Answer :
Sure! Let's solve the problems step by step.
### Problem 1
[tex]\[ \frac{1}{4} = \frac{0}{12} \][/tex]
Here, [tex]\(\frac{0}{12}\)[/tex] simplifies to [tex]\(0\)[/tex]. Since [tex]\(\frac{1}{4}\)[/tex] does not equal [tex]\(0\)[/tex], this proportion does not have a solution.
### Problem 2
[tex]\[ \frac{14}{y} = \frac{7}{2} \][/tex]
To solve for [tex]\(y\)[/tex], we can use cross multiplication:
[tex]\[ 14 \times 2 = 7 \times y \][/tex]
[tex]\[ 28 = 7y \][/tex]
Now, divide both sides by 7:
[tex]\[ y = \frac{28}{7} \][/tex]
[tex]\[ y = 4 \][/tex]
So, [tex]\( y = 4 \)[/tex].
### Problem 3
[tex]\[ \frac{5}{3} = \frac{8}{2} \][/tex]
Simplify [tex]\(\frac{8}{2}\)[/tex] to [tex]\(4\)[/tex]:
[tex]\[ \frac{5}{3} \neq 4 \][/tex]
Since [tex]\( \frac{5}{3} \approx 1.67 \)[/tex] and [tex]\( 4 \)[/tex], the two fractions are not equal. Therefore, this proportion does not have a solution.
### Problem 6
You are 5 feet 4 inches tall, and your shadow is 3 feet long. A nearby water tower casts a shadow 74 feet 3 inches long.
#### a. Find the height of the water tower.
First, convert the heights and shadow lengths to feet as a single unit.
[tex]\[ 5 \text{ feet } 4 \text{ inches} = 5 \frac{4}{12} \text{ feet} = 5.33 \text{ feet} \][/tex]
[tex]\[ 74 \text{ feet } 3 \text{ inches} = 74 \frac{3}{12} \text{ feet} = 74.25 \text{ feet} \][/tex]
The ratios of the heights to shadows are equal, so we set up the proportion:
[tex]\[ \frac{\text{Height of water tower}}{74.25} = \frac{5.33}{3} \][/tex]
Cross-multiply to solve for the height of the water tower:
[tex]\[ \text{Height of water tower} = \frac{5.33}{3} \times 74.25 \][/tex]
[tex]\[ \text{Height of water tower} = 1.7767 \times 74.25 \][/tex]
[tex]\[ \text{Height of water tower} \approx 132 \text{ feet} \][/tex]
#### b. Estimate the water pressure at the base of the tower.
Each additional foot in tower height increases the water pressure at the base of the tower by 0.43 pounds per square inch.
So, we calculate the total increase in water pressure:
[tex]\[ \text{Water pressure increase} = 132 \times 0.43 \][/tex]
[tex]\[ \text{Water pressure increase} \approx 56.76 \text{ pounds per square inch} \][/tex]
### Summary
1. No solution for [tex]\(\frac{1}{4} = \frac{0}{12}\)[/tex].
2. [tex]\(y\)[/tex] is 4 for [tex]\(\frac{14}{y} = \frac{7}{2}\)[/tex].
3. No solution for [tex]\(\frac{5}{3} = \frac{8}{2}\)[/tex].
4. Water tower height is approximately 132 feet.
5. Estimated water pressure increase at the base of the tower is approximately 56.76 pounds per square inch.
### Problem 1
[tex]\[ \frac{1}{4} = \frac{0}{12} \][/tex]
Here, [tex]\(\frac{0}{12}\)[/tex] simplifies to [tex]\(0\)[/tex]. Since [tex]\(\frac{1}{4}\)[/tex] does not equal [tex]\(0\)[/tex], this proportion does not have a solution.
### Problem 2
[tex]\[ \frac{14}{y} = \frac{7}{2} \][/tex]
To solve for [tex]\(y\)[/tex], we can use cross multiplication:
[tex]\[ 14 \times 2 = 7 \times y \][/tex]
[tex]\[ 28 = 7y \][/tex]
Now, divide both sides by 7:
[tex]\[ y = \frac{28}{7} \][/tex]
[tex]\[ y = 4 \][/tex]
So, [tex]\( y = 4 \)[/tex].
### Problem 3
[tex]\[ \frac{5}{3} = \frac{8}{2} \][/tex]
Simplify [tex]\(\frac{8}{2}\)[/tex] to [tex]\(4\)[/tex]:
[tex]\[ \frac{5}{3} \neq 4 \][/tex]
Since [tex]\( \frac{5}{3} \approx 1.67 \)[/tex] and [tex]\( 4 \)[/tex], the two fractions are not equal. Therefore, this proportion does not have a solution.
### Problem 6
You are 5 feet 4 inches tall, and your shadow is 3 feet long. A nearby water tower casts a shadow 74 feet 3 inches long.
#### a. Find the height of the water tower.
First, convert the heights and shadow lengths to feet as a single unit.
[tex]\[ 5 \text{ feet } 4 \text{ inches} = 5 \frac{4}{12} \text{ feet} = 5.33 \text{ feet} \][/tex]
[tex]\[ 74 \text{ feet } 3 \text{ inches} = 74 \frac{3}{12} \text{ feet} = 74.25 \text{ feet} \][/tex]
The ratios of the heights to shadows are equal, so we set up the proportion:
[tex]\[ \frac{\text{Height of water tower}}{74.25} = \frac{5.33}{3} \][/tex]
Cross-multiply to solve for the height of the water tower:
[tex]\[ \text{Height of water tower} = \frac{5.33}{3} \times 74.25 \][/tex]
[tex]\[ \text{Height of water tower} = 1.7767 \times 74.25 \][/tex]
[tex]\[ \text{Height of water tower} \approx 132 \text{ feet} \][/tex]
#### b. Estimate the water pressure at the base of the tower.
Each additional foot in tower height increases the water pressure at the base of the tower by 0.43 pounds per square inch.
So, we calculate the total increase in water pressure:
[tex]\[ \text{Water pressure increase} = 132 \times 0.43 \][/tex]
[tex]\[ \text{Water pressure increase} \approx 56.76 \text{ pounds per square inch} \][/tex]
### Summary
1. No solution for [tex]\(\frac{1}{4} = \frac{0}{12}\)[/tex].
2. [tex]\(y\)[/tex] is 4 for [tex]\(\frac{14}{y} = \frac{7}{2}\)[/tex].
3. No solution for [tex]\(\frac{5}{3} = \frac{8}{2}\)[/tex].
4. Water tower height is approximately 132 feet.
5. Estimated water pressure increase at the base of the tower is approximately 56.76 pounds per square inch.