Question

consider that $i = \sqrt{-1}$

use the imaginary number $i$ to rewrite the expression
$\left(\sqrt{-9}\
ight)\left(i^2\
ight)$

options for the first (partial) question (not fully visible):
a -81i
b -9
c -9i
d 9i

options for the main question:
a -3
b -3i
c -9i
d 3i

Explanation:

Step1: Simplify \(\sqrt{-9}\)

Recall that \(\sqrt{-a}=\sqrt{a}\cdot i\) for \(a>0\). So, \(\sqrt{-9}=\sqrt{9}\cdot i = 3i\).

Step2: Simplify \(i^{2}\)

By definition, \(i^{2}=-1\).

Step3: Multiply the two results

Now, multiply \(\sqrt{-9}\) (which is \(3i\)) and \(i^{2}\) (which is \(-1\)): \((3i)\times(-1)= - 3i\)? Wait, no, wait. Wait, the expression is \((\sqrt{-9})(i^{2})\). Wait, \(\sqrt{-9}=3i\), \(i^{2}=-1\), so \(3i\times(-1)=-3i\)? But let's check again. Wait, maybe I made a mistake. Wait, \(\sqrt{-9}=\sqrt{9}\times\sqrt{-1}=3i\), \(i^{2}=-1\), so \(3i\times(-1)= - 3i\). But looking at the options, option C is \(-9i\)? No, wait the options for the second problem (the one with \((\sqrt{-9})(i^{2})\)): options are A: -3, B: -3i, C: -9i, D: 3i. Wait, let's recalculate. \(\sqrt{-9}=3i\), \(i^{2}=-1\), so \(3i\times(-1)=-3i\), which is option B. Wait, maybe the first problem is missing, but the second problem is \((\sqrt{-9})(i^{2})\). Let's focus on that.

Wait, the problem is to rewrite \((\sqrt{-9})(i^{2})\) using \(i\). So step by step:

1. Simplify \(\sqrt{-9}\): \(\sqrt{-9}=\sqrt{9\times(-1)}=\sqrt{9}\times\sqrt{-1}=3i\) (since \(\sqrt{-1}=i\)).
2. Simplify \(i^{2}\): \(i^{2}=-1\) (by definition of imaginary unit).
3. Multiply the two: \(3i\times(-1)=-3i\).

So the correct option is B. -3i.

Answer:

B. -3i