12th Grade Math Review Worksheet featuring algebra, trigonometry, and polar coordinates problems.

Let's solve each problem step by step.

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Problem 1: Solve the inequality

$$
-4 \leq \frac{x + 5}{3} \leq 7
$$

#### Step 1: Eliminate the fraction by multiplying through by 3
Multiply all parts of the inequality by 3 to clear the denominator:
$$
3 \cdot (-4) \leq x + 5 \leq 3 \cdot 7
$$
$$
-12 \leq x + 5 \leq 21
$$

#### Step 2: Isolate \( x \)
Subtract 5 from all parts of the inequality:
$$
-12 - 5 \leq x \leq 21 - 5
$$
$$
-17 \leq x \leq 16
$$

#### Final Answer for Problem 1:
$$
\boxed{-17 \leq x \leq 16}
$$

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Problem 2: Simplify the expression

$$
\cos(\theta) + \sin(\theta) \tan(\theta)
$$

#### Step 1: Recall the definition of \(\tan(\theta)\)
The tangent function is defined as:
$$
\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}
$$

#### Step 2: Substitute \(\tan(\theta)\) into the expression
Replace \(\tan(\theta)\) in the given expression:
$$
\cos(\theta) + \sin(\theta) \cdot \frac{\sin(\theta)}{\cos(\theta)}
$$

#### Step 3: Simplify the second term
The second term becomes:
$$
\sin(\theta) \cdot \frac{\sin(\theta)}{\cos(\theta)} = \frac{\sin^2(\theta)}{\cos(\theta)}
$$

So the expression now is:
$$
\cos(\theta) + \frac{\sin^2(\theta)}{\cos(\theta)}
$$

#### Step 4: Combine the terms into a single fraction
To combine the terms, write \(\cos(\theta)\) with a common denominator:
$$
\cos(\theta) = \frac{\cos^2(\theta)}{\cos(\theta)}
$$
Thus, the expression becomes:
$$
\frac{\cos^2(\theta)}{\cos(\theta)} + \frac{\sin^2(\theta)}{\cos(\theta)} = \frac{\cos^2(\theta) + \sin^2(\theta)}{\cos(\theta)}
$$

#### Step 5: Use the Pythagorean identity
Recall the Pythagorean identity:
$$
\cos^2(\theta) + \sin^2(\theta) = 1
$$
Substitute this into the numerator:
$$
\frac{\cos^2(\theta) + \sin^2(\theta)}{\cos(\theta)} = \frac{1}{\cos(\theta)}
$$

#### Final Answer for Problem 2:
$$
\boxed{\sec(\theta)}
$$

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Problem 3: Convert to polar form

Convert the equation:
$$
x^2 + (y - 6)^2 = 35
$$
to polar coordinates.

#### Step 1: Recall the relationships between Cartesian and polar coordinates
In polar coordinates:
$$
x = r \cos(\theta), \quad y = r \sin(\theta)
$$

#### Step 2: Substitute \( x \) and \( y \) into the equation
The given equation is:
$$
x^2 + (y - 6)^2 = 35
$$
Substitute \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \):
$$
(r \cos(\theta))^2 + (r \sin(\theta) - 6)^2 = 35
$$

#### Step 3: Expand and simplify
First, expand \( (r \sin(\theta) - 6)^2 \):
$$
(r \sin(\theta) - 6)^2 = r^2 \sin^2(\theta) - 12r \sin(\theta) + 36
$$
So the equation becomes:
$$
r^2 \cos^2(\theta) + r^2 \sin^2(\theta) - 12r \sin(\theta) + 36 = 35
$$

#### Step 4: Combine like terms
Use the Pythagorean identity \( \cos^2(\theta) + \sin^2(\theta) = 1 \):
$$
r^2 (\cos^2(\theta) + \sin^2(\theta)) - 12r \sin(\theta) + 36 = 35
$$
$$
r^2 \cdot 1 - 12r \sin(\theta) + 36 = 35
$$
$$
r^2 - 12r \sin(\theta) + 36 = 35
$$

#### Step 5: Simplify further
Subtract 35 from both sides:
$$
r^2 - 12r \sin(\theta) + 1 = 0
$$

#### Final Answer for Problem 3:
$$
\boxed{r^2 - 12r \sin(\theta) + 1 = 0}
$$

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Final Answers:

1. \(\boxed{-17 \leq x \leq 16}\)
2. \(\boxed{\sec(\theta)}\)
3. \(\boxed{r^2 - 12r \sin(\theta) + 1 = 0}\)