strengthening concepts written practice 1. convert the improper fraction $\frac{29}{8}$ to a mixed number. remember to reduce the fraction part of the mixed number. 2. a fathom is 6 feet. how many feet deep is water that is $2\frac{1}{2}$ fathoms deep? 3. in 3 hours 3769 cars passed through a tollbooth. what is the average number of cars that pass through the tollbooth per hour? * 4. $5\frac{1}{2} + 2\frac{4}{5}$ * 6. $5\frac{1}{3} - 1\frac{2}{2}$ * 6. $5\frac{1}{2} + 2\frac{3}{5} + 3\frac{1}{4}$ * 7. $3\frac{9}{4} + 3\frac{1}{5}$ 8. compare a. $\frac{3}{5} \circ \frac{3}{8}$ b. $4^2 \circ \sqrt{144}$ 9. $\frac{5}{8} \div 6^2$ 10. $\frac{3}{8} \div \frac{2}{3}$ 11. how many $\frac{3}{8}$s are in $3\frac{3}{4}$? 12. $(4 - 0.4) \div 4$ 13. $4 - (0.4 \div 4)$ 14. which digit in 49.63 has the same place value as the 7 in 8.77? 15. estimate: find the sum of \$642.23 and \$861.17 to the nearest hundred dollars. explain how you arrived at your answer. 16. elizabeth used a compass to draw a circle with a radius of 4 cm. a. what was the diameter of the circle? describe how the radius and diameter are related. b. what was the circumference of the circle? (use 3.14 for $\pi$) * 17. predict: what is the next number in this sequence? $\dots, 100, 10, 1, \dots$ 18. the perimeter of a square is 1 foot. how many square inches cover its area? 19. connect: what is the ratio of the value of a dime to the value of a quarter? find each unknown number. 20. $15m = 3 \cdot 10^2$ 21. $\frac{1}{10} = \frac{n}{100}$ 22. by what fraction name for 1 must $\frac{5}{8}$ be multiplied to form a fraction with a denominator of 15? 23. what time is 5 hours 15 minutes after 9:50 a.m.? saxon math course 1

Question

Accepted Answer

Let's solve some of these problems one by one. We'll start with problem 16a as it's about circles, radius, and diameter.

### Problem 16a
#### Explanation:
## Step 1: Recall the relationship between radius and diameter
The diameter $ d $ of a circle is twice the radius $ r $, so the formula is $ d = 2r $.
## Step 2: Substitute the given radius
Given that the radius $ r = 4 $ cm, we substitute into the formula: $ d = 2\times4 = 8 $ cm. The radius is a line segment from the center of the circle to any point on the circle, and the diameter is a line segment that passes through the center and has endpoints on the circle, so the diameter is twice as long as the radius.
# Answer:
The diameter of the circle is 8 cm. The radius is half of the diameter (or the diameter is twice the radius), so if the radius is 4 cm, multiplying by 2 gives the diameter.

### Problem 16b
#### Explanation:
## Step 1: Recall the formula for the circumference of a circle
The circumference $ C $ of a circle can be calculated using the formula $ C = \pi d $ (or $ C = 2\pi r $). We are given to use $ \pi = 3.14 $ and we found the diameter $ d = 8 $ cm from part (a).
## Step 2: Substitute the values into the formula
Substitute $ d = 8 $ cm and $ \pi = 3.14 $ into $ C = \pi d $: $ C = 3.14\times8 $.
## Step 3: Calculate the product
$ 3.14\times8 = 25.12 $ cm.
# Answer:
The circumference of the circle is 25.12 cm. Using the formula $ C=\pi d $ with $ d = 8 $ cm (from part a) and $ \pi = 3.14 $, we multiply to get the circumference.

### Problem 17 (Sequence)
#### Explanation:
## Step 1: Identify the pattern in the sequence
Looking at the sequence $ 100, 10, 1, \dots $, we can see that each term is divided by 10 to get the next term. $ 100\div10 = 10 $, $ 10\div10 = 1 $.
## Step 2: Find the next term
To find the next term, we divide the last given term (1) by 10: $ 1\div10 = 0.1 $ (or $ \frac{1}{10} $).
# Answer:
The next number in the sequence is 0.1 (or $ \frac{1}{10} $). The pattern is dividing each term by 10.

### Problem 18 (Perimeter and Area of a Square)
#### Explanation:
## Step 1: Recall the formula for the perimeter of a square
The perimeter $ P $ of a square is $ P = 4s $, where $ s $ is the length of a side. We know the perimeter is 1 foot. First, we need to find the length of a side in inches because we need the area in square inches.
## Step 2: Convert feet to inches and find the side length
Since 1 foot = 12 inches, and $ P = 4s = 12 $ inches (because the perimeter is 1 foot = 12 inches). Solving for $ s $: $ s=\frac{12}{4}=3 $ inches.
## Step 3: Recall the formula for the area of a square
The area $ A $ of a square is $ A = s^2 $. Substituting $ s = 3 $ inches: $ A = 3^2 = 9 $ square inches.
# Answer:
9 square inches cover the area. The perimeter of 1 foot (12 inches) gives a side length of 3 inches, and the area is $ 3^2 = 9 $ square inches.

### Problem 19 (Ratio of a Dime to a Quarter)
#### Explanation:
## Step 1: Recall the values of a dime and a quarter
A dime is worth 10 cents, and a quarter is worth 25 cents.
## Step 2: Find the ratio of the value of a dime to a quarter
The ratio is $ \frac{\text{Value of Dime}}{\text{Value of Quarter}}=\frac{10}{25} $. Simplify this fraction by dividing numerator and denominator by 5: $ \frac{10\div5}{25\div5}=\frac{2}{5} $.
# Answer:
The ratio of the value of a dime to the value of a quarter is $ \frac{2}{5} $. A dime is 10 cents and a quarter is 25 cents, so the ratio is $ \frac{10}{25}=\frac{2}{5} $.

### Problem 20 ($ 15m = 3\times10^2 $)
#### Explanation:
## Step 1: Simplify the right - hand side
First, calculate $ 3\times10^2 $. $ 10^2 = 100 $, so $ 3\times100 = 300 $. The equation becomes $ 15m = 300 $.
## Step 2: Solve for $ m $
Divide both sides of the equation by 15: $ m=\frac{300}{15}=20 $.
# Answer:
$ m = 20 $. We first simplify $ 3\times10^2 = 300 $, then divide both sides of $ 15m = 300 $ by 15.

### Problem 21 ($ \frac{1}{10}=\frac{n}{100} $)
#### Explanation:
## Step 1: Cross - multiply to solve for $ n $
If $ \frac{1}{10}=\frac{n}{100} $, then cross - multiplying gives $ 10\times n = 1\times100 $.
## Step 2: Solve for $ n $
Simplify the equation: $ 10n = 100 $. Divide both sides by 10: $ n=\frac{100}{10}=10 $.
# Answer:
$ n = 10 $. By cross - multiplying $ \frac{1}{10}=\frac{n}{100} $ (i.e., $ 10n = 100 $) and then dividing by 10, we find $ n = 10 $.

### Problem 22 (Fraction Multiplication)
#### Explanation:
## Step 1: Let the fraction be $ x $
We know that $ \frac{2}{3}\times x=\frac{y}{15} $, and we want the denominator to be 15. First, find what we multiply 3 (the denominator of $ \frac{2}{3} $) by to get 15. Let that number be $ k $, so $ 3\times k = 15 $, so $ k = 5 $.
## Step 2: Multiply the numerator and denominator by $ k $
To keep the fraction equivalent, we multiply both the numerator and denominator of $ \frac{2}{3} $ by 5. So $ \frac{2\times5}{3\times5}=\frac{10}{15} $. So the fraction $ \frac{2}{3} $ must be multiplied by $ \frac{5}{5} $ (which is 1, so it's equivalent) but actually, we can think of it as: let $ \frac{2}{3}\times x=\frac{a}{15} $, and since $ 3\times5 = 15 $, we multiply numerator and denominator by 5, so $ x=\frac{5}{5} $? Wait, no. Wait, the question is: "By what fraction name for 1 must $ \frac{2}{3} $ be multiplied to form a fraction with a denominator of 15?" A fraction name for 1 is a fraction where numerator and denominator are equal, like $ \frac{5}{5} $, $ \frac{6}{6} $, etc. To get from denominator 3 to 15, we multiply by 5, so we multiply $ \frac{2}{3} $ by $ \frac{5}{5} $ (since $ \frac{5}{5}=1 $) to get $ \frac{2\times5}{3\times5}=\frac{10}{15} $. So the fraction is $ \frac{5}{5} $, but we can say the fraction is $ \frac{5}{5} $ (or just 5/5, but as a fraction name for 1, it's $ \frac{5}{5} $).
# Answer:
The fraction is $ \frac{5}{5} $. To get a denominator of 15 from 3, we multiply by 5, so we use the fraction name for 1, $ \frac{5}{5} $, to multiply $ \frac{2}{3} $ and get $ \frac{10}{15} $.

### Problem 23 (Time Calculation)
#### Explanation:
## Step 1: Add the hours and minutes separately
We have 5 hours and 15 minutes to add to 9:50 a.m. First, add the minutes: 50 minutes+15 minutes = 65 minutes.
## Step 2: Convert 65 minutes to hours and minutes
Since 60 minutes = 1 hour, 65 minutes = 1 hour and 5 minutes.
## Step 3: Add the hours
Now add the hours: 9 hours+5 hours+1 hour (from the 65 minutes) = 15 hours. But 15 hours in 12 - hour time is 3 p.m. (since 15 - 12 = 3), and we have 5 minutes left. Wait, wait, let's do it step by step. Start with 9:50 a.m. Add 5 hours: 9 + 5 = 14, so 14:50, which is 2:50 p.m. Then add 15 minutes: 50+15 = 65 minutes = 1 hour and 5 minutes. So 2 + 1 = 3 hours, and 5 minutes. So 3:05 p.m. Wait, no, let's correct. 9:50 a.m. Add 5 hours: 9 + 5 = 14, so 14:50 (which is 2:50 p.m.). Then add 15 minutes: 50+15 = 65 minutes. 65 minutes is 1 hour and 5 minutes. So 2:50 p.m. + 1 hour = 3:50 p.m.? Wait, no, I messed up. Wait, 9:50 a.m. + 5 hours: 9 + 5 = 14, 50 minutes, so 14:50 (2:50 p.m.). Then +15 minutes: 50 + 15 = 65 minutes. 65 minutes - 60 minutes = 5 minutes, and 14 + 1 = 15 hours. 15 hours is 3 p.m. (15 - 12 = 3) and 5 minutes. So 3:05 p.m.? Wait, no, 9:50 a.m. + 5 hours: 9 + 5 = 14, 50 minutes: 14:50 (2:50 p.m.). Then +15 minutes: 50 + 15 = 65 minutes. 65 minutes is 1 hour and 5 minutes. So 2:50 p.m. + 1 hour = 3:50 p.m.? No, 2:50 p.m. + 15 minutes: 50 + 15 = 65, so 65 minutes is 1 hour and 5 minutes, so 2 + 1 = 3 hours, 5 minutes. So 3:05 p.m.? Wait, no, let's use a better method. 9:50 a.m. Add 15 minutes first: 9:50 + 0:15 = 10:05 a.m. Then add 5 hours: 10 + 5 = 15, so 15:05, which is 3:05 p.m. Ah, that's better. So first add the 15 minutes to 9:50 a.m.: 50 + 15 = 65 minutes = 1 hour and 5 minutes, so 9 + 1 = 10, 5 minutes: 10:05 a.m. Then add 5 hours: 10 + 5 = 15, so 15:05, which is 3:05 p.m.
# Answer:
The time is 3:05 p.m. (or 15:05). First, add 15 minutes to 9:50 a.m. to get 10:05 a.m., then add 5 hours to get 3:05 p.m.

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QUESTION IMAGE

Question

Explanation:

Problem 16a

Step 1: Recall the relationship between radius and diameter

Step 2: Substitute the given radius

Step 1: Recall the formula for the circumference of a circle

Step 2: Substitute the values into the formula

Step 3: Calculate the product

Step 1: Identify the pattern in the sequence

Step 2: Find the next term

Answer:

Problem 16b