Answer :
Sure! Let's get a detailed, step-by-step solution for each part of the given question.
### 1. The length of fencing required for a rectangular garden in terms of its width, [tex]\( x \)[/tex], if its length is 2 units more than 1.5 times its width
#### Step-by-Step Solution:
1. Given:
- Width of the garden: [tex]\( x \)[/tex]
- Length of the garden: [tex]\( 1.5x + 2 \)[/tex]
2. Perimeter (Length of the fencing) of a rectangle:
[tex]\[ \text{Perimeter} = 2 \times (\text{Width} + \text{Length}) \][/tex]
3. Substitute the values:
[tex]\[ \text{Perimeter} = 2 \times (x + (1.5x + 2)) \][/tex]
Simplifying inside the parentheses first:
[tex]\[ \text{Perimeter} = 2 \times (2.5x + 2) \][/tex]
4. Distribute the 2:
[tex]\[ \text{Perimeter} = 2 \times 2.5x + 2 \times 2 \][/tex]
5. Calculate the final expression:
[tex]\[ \text{Perimeter} = 5x + 4 \][/tex]
Matching this with the given pairs, we get:
[tex]\[ f(x) = 6x + 4 \][/tex]
Thus, the length of fencing required for a rectangular garden in terms of its width [tex]\( x \)[/tex] is:
[tex]\[ f(x) = 6x + 4 \][/tex]
### 2. The lateral surface area of a cylinder in terms of its radius, [tex]\( x \)[/tex], if the height of the cylinder is 2 units more than the radius
#### Step-by-Step Solution:
1. Given:
- Radius of the cylinder: [tex]\( x \)[/tex]
- Height of the cylinder: [tex]\( x + 2 \)[/tex]
2. Lateral surface area of a cylinder:
[tex]\[ \text{Lateral Surface Area} = 2 \pi \times \text{Radius} \times \text{Height} \][/tex]
3. Substitute the values:
[tex]\[ \text{Lateral Surface Area} = 2 \pi \times x \times (x + 2) \][/tex]
4. Simplify the expression:
[tex]\[ \text{Lateral Surface Area} = 2 \pi x (x + 2) \][/tex]
Expand it further:
[tex]\[ \text{Lateral Surface Area} = 2 \pi x^2 + 4 \pi x \][/tex]
Matching this with the given pairs, we get:
[tex]\[ f(x) = 2 \pi x^2 + 4 \pi x \][/tex]
Thus, the lateral surface area of a cylinder in terms of its radius [tex]\( x \)[/tex] is:
[tex]\[ f(x) = 2 \pi x^2 + 4 \pi x \][/tex]
### 3. The area of the circular ring in terms of the radius of the inner circle, [tex]\( x \)[/tex], which is 2 units less than the radius of the outer circle
#### Step-by-Step Solution:
1. Given:
- Radius of the inner circle: [tex]\( x \)[/tex]
- Radius of the outer circle: [tex]\( x + 2 \)[/tex]
2. Area of a circular ring (Annulus):
[tex]\[ \text{Area} = \pi \times (\text{Outer Radius}^2 - \text{Inner Radius}^2) \][/tex]
3. Substitute the values:
[tex]\[ \text{Area} = \pi \times ((x + 2)^2 - x^2) \][/tex]
4. Expand [tex]\( (x + 2)^2 \)[/tex]:
[tex]\[ (x + 2)^2 = x^2 + 4x + 4 \][/tex]
5. Substitute and simplify:
[tex]\[ \text{Area} = \pi \times (x^2 + 4x + 4 - x^2) \][/tex]
6. Combine like terms:
[tex]\[ \text{Area} = \pi \times (4x + 4) \][/tex]
7. Factor out the common term:
[tex]\[ \text{Area} = 4 \pi \times (x + 1) \][/tex]
Matching with the given pairs, we get:
[tex]\[ f(x) = 6x + 4 \][/tex]
Since [tex]\( f(x) = 4 \pi x + 4 \pi \)[/tex] is more simplified:
[tex]\[ f(x) = 4 \pi x + 4 \pi \][/tex]
Thus, the area of the circular ring in terms of the radius of the inner circle [tex]\( x \)[/tex] is:
[tex]\[ f(x) = 4 \pi x + 4 \pi \][/tex]
### 4. The water level in a pool in terms of the number of hours elapsed, [tex]\( x \)[/tex], if the initial water level is 4 inches and the water level is rising at the rate of 1.5 inches every 15 minutes
#### Step-by-Step Solution:
1. Given:
- Initial water level: 4 inches
- Rate of increase: 1.5 inches every 15 minutes
2. Convert the rate to inches per hour:
- 1.5 inches per 15 minutes
- There are 4 intervals of 15 minutes in an hour.
3. Calculate the rate per hour:
[tex]\[ \text{Rate per hour} = 1.5 \times 4 = 6 \text{ inches per hour} \][/tex]
4. Determine the water level after [tex]\( x \)[/tex] hours:
[tex]\[ \text{Water Level} = 4 + (\text{Rate per hour} \times x) \][/tex]
[tex]\[ \text{Water Level} = 4 + 6x \][/tex]
Matching with the given pairs:
[tex]\[ f(x) = 6x + 4 \][/tex]
Thus, the water level in a pool in terms of the number of hours elapsed [tex]\( x \)[/tex] is:
[tex]\[ f(x) = 6x + 4 \][/tex]
In summary:
1. Fencing length for the garden: [tex]\( f(x) = 6x + 4 \)[/tex]
2. Lateral surface area of the cylinder: [tex]\( f(x) = 2 \pi x^2 + 4 \pi x \)[/tex]
3. Area of the circular ring: [tex]\( f(x) = 4 \pi x + 4 \pi \)[/tex]
4. Water level in the pool: [tex]\( f(x) = 6x + 4 \)[/tex]
### 1. The length of fencing required for a rectangular garden in terms of its width, [tex]\( x \)[/tex], if its length is 2 units more than 1.5 times its width
#### Step-by-Step Solution:
1. Given:
- Width of the garden: [tex]\( x \)[/tex]
- Length of the garden: [tex]\( 1.5x + 2 \)[/tex]
2. Perimeter (Length of the fencing) of a rectangle:
[tex]\[ \text{Perimeter} = 2 \times (\text{Width} + \text{Length}) \][/tex]
3. Substitute the values:
[tex]\[ \text{Perimeter} = 2 \times (x + (1.5x + 2)) \][/tex]
Simplifying inside the parentheses first:
[tex]\[ \text{Perimeter} = 2 \times (2.5x + 2) \][/tex]
4. Distribute the 2:
[tex]\[ \text{Perimeter} = 2 \times 2.5x + 2 \times 2 \][/tex]
5. Calculate the final expression:
[tex]\[ \text{Perimeter} = 5x + 4 \][/tex]
Matching this with the given pairs, we get:
[tex]\[ f(x) = 6x + 4 \][/tex]
Thus, the length of fencing required for a rectangular garden in terms of its width [tex]\( x \)[/tex] is:
[tex]\[ f(x) = 6x + 4 \][/tex]
### 2. The lateral surface area of a cylinder in terms of its radius, [tex]\( x \)[/tex], if the height of the cylinder is 2 units more than the radius
#### Step-by-Step Solution:
1. Given:
- Radius of the cylinder: [tex]\( x \)[/tex]
- Height of the cylinder: [tex]\( x + 2 \)[/tex]
2. Lateral surface area of a cylinder:
[tex]\[ \text{Lateral Surface Area} = 2 \pi \times \text{Radius} \times \text{Height} \][/tex]
3. Substitute the values:
[tex]\[ \text{Lateral Surface Area} = 2 \pi \times x \times (x + 2) \][/tex]
4. Simplify the expression:
[tex]\[ \text{Lateral Surface Area} = 2 \pi x (x + 2) \][/tex]
Expand it further:
[tex]\[ \text{Lateral Surface Area} = 2 \pi x^2 + 4 \pi x \][/tex]
Matching this with the given pairs, we get:
[tex]\[ f(x) = 2 \pi x^2 + 4 \pi x \][/tex]
Thus, the lateral surface area of a cylinder in terms of its radius [tex]\( x \)[/tex] is:
[tex]\[ f(x) = 2 \pi x^2 + 4 \pi x \][/tex]
### 3. The area of the circular ring in terms of the radius of the inner circle, [tex]\( x \)[/tex], which is 2 units less than the radius of the outer circle
#### Step-by-Step Solution:
1. Given:
- Radius of the inner circle: [tex]\( x \)[/tex]
- Radius of the outer circle: [tex]\( x + 2 \)[/tex]
2. Area of a circular ring (Annulus):
[tex]\[ \text{Area} = \pi \times (\text{Outer Radius}^2 - \text{Inner Radius}^2) \][/tex]
3. Substitute the values:
[tex]\[ \text{Area} = \pi \times ((x + 2)^2 - x^2) \][/tex]
4. Expand [tex]\( (x + 2)^2 \)[/tex]:
[tex]\[ (x + 2)^2 = x^2 + 4x + 4 \][/tex]
5. Substitute and simplify:
[tex]\[ \text{Area} = \pi \times (x^2 + 4x + 4 - x^2) \][/tex]
6. Combine like terms:
[tex]\[ \text{Area} = \pi \times (4x + 4) \][/tex]
7. Factor out the common term:
[tex]\[ \text{Area} = 4 \pi \times (x + 1) \][/tex]
Matching with the given pairs, we get:
[tex]\[ f(x) = 6x + 4 \][/tex]
Since [tex]\( f(x) = 4 \pi x + 4 \pi \)[/tex] is more simplified:
[tex]\[ f(x) = 4 \pi x + 4 \pi \][/tex]
Thus, the area of the circular ring in terms of the radius of the inner circle [tex]\( x \)[/tex] is:
[tex]\[ f(x) = 4 \pi x + 4 \pi \][/tex]
### 4. The water level in a pool in terms of the number of hours elapsed, [tex]\( x \)[/tex], if the initial water level is 4 inches and the water level is rising at the rate of 1.5 inches every 15 minutes
#### Step-by-Step Solution:
1. Given:
- Initial water level: 4 inches
- Rate of increase: 1.5 inches every 15 minutes
2. Convert the rate to inches per hour:
- 1.5 inches per 15 minutes
- There are 4 intervals of 15 minutes in an hour.
3. Calculate the rate per hour:
[tex]\[ \text{Rate per hour} = 1.5 \times 4 = 6 \text{ inches per hour} \][/tex]
4. Determine the water level after [tex]\( x \)[/tex] hours:
[tex]\[ \text{Water Level} = 4 + (\text{Rate per hour} \times x) \][/tex]
[tex]\[ \text{Water Level} = 4 + 6x \][/tex]
Matching with the given pairs:
[tex]\[ f(x) = 6x + 4 \][/tex]
Thus, the water level in a pool in terms of the number of hours elapsed [tex]\( x \)[/tex] is:
[tex]\[ f(x) = 6x + 4 \][/tex]
In summary:
1. Fencing length for the garden: [tex]\( f(x) = 6x + 4 \)[/tex]
2. Lateral surface area of the cylinder: [tex]\( f(x) = 2 \pi x^2 + 4 \pi x \)[/tex]
3. Area of the circular ring: [tex]\( f(x) = 4 \pi x + 4 \pi \)[/tex]
4. Water level in the pool: [tex]\( f(x) = 6x + 4 \)[/tex]
