9. write each group of numbers in order from least to greatest:
a) 0.4, 0.4\overline{2}, 0.\overline{42}, 0.42
b) 0.16, 0.\overline{1}, 0.1\overline{6}, 0.\overline{16}
c) 0.387, 0. 38\overline{7}, 0. \overline{387}, 0. \overline{387}
d) 0.546, 0. 54\overline{6}, 0. \overline{546}, 0. \overline{546}
e) 0.383, 0. 38\overline{3}, 0. \overline{383}, 0. \overline{383}
f) 0.786, 0. 78\overline{6}, 0. \overline{786}, 0. \overline{786}
10. find the decimal representations of the fractions by long division:
a) $\frac{1}{11}$
b) $\frac{2}{11}$
c) $\frac{3}{11}$
d) $\frac{4}{11}$
11. use the pattern you found in question 10 to find...
a) $\frac{5}{11}$
b) $\frac{6}{11}$
c) $\frac{7}{11}$
d) $\frac{8}{11}$
e) $\frac{9}{11}$
12. bilal wants to represent $\frac{1}{11}$ as a decimal using base ten materials. what problem will he run into?
13. explain why $.6 \leq.\overline{6} \leq.7$.

Question

Accepted Answer

### Question 10a: Find the decimal representation of $\boldsymbol{\frac{1}{11}}$ by long division.

# Explanation:
## Step 1: Set up the long division
We divide 1 by 11. Since 1 is less than 11, we write it as 1.0000... and divide.
$$1 \div 11$$

## Step 2: Perform the division
11 goes into 10 zero times, so we write 0 above the decimal point. Then we consider 100 (bring down a 0). 11 goes into 100 nine times (9×11 = 99). Subtract 99 from 100, we get 1.
$$0.09...$$
Now we have a remainder of 1, which is the same as the original dividend. So the division will repeat.

## Step 3: Identify the repeating pattern
Since we get a remainder of 1 again, the decimal will repeat the 09. So $\frac{1}{11} = 0.\overline{09}$.

# Answer:
$0.\overline{09}$

### Question 10b: Find the decimal representation of $\boldsymbol{\frac{2}{11}}$ by long division.

# Explanation:
## Step 1: Set up the long division
Divide 2 by 11. 2 is less than 11, so we write it as 2.0000...
$$2 \div 11$$

## Step 2: Perform the division
11 goes into 20 once (1×11 = 11). Subtract 11 from 20, we get 9. Bring down a 0 to make 90. 11 goes into 90 eight times (8×11 = 88). Subtract 88 from 90, we get 2.
$$0.18...$$
Now we have a remainder of 2, which is the original dividend. So the division will repeat.

## Step 3: Identify the repeating pattern
Since the remainder is 2 again, the decimal will repeat 18. So $\frac{2}{11} = 0.\overline{18}$.

# Answer:
$0.\overline{18}$

### Question 10c: Find the decimal representation of $\boldsymbol{\frac{3}{11}}$ by long division.

# Explanation:
## Step 1: Set up the long division
Divide 3 by 11. 3 is less than 11, so we write it as 3.0000...
$$3 \div 11$$

## Step 2: Perform the division
11 goes into 30 two times (2×11 = 22). Subtract 22 from 30, we get 8. Bring down a 0 to make 80. 11 goes into 80 seven times (7×11 = 77). Subtract 77 from 80, we get 3.
$$0.27...$$
Now we have a remainder of 3, which is the original dividend. So the division will repeat.

## Step 3: Identify the repeating pattern
Since the remainder is 3 again, the decimal will repeat 27. So $\frac{3}{11} = 0.\overline{27}$.

# Answer:
$0.\overline{27}$

### Question 10d: Find the decimal representation of $\boldsymbol{\frac{4}{11}}$ by long division.

# Explanation:
## Step 1: Set up the long division
Divide 4 by 11. 4 is less than 11, so we write it as 4.0000...
$$4 \div 11$$

## Step 2: Perform the division
11 goes into 40 three times (3×11 = 33). Subtract 33 from 40, we get 7. Bring down a 0 to make 70. 11 goes into 70 six times (6×11 = 66). Subtract 66 from 70, we get 4.
$$0.36...$$
Now we have a remainder of 4, which is the original dividend. So the division will repeat.

## Step 3: Identify the repeating pattern
Since the remainder is 4 again, the decimal will repeat 36. So $\frac{4}{11} = 0.\overline{36}$.

# Answer:
$0.\overline{36}$

### Question 11a: Use the pattern from Question 10 to find the decimal representation of $\boldsymbol{\frac{5}{11}}$.

# Explanation:
From Question 10, we see the pattern:
- $\frac{1}{11} = 0.\overline{09}$
- $\frac{2}{11} = 0.\overline{18}$
- $\frac{3}{11} = 0.\overline{27}$
- $\frac{4}{11} = 0.\overline{36}$

The repeating part increases by 09 each time. So for $\frac{5}{11}$, we add 09 to the repeating part of $\frac{4}{11}$ (36 + 09 = 45). So $\frac{5}{11} = 0.\overline{45}$.

# Answer:
$0.\overline{45}$

### Question 11b: Use the pattern from Question 10 to find the decimal representation of $\boldsymbol{\frac{6}{11}}$.

# Explanation:
Using the pattern from Question 10:
- $\frac{5}{11} = 0.\overline{45}$
- So for $\frac{6}{11}$, we add 09 to 45 (45 + 09 = 54). Thus, $\frac{6}{11} = 0.\overline{54}$.

# Answer:
$0.\overline{54}$

### Question 11c: Use the pattern from Question 10 to find the decimal representation of $\boldsymbol{\frac{7}{11}}$.

# Explanation:
Using the pattern:
- $\frac{6}{11} = 0.\overline{54}$
- Add 09 to 54 (54 + 09 = 63). So $\frac{7}{11} = 0.\overline{63}$.

# Answer:
$0.\overline{63}$

### Question 11d: Use the pattern from Question 10 to find the decimal representation of $\boldsymbol{\frac{8}{11}}$.

# Explanation:
Using the pattern:
- $\frac{7}{11} = 0.\overline{63}$
- Add 09 to 63 (63 + 09 = 72). So $\frac{8}{11} = 0.\overline{72}$.

# Answer:
$0.\overline{72}$

### Question 11e: Use the pattern from Question 10 to find the decimal representation of $\boldsymbol{\frac{9}{11}}$.

# Explanation:
Using the pattern:
- $\frac{8}{11} = 0.\overline{72}$
- Add 09 to 72 (72 + 09 = 81). So $\frac{9}{11} = 0.\overline{81}$.

# Answer:
$0.\overline{81}$

### Question 12: Bilal wants to represent $\boldsymbol{\frac{1}{11}}$ as a decimal using base ten materials. What problem will he run into?

# Brief Explanations:
$\frac{1}{11}$ as a decimal is a repeating decimal ($0.\overline{09}$), which means it has an infinite number of digits that repeat indefinitely. Base ten materials (like base ten blocks) are designed to represent finite decimals or whole numbers. Since the decimal for $\frac{1}{11}$ never terminates and repeats, Bilal cannot represent it exactly using base ten materials because they can only represent a finite number of decimal places.

# Answer:
$\frac{1}{11}$ is a repeating decimal ($0.\overline{09}$) with an infinite number of repeating digits, so Bilal cannot represent it exactly using base ten materials (which represent finite quantities).

### Question 13: Explain why $\boldsymbol{.6 \leq. \overline{6} \leq.7}$.

# Brief Explanations:
First, $0.6$ is a finite decimal. $0.\overline{6}$ is a repeating decimal where the 6 repeats indefinitely, so $0.\overline{6} = 0.666...$. Now, compare the numbers:
- $0.6$ is less than $0.666...$ because the tenths place is the same (6), but the hundredths place of $0.6$ is 0 and of $0.\overline{6}$ is 6 (so $0.6 < 0.\overline{6}$).
- $0.666...$ is less than $0.7$ because the tenths place of $0.666...$ is 6 and of $0.7$ is 7 (so $0.\overline{6} < 0.7$).

Combining these, we get $0.6 \leq 0.\overline{6} \leq 0.7$.

# Answer:
$0.6 = 0.600...$, $0.\overline{6} = 0.666...$, and $0.7 = 0.700...$. Since $0.600... \leq 0.666... \leq 0.700...$, we have $0.6 \leq 0.\overline{6} \leq 0.7$.

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QUESTION IMAGE

Question

Explanation:

Question 10a: Find the decimal representation of $\boldsymbol{\frac{1}{11}}$ by long division.

Step 1: Set up the long division

Step 2: Perform the division

Step 3: Identify the repeating pattern

Step 1: Set up the long division

Step 2: Perform the division

Step 3: Identify the repeating pattern

Step 1: Set up the long division

Step 2: Perform the division

Step 3: Identify the repeating pattern

Answer:

Question 10b: Find the decimal representation of $\boldsymbol{\frac{2}{11}}$ by long division.