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Let's work through the problems on your homework one by one.

### 1. Graph and Fully Describe the Function \( y = \sqrt{x} - 2 \)

**Description:**
- **Domain**: The function \( y = \sqrt{x} - 2 \) is defined for \( x \geq 0 \) because the square root function only takes non-negative values.
- **Range**: The output of \( y \) is \( y \geq -2 \), since the minimum value of \( \sqrt{x} \) is 0, making \( y \) start from \(-2\) and increase as \( x \) increases.
- **Graph**: The graph starts at the point \( (0, -2) \) and increases as \( x \) increases, following the shape of the square root function but shifted downward by 2 units.

### 2. Line Passing Through Points

**Given Points:**
- \( A(-3, -2) \)
- \( B(2, 1) \)

**Slope Calculation**:
The slope \( m \) between the two points \( A \) and \( B \) is given by:
\[
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - (-2)}{2 - (-3)} = \frac{3}{5}
\]

**Equation of the Line**:
Using point-slope form, the equation of the line is:
\[
y - y_1 = m(x - x_1)
\]
Substituting \( A(-3, -2) \) and \( m = \frac{3}{5} \):
\[
y + 2 = \frac{3}{5}(x + 3)
\]
Simplifying:
\[
y = \frac{3}{5}x + \frac{9}{5} - 2
\]
\[
y = \frac{3}{5}x - \frac{1}{5}
\]

**Checking Point \( C(5, 3) \)**:
Substitute \( x = 5 \) into the equation \( y = \frac{3}{5}x - \frac{1}{5} \):
\[
y = \frac{3}{5}(5) - \frac{1}{5} = 3 - \frac{1}{5} = \frac{14}{5} = 2.8
\]
Since \( y = 2.8 \) and not 3, the point \( C(5, 3) \) does **not** lie on the line.

### 3. Simplify Each Expression

a. \( (5x)^2 \times 5x^3 \)
\[
= 25x^2 \times 5x^3 = 125x^{5}
\]

b. \( \left(\frac{2a^5b^{-4}}{4a^{-2}b^3}\right)^2 \)
\[
= \left(\frac{2a^5b^{-4}}{4a^{-2}b^3}\right)^2 = \left(\frac{a^7}{2b^7}\right)^2 = \frac{a^{14}}{4b^{14}}
\]

c. \( \frac{7b^{-2}a^4}{\frac{7b^2}{a^6}} \)
\[
= \frac{7b^{-2}a^4}{7b^2a^{-6}} = \frac{a^{10}}{b^4}
\]

d. \( (4x^0)^2 \)
\[
= 4^2 \times (x^0)^2 = 16 \times 1 = 16
\]

e. \( \left(\frac{c^{-2}}{d^4}\right)^{-3} \)
\[
= \frac{d^{12}}{c^6}
\]

f. \( \left(4m^2n^4\right)^{-3/2} \)
\[
= \frac{1}{8m^3n^6}
\]

g. \( m^{-3} \times m^2 \)
\[
= m^{-1} = \frac{1}{m}
\]

### 4. Solve Each Inequality

a. \( 4|2x - 1| \geq 10 \)
\[
|2x - 1| \geq \frac{10}{4} = 2.5
\]
\[
2x - 1 \geq 2.5 \quad \text{or} \quad 2x - 1 \leq -2.5
\]
\[
x \geq 1.75 \quad \text{or} \quad x \leq -0.75
\]

b. \( |2x - 9| > -8 \)
\[
Since absolute value is always non-negative, this inequality is always true for all \( x \).

c. \( |4x + 15| < -2 \)
\[
No solution, because the absolute value cannot be negative.

d. \( |3x + 2| < 0 \)
\[
No solution, because the absolute value cannot be negative.

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Would you like a detailed explanation for any specific problem? Here are some related questions you might explore:

1. How does shifting a graph affect its equation?
2. What are the properties of absolute value inequalities?
3. How do you find the equation of a line given two points?
4. How can we simplify expressions involving exponents and radicals?
5. What are the implications when an absolute value inequality has no solution?

**Tip**: When simplifying expressions, always pay attention to negative exponents and how they transform when the expression is simplified.

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