Math worksheet with questions on distributivity, reciprocals, and number line representation.

Problem 1: Find using distributivity

$$
\left[\frac{9}{16} \times \frac{4}{12}\right] + \left[\frac{9}{16} \times \frac{-3}{9}\right]
$$

#### Solution:
We will use the distributive property of multiplication over addition, which states:
$$
a \times (b + c) = a \times b + a \times c
$$

Here, we can factor out $\frac{9}{16}$ from both terms:
$$
\left[\frac{9}{16} \times \frac{4}{12}\right] + \left[\frac{9}{16} \times \frac{-3}{9}\right] = \frac{9}{16} \times \left(\frac{4}{12} + \frac{-3}{9}\right)
$$

#### Step 1: Simplify the fractions inside the parentheses.
- Simplify $\frac{4}{12}$:
$$
\frac{4}{12} = \frac{1}{3}
$$

- Simplify $\frac{-3}{9}$:
$$
\frac{-3}{9} = \frac{-1}{3}
$$

So, the expression becomes:
$$
\frac{9}{16} \times \left(\frac{1}{3} + \frac{-1}{3}\right)
$$

#### Step 2: Add the fractions inside the parentheses.
$$
\frac{1}{3} + \frac{-1}{3} = \frac{1 - 1}{3} = \frac{0}{3} = 0
$$

#### Step 3: Multiply by $\frac{9}{16}$.
$$
\frac{9}{16} \times 0 = 0
$$

#### Final Answer:
$$
\boxed{0}
$$

---

Problem 2: Using appropriate properties, find

$$
\frac{2}{5} \times \left(\frac{-3}{7}\right) - \frac{1}{6} \times \frac{3}{2} + \frac{1}{14} \times \frac{2}{5}
$$

#### Solution:
We will simplify each term step by step.

#### Step 1: Simplify each multiplication.
1. $\frac{2}{5} \times \left(\frac{-3}{7}\right)$:
$$
\frac{2}{5} \times \left(\frac{-3}{7}\right) = \frac{2 \times (-3)}{5 \times 7} = \frac{-6}{35}
$$

2. $\frac{1}{6} \times \frac{3}{2}$:
$$
\frac{1}{6} \times \frac{3}{2} = \frac{1 \times 3}{6 \times 2} = \frac{3}{12} = \frac{1}{4}
$$

3. $\frac{1}{14} \times \frac{2}{5}$:
$$
\frac{1}{14} \times \frac{2}{5} = \frac{1 \times 2}{14 \times 5} = \frac{2}{70} = \frac{1}{35}
$$

#### Step 2: Substitute the simplified terms back into the expression.
$$
\frac{-6}{35} - \frac{1}{4} + \frac{1}{35}
$$

#### Step 3: Combine the fractions.
To combine these fractions, we need a common denominator. The denominators are $35$, $4$, and $35$. The least common multiple (LCM) of $35$ and $4$ is $140$.

1. Convert $\frac{-6}{35}$ to a fraction with denominator $140$:
$$
\frac{-6}{35} = \frac{-6 \times 4}{35 \times 4} = \frac{-24}{140}
$$

2. Convert $\frac{1}{4}$ to a fraction with denominator $140$:
$$
\frac{1}{4} = \frac{1 \times 35}{4 \times 35} = \frac{35}{140}
$$

3. Convert $\frac{1}{35}$ to a fraction with denominator $140$:
$$
\frac{1}{35} = \frac{1 \times 4}{35 \times 4} = \frac{4}{140}
$$

Now, substitute these back into the expression:
$$
\frac{-24}{140} - \frac{35}{140} + \frac{4}{140}
$$

Combine the numerators:
$$
\frac{-24 - 35 + 4}{140} = \frac{-55}{140}
$$

Simplify the fraction:
$$
\frac{-55}{140} = \frac{-11}{28}
$$

#### Final Answer:
$$
\boxed{\frac{-11}{28}}
$$

---

Problem 3: Write

#### Part (a): The rational number that does not have a reciprocal.
A rational number does not have a reciprocal if it is equal to zero. This is because division by zero is undefined.

#### Answer:
$$
\boxed{0}
$$

#### Part (b): The rational numbers that are equal to their reciprocals.
A rational number $x$ is equal to its reciprocal if:
$$
x = \frac{1}{x}
$$
Multiplying both sides by $x$ (assuming $x \neq 0$):
$$
x^2 = 1
$$
Taking the square root of both sides:
$$
x = \pm 1
$$

#### Answer:
$$
\boxed{1 \text{ and } -1}
$$

#### Part (c): The rational number that is equal to its negative.
A rational number $x$ is equal to its negative if:
$$
x = -x
$$
Adding $x$ to both sides:
$$
2x = 0
$$
Dividing by 2:
$$
x = 0
$$

#### Answer:
$$
\boxed{0}
$$

---

Problem 4: Fill in the blanks

#### Part (a): Zero has ________ reciprocal.
Zero does not have a reciprocal because division by zero is undefined.

#### Answer:
$$
\boxed{\text{no}}
$$

#### Part (b): The reciprocal of $-5$ is ________.
The reciprocal of a number $x$ is $\frac{1}{x}$. For $-5$:
$$
\text{Reciprocal of } -5 = \frac{1}{-5} = -\frac{1}{5}
$$

#### Answer:
$$
\boxed{-\frac{1}{5}}
$$

#### Part (c): The reciprocal of $\frac{1}{x}$, where $x \neq 0$, is ________.
The reciprocal of $\frac{1}{x}$ is:
$$
\text{Reciprocal of } \frac{1}{x} = x
$$

#### Answer:
$$
\boxed{x}
$$

#### Part (d): The product of two rational numbers is always a ________.
The product of two rational numbers is always a rational number because the set of rational numbers is closed under multiplication.

#### Answer:
$$
\boxed{\text{rational number}}
$$

#### Part (e): The reciprocal of a positive rational number is ________.
The reciprocal of a positive rational number is also positive because the reciprocal of a positive number is obtained by dividing 1 by that number, which remains positive.

#### Answer:
$$
\boxed{\text{positive}}
$$

---

Problem 5: Represent these numbers on the number line

#### Part (a): $\frac{7}{4}$
$\frac{7}{4} = 1.75$. On the number line, this is located between 1 and 2, closer to 2.

#### Part (b): $\frac{-5}{6}$
$\frac{-5}{6} \approx -0.833$. On the number line, this is located between 0 and -1, closer to -1.

#### Answer:
$$
\boxed{\text{Represent } \frac{7}{4} \text{ at 1.75 and } \frac{-5}{6} \text{ at -0.833 on the number line.}}
$$

---

Final Summary of Answers:

1. $\boxed{0}$
2. $\boxed{\frac{-11}{28}}$
3. (a) $\boxed{0}$, (b) $\boxed{1 \text{ and } -1}$, (c) $\boxed{0}$
4. (a) $\boxed{\text{no}}$, (b) $\boxed{-\frac{1}{5}}$, (c) $\boxed{x}$, (d) $\boxed{\text{rational number}}$, (e) $\boxed{\text{positive}}$
5. $\boxed{\text{Represent } \frac{7}{4} \text{ at 1.75 and } \frac{-5}{6} \text{ at -0.833 on the number line.}}$