Question

1. fill in the boxes to make a true statement:

(\square - 3i) - (15 + \square i) = 7 - 12i

1. plot each number on the real number line, or explain why the number is not on the real number line.

a. \sqrt{16}

b. -\sqrt{16}

c. \sqrt{-16}

d. 56^{1/2}

e. -56^{1/2}

f. (-56)^{1/2}

\underline{\qquad -10 \quad -8 \quad -6 \quad -4 \quad -2 \quad 0 \quad 2 \quad 4 \quad 6 \quad 8 \quad 10 \qquad}

(from unit 3, lesson 10.)

which expression is equivalent to \sqrt{-4}?

a. -2i

b. -4i

c. 2i

d. 4i

Explanation:

Question 4

Step1: Separate real and imaginary parts

Let the first box be \( x \) (real part) and the second box be \( y \) (coefficient of imaginary part). The equation is \((x - 3i)-(15 + yi)=7 - 12i\). Simplify the left - hand side: \(x-15+(-3 - y)i=7 - 12i\).

Step2: Equate real parts

For the real parts: \(x - 15=7\). Solve for \(x\): \(x=7 + 15=22\).

Step3: Equate imaginary parts

For the imaginary parts: \(-3-y=-12\). Solve for \(y\): \(y=-3 + 12 = 9\).

\(\sqrt{16} = 4\), and \(4\) is a real number. On the real number line, we can plot it at the position corresponding to \(4\).

\(-\sqrt{16}=- 4\), and \(-4\) is a real number. On the real number line, we can plot it at the position corresponding to \(-4\).

Answer:

The first box is \(22\) and the second box is \(9\)

Question 5

Part a: \(\boldsymbol{\sqrt{16}}\)