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.