The problem consists of several sub-problems: 1. Classify numbers as rational or irrational. The numbers are $2\pi$, $-\frac{4}{2}$, $\sqrt{20}$, and $7.04339265$.

1. Problem Description

The problem consists of several sub-problems:

1. Classify numbers as rational or irrational. The numbers are $2\pi$, $-\frac{4}{2}$, $\sqrt{20}$, and $7.04339265$.

2. Write each repeating decimal as a fraction in simplest form. These problems are crossed out.

3. Simplify expressions, writing only positive exponents. The expressions are $3^4 \cdot 3^8$, $25^{-6} \cdot 5^{-6}$, $(2x^5)^4$, and $6x^0y^{-9}$.

4. Solve the equation $x^2 + 29 = 225$.

5. Approximate $\sqrt{13}$ to the nearest tenth and plot it on the number line.

2. Solution Steps

Problem 1:

a. $2\pi$ is irrational because $\pi$ is irrational, and any non-zero rational number times an irrational number is irrational.

b. $-\frac{4}{2} = -2$ which is an integer and thus rational.

c. $\sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}$. Since $\sqrt{5}$ is irrational, $2\sqrt{5}$ is also irrational.

d. $7.04339265$ is a terminating decimal, so it is rational.

Problem 3:

a. $3^4 \cdot 3^8 = 3^{4+8} = 3^{12}$

b. $25^{-6} \cdot 5^{-6} = (25 \cdot 5)^{-6} = (125)^{-6} = (5^3)^{-6} = 5^{-18} = \frac{1}{5^{18}}$

c. $(2x^5)^4 = 2^4 \cdot (x^5)^4 = 16x^{5 \cdot 4} = 16x^{20}$

d. $6x^0y^{-9} = 6 \cdot 1 \cdot y^{-9} = \frac{6}{y^9}$

Problem 4:

$x^2 + 29 = 225$

$x^2 = 225 - 29$

$x^2 = 196$

$x = \pm \sqrt{196}$

$x = \pm 14$

Problem 5:

Since $3^2 = 9$ and $4^2 = 16$, $\sqrt{13}$ is between 3 and

4. $3.5^2 = 12.25$

$3.6^2 = 12.96$

$3.7^2 = 13.69$

So, $\sqrt{13}$ is closer to 3.6 than 3.

7. We want the nearest tenth, so we check $3.6$.

$\sqrt{13} \approx 3.6$

3. Final Answer

1. a. Irrational

b. Rational

c. Irrational

d. Rational

2. Not Applicable

3. a. $3^{12}$

b. $\frac{1}{5^{18}}$

c. $16x^{20}$

d. $\frac{6}{y^9}$

4. $x = \pm 14$

5. $\sqrt{13} \approx 3.6$ (Plot 3.6 on the number line between 3 and 4)