Let's go over the questions from your uploaded image:

Simplify each expression:

3) $6(5n - 9) - 3(5n - 8)$

• Distribute the constants:
  $6 \cdot 5n = 30n$, $6 \cdot (-9) = -54$, $3 \cdot 5n = 15n$, and $3 \cdot (-8) = -24$.
  This gives:
  $30n - 54 - 15n + 24$.
• Combine like terms:
  $30n - 15n = 15n$, and $-54 + 24 = -30$.
  So the simplified expression is:
  $15n - 30.$

4) $-4(1 + v) - 10(8v + 8)$

• Distribute the constants:
  $-4 \cdot 1 = -4$, $-4 \cdot v = -4v$, $-10 \cdot 8v = -80v$, and $-10 \cdot 8 = -80$.
  This gives:
  $-4 - 4v - 80v - 80$.
• Combine like terms:
  $-4v - 80v = -84v$, and $-4 - 80 = -84$.
  So the simplified expression is:
  $-84v - 84.$

Evaluate. Given $a = -4$ and $c = 5$:

5) $|4c - a - 10| + 3$

• Substitute the values:
  $4c = 4 \cdot 5 = 20$ and $a = -4$, so:
  $|20 - (-4) - 10| + 3 = |20 + 4 - 10| + 3 = |14| + 3 = 14 + 3 = 17$.

6) $4|-c + a| - 5$

• Substitute the values:
  $c = 5$ and $a = -4$, so:
  $4|-(5) + (-4)| - 5 = 4| -5 - 4 | - 5 = 4| -9 | - 5 = 4 \cdot 9 - 5 = 36 - 5 = 31$.

Word Problems:

7) Two times the sum of a number and three is equal to eleven.

• Set up the equation:
  $2(x + 3) = 11$.
• Solve for $x$:
  $2x + 6 = 11$, so $2x = 5$, and $x = \frac{5}{2} = 2.5$.

8) The voltage $V$ is equal to amperage $I$ times the resistance $R$.

• This describes Ohm's Law:
  $V = IR.$

9) Distance is equal to the product of rate and time.

• The formula for distance is:
  $d = r \cdot t.$

10) The perimeter $P$ of an isosceles triangle is equal to the base $b$ added to twice the length of one of the legs $l$.

• The formula for the perimeter of an isosceles triangle is:
  $P = b + 2l.$

Let me know if you'd like more detailed steps or explanations on any of these!

Here are 5 questions based on this material:

1. How do you distribute a negative constant when simplifying expressions?
2. What happens to the absolute value when evaluating expressions like in problem 5?
3. What kind of triangle does the formula in problem 10 apply to?
4. How does combining like terms work when both constants and variables are involved?
5. How would you solve for the current $I$ in Ohm’s law if $V$ and $R$ are known?

Tip: Always start by simplifying inside parentheses first before distributing or solving any expressions!