1: Simplify square root of negative number

Step2: Calculate \(\sqrt{100}\) and simplify

Answer:

2. Solve \(i^2\)