5-A-Day Math Review: Week 2 worksheet featuring Monday and Tuesday math exercises.

Let's solve the problems step by step.

---

Monday

#### Problem 1: Write and solve an expression for the following situation
"7 more than the product of 6 and 9"

- Expression: The product of 6 and 9 is \(6 \times 9\). Adding 7 to this product gives:
\[
6 \times 9 + 7
\]
- Solution:
\[
6 \times 9 = 54
\]
\[
54 + 7 = 61
\]
So, the answer is:
\[
\boxed{61}
\]

#### Problem 2: Model and solve the problem
\[ 0.2 \times 0.3 \]

- Model: Use a grid to represent the multiplication of decimals.
- A 10x10 grid represents 1 whole.
- Shade 2 rows (representing 0.2) and 3 columns (representing 0.3).
- The overlapping shaded area represents \(0.2 \times 0.3\).

- Solution:
\[
0.2 \times 0.3 = 0.06
\]
So, the answer is:
\[
\boxed{0.06}
\]

#### Problem 3: Solve using area models
- Part 1: \(\frac{1}{2}\)
- Draw a rectangle and divide it into 2 equal parts. Shade 1 part.
- This represents \(\frac{1}{2}\).

- Part 2: \(\frac{1}{4}\)
- Draw a rectangle and divide it into 4 equal parts. Shade 1 part.
- This represents \(\frac{1}{4}\).

#### Problem 4: Which number is equal to \(10^3\)?
- Solution:
\[
10^3 = 10 \times 10 \times 10 = 1,000
\]
So, the answer is:
\[
\boxed{C}
\]

#### Problem 5: Convert units
- Part 1: \(2 \text{ m} = \_\_\_\_ \text{ cm}\)
- 1 meter = 100 centimeters.
- Therefore:
\[
2 \text{ m} = 2 \times 100 = 200 \text{ cm}
\]
- Expression:
\[
2 \times 100 = 200
\]

- Part 2: \(3,000 \text{ g} = \_\_\_\_ \text{ kg}\)
- 1 kilogram = 1,000 grams.
- Therefore:
\[
3,000 \text{ g} = 3,000 \div 1,000 = 3 \text{ kg}
\]
- Expression:
\[
3,000 \div 1,000 = 3
\]

So, the answers are:
\[
\boxed{200 \text{ cm}, 3 \text{ kg}}
\]

---

Tuesday

#### Problem 1: Multiply the numbers
- Part 1: \(59 \times 79\)
- Use the standard multiplication algorithm:
\[
\begin{array}{r}
59 \\
\times 79 \\
\hline
531 \quad (\text{from } 59 \times 9) \\
4130 \quad (\text{from } 59 \times 70, \text{ shifted one place to the left}) \\
\hline
4661 \\
\end{array}
\]
- So:
\[
59 \times 79 = 4661
\]

- Part 2: \(39 \times 73\)
- Use the standard multiplication algorithm:
\[
\begin{array}{r}
39 \\
\times 73 \\
\hline
117 \quad (\text{from } 39 \times 3) \\
2730 \quad (\text{from } 39 \times 70, \text{ shifted one place to the left}) \\
\hline
2847 \\
\end{array}
\]
- So:
\[
39 \times 73 = 2847
\]

So, the answers are:
\[
\boxed{4661, 2847}
\]

#### Problem 2: Complete the table and then graph the coordinates
- Table:
- Given: \(x + 1\) and \(y + 2\).
- Choose values for \(x\) and calculate corresponding \(y\):
\[
\begin{array}{c|c}
x & y \\
\hline
1 & 3 \\
2 & 4 \\
3 & 5 \\
4 & 6 \\
\end{array}
\]
- Graph these points on the coordinate plane.

#### Problem 3: Model and solve the problem
- Problem: \(\frac{1}{2}\) of 6
- Draw 6 bears and shade half of them.
- Half of 6 is:
\[
\frac{1}{2} \times 6 = 3
\]
- So, the answer is:
\[
\boxed{3}
\]

#### Problem 4: Order from least to greatest
- Numbers: \(0.4, 0.5, 0.45, 0.04\)
- Order:
\[
0.04 < 0.4 < 0.45 < 0.5
\]
So, the answer is:
\[
\boxed{0.04, 0.4, 0.45, 0.5}
\]

#### Problem 5: Solve. Show your work
- Part 1: \(7.4 + 1.2\)
- Add the numbers:
\[
7.4 + 1.2 = 8.6
\]
- So:
\[
\boxed{8.6}
\]

- Part 2: \(0.93 - 0.77\)
- Subtract the numbers:
\[
0.93 - 0.77 = 0.16
\]
- So:
\[
\boxed{0.16}
\]

---

Final Answers

- Monday:
1. \(\boxed{61}\)
2. \(\boxed{0.06}\)
3. Area models for \(\frac{1}{2}\) and \(\frac{1}{4}\)
4. \(\boxed{C}\)
5. \(\boxed{200 \text{ cm}, 3 \text{ kg}}\)

- Tuesday:
1. \(\boxed{4661, 2847}\)
2. Table and graph
3. \(\boxed{3}\)
4. \(\boxed{0.04, 0.4, 0.45, 0.5}\)
5. \(\boxed{8.6, 0.16}\)