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The image contains the following mathematical problems that need solutions:

1. \( g(x) = 5x - 7; g(x + 1) \)
2. \( h(x) = \sqrt{x^2 + 2x + 4}; h(2) \)
3. \( k(x) = \frac{3x - 2}{2x + 4}; k(-3) \)
4. \( f(x) = 2x^2 + 5x - 9; f(5x - 2) \)
5. \( g(p) = 4p; p\left(\frac{3}{2}\right) \)

### Solutions:

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#### 1. \( g(x) = 5x - 7 \), find \( g(x + 1) \).

To solve \( g(x + 1) \), substitute \( x + 1 \) into the function \( g(x) \):
\[
g(x + 1) = 5(x + 1) - 7 = 5x + 5 - 7 = 5x - 2
\]
So, \( g(x + 1) = 5x - 2 \).

---

#### 2. \( h(x) = \sqrt{x^2 + 2x + 4} \), find \( h(2) \).

To solve \( h(2) \), substitute \( x = 2 \) into the function \( h(x) \):
\[
h(2) = \sqrt{2^2 + 2(2) + 4} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3}
\]
So, \( h(2) = 2\sqrt{3} \).

---

#### 3. \( k(x) = \frac{3x - 2}{2x + 4} \), find \( k(-3) \).

To solve \( k(-3) \), substitute \( x = -3 \) into the function \( k(x) \):
\[
k(-3) = \frac{3(-3) - 2}{2(-3) + 4} = \frac{-9 - 2}{-6 + 4} = \frac{-11}{-2} = \frac{11}{2}
\]
So, \( k(-3) = \frac{11}{2} \).

---

#### 4. \( f(x) = 2x^2 + 5x - 9 \), find \( f(5x - 2) \).

We need to substitute \( 5x - 2 \) for \( x \) in the function \( f(x) \):
\[
f(5x - 2) = 2(5x - 2)^2 + 5(5x - 2) - 9
\]
First, expand \( (5x - 2)^2 \):
\[
(5x - 2)^2 = 25x^2 - 20x + 4
\]
Now substitute:
\[
f(5x - 2) = 2(25x^2 - 20x + 4) + 5(5x - 2) - 9 = 50x^2 - 40x + 8 + 25x - 10 - 9
\]
Simplify:
\[
f(5x - 2) = 50x^2 - 15x - 11
\]

---

#### 5. \( g(p) = 4p \), find \( p\left(\frac{3}{2}\right) \).

Since \( g(p) = 4p \), we can evaluate \( g\left(\frac{3}{2}\right) \):
\[
g\left(\frac{3}{2}\right) = 4 \times \frac{3}{2} = 6
\]
So, \( g\left(\frac{3}{2}\right) = 6 \).

---

### Summary of Results:
1. \( g(x + 1) = 5x - 2 \)
2. \( h(2) = 2\sqrt{3} \)
3. \( k(-3) = \frac{11}{2} \)
4. \( f(5x - 2) = 50x^2 - 15x - 11 \)
5. \( g\left(\frac{3}{2}\right) = 6 \)

Would you like further details on any step? Here are five related questions for practice:

1. How would you solve \( g(x + 2) \) if \( g(x) = 5x - 7 \)?
2. What is the domain of \( h(x) = \sqrt{x^2 + 2x + 4} \)?
3. Can you simplify the expression for \( k(x) = \frac{3x - 2}{2x + 4} \) in general?
4. How would \( f(x) = 2x^2 + 5x - 9 \) change if \( f(3x - 4) \) was required?
5. What are the intercepts of \( g(p) = 4p \)?

**Tip**: Always start by simplifying your algebraic expressions before substituting the values.

Evaluate the following functions: 1. g(x) = 5x - 7; g(x + 1) 2. h(x) = √(x² + 2x + 4); h(2) 3. k(x) = (3x - 2)/(2x + 4); k(-3) 4. f(x) = 2x² + 5x - 9; f(5x - 2) 5. g(p) = 4p; p(3/2)

Learn how to evaluate functions such as g(x) = 5x - 7, h(x) = √(x² + 2x + 4), and k(x) = (3x - 2)/(2x + 4). This guide provides step-by-step solutions for finding g(x + 1), h(2), k(-3), and more. Ideal for students in Grades 9-11.

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