Answer :
Sure, let's tackle this problem step-by-step.
### Part (a): Describe the Domain and Range of the Function
The given function is:
[tex]\[ z = 8900d^2 \][/tex]
Where [tex]\( z \)[/tex] is the breaking strength in pounds and [tex]\( d \)[/tex] is the diameter in inches.
- Domain: The domain represents all possible values of [tex]\( d \)[/tex].
- Since [tex]\( d \)[/tex] represents a diameter, it must be positive.
- Therefore, the domain is all positive real numbers: [tex]\( d > 0 \)[/tex].
- Range: The range represents all possible values of [tex]\( z \)[/tex].
- Since [tex]\( z = 8900d^2 \)[/tex] and [tex]\( 8900 \)[/tex] is a positive constant, the smallest value of [tex]\( z \)[/tex] occurs when [tex]\( d = 0 \)[/tex], which gives [tex]\( z = 0 \)[/tex].
- As [tex]\( d \)[/tex] increases, [tex]\( z \)[/tex] increases without bound.
- Therefore, the range is all non-negative real numbers: [tex]\( z \geq 0 \)[/tex].
So we have:
- Domain: [tex]\( (0, \infty) \)[/tex]
- Range: [tex]\( [0, \infty) \)[/tex]
### Part (b): Graph the Function
To graph the function [tex]\( z = 8900d^2 \)[/tex], we will plot several points and sketch the curve.
1. Choose some sample values for [tex]\( d \)[/tex]:
- [tex]\( d = 0.5 \)[/tex], [tex]\( d = 1 \)[/tex], [tex]\( d = 1.5 \)[/tex], [tex]\( d = 2 \)[/tex], [tex]\( d = 2.5 \)[/tex], [tex]\( d = 3 \)[/tex]
2. Compute the corresponding [tex]\( z \)[/tex] values:
- For [tex]\( d = 0.5 \)[/tex]: [tex]\( z = 8900 \times (0.5)^2 = 8900 \times 0.25 = 2225 \)[/tex]
- For [tex]\( d = 1 \)[/tex]: [tex]\( z = 8900 \times (1)^2 = 8900 \)[/tex]
- For [tex]\( d = 1.5 \)[/tex]: [tex]\( z = 8900 \times (1.5)^2 = 8900 \times 2.25 = 20025 \)[/tex]
- For [tex]\( d = 2 \)[/tex]: [tex]\( z = 8900 \times (2)^2 = 8900 \times 4 = 35600 \)[/tex]
- For [tex]\( d = 2.5 \)[/tex]: [tex]\( z = 8900 \times (2.5)^2 = 8900 \times 6.25 = 55625 \)[/tex]
- For [tex]\( d = 3 \)[/tex]: [tex]\( z = 8900 \times (3)^2 = 8900 \times 9 = 80100 \)[/tex]
3. Plot these points on a graph and draw the curve [tex]\( z = 8900d^2 \)[/tex]. The curve will be a parabola opening upwards, starting from the origin.
### Part (c): Comparison of Diameters for Different Breaking Strengths
Suppose we have two ropes with diameters [tex]\( d_1 \)[/tex] and [tex]\( d_2 \)[/tex], and the breaking strength of the second rope is four times that of the first rope. Let's determine whether [tex]\( d_2 = 4d_1 \)[/tex]:
1. Let [tex]\( z_1 \)[/tex] be the breaking strength of the rope with diameter [tex]\( d_1 \)[/tex]:
[tex]\[ z_1 = 8900 \cdot d_1^2 \][/tex]
2. Let [tex]\( z_2 \)[/tex] be the breaking strength of the rope with diameter [tex]\( d_2 \)[/tex], and given [tex]\( z_2 = 4z_1 \)[/tex]:
[tex]\[ z_2 = 4z_1 = 4 \cdot 8900 \cdot d_1^2 = 35600 \cdot d_1^2 \][/tex]
3. The breaking strength of the rope with diameter [tex]\( d_2 \)[/tex] is also given by:
[tex]\[ z_2 = 8900 \cdot d_2^2 \][/tex]
4. Equate the two expressions for [tex]\( z_2 \)[/tex]:
[tex]\[ 8900 \cdot d_2^2 = 35600 \cdot d_1^2 \][/tex]
5. Solve for [tex]\( d_2 \)[/tex]:
[tex]\[ d_2^2 = 4 \cdot d_1^2 \][/tex]
[tex]\[ d_2 = 2d_1 \][/tex]
Therefore, if one rope has four times the breaking strength of another, the diameter of the stronger rope is twice the diameter of the other rope, not four times.
### Conclusion:
- Domain of [tex]\( z = 8900d^2 \)[/tex]: [tex]\( (0, \infty) \)[/tex]
- Range of [tex]\( z = 8900d^2 \)[/tex]: [tex]\( [0, \infty) \)[/tex]
- The stronger rope has a diameter twice the size of the weaker rope, not four times the size.
### Part (a): Describe the Domain and Range of the Function
The given function is:
[tex]\[ z = 8900d^2 \][/tex]
Where [tex]\( z \)[/tex] is the breaking strength in pounds and [tex]\( d \)[/tex] is the diameter in inches.
- Domain: The domain represents all possible values of [tex]\( d \)[/tex].
- Since [tex]\( d \)[/tex] represents a diameter, it must be positive.
- Therefore, the domain is all positive real numbers: [tex]\( d > 0 \)[/tex].
- Range: The range represents all possible values of [tex]\( z \)[/tex].
- Since [tex]\( z = 8900d^2 \)[/tex] and [tex]\( 8900 \)[/tex] is a positive constant, the smallest value of [tex]\( z \)[/tex] occurs when [tex]\( d = 0 \)[/tex], which gives [tex]\( z = 0 \)[/tex].
- As [tex]\( d \)[/tex] increases, [tex]\( z \)[/tex] increases without bound.
- Therefore, the range is all non-negative real numbers: [tex]\( z \geq 0 \)[/tex].
So we have:
- Domain: [tex]\( (0, \infty) \)[/tex]
- Range: [tex]\( [0, \infty) \)[/tex]
### Part (b): Graph the Function
To graph the function [tex]\( z = 8900d^2 \)[/tex], we will plot several points and sketch the curve.
1. Choose some sample values for [tex]\( d \)[/tex]:
- [tex]\( d = 0.5 \)[/tex], [tex]\( d = 1 \)[/tex], [tex]\( d = 1.5 \)[/tex], [tex]\( d = 2 \)[/tex], [tex]\( d = 2.5 \)[/tex], [tex]\( d = 3 \)[/tex]
2. Compute the corresponding [tex]\( z \)[/tex] values:
- For [tex]\( d = 0.5 \)[/tex]: [tex]\( z = 8900 \times (0.5)^2 = 8900 \times 0.25 = 2225 \)[/tex]
- For [tex]\( d = 1 \)[/tex]: [tex]\( z = 8900 \times (1)^2 = 8900 \)[/tex]
- For [tex]\( d = 1.5 \)[/tex]: [tex]\( z = 8900 \times (1.5)^2 = 8900 \times 2.25 = 20025 \)[/tex]
- For [tex]\( d = 2 \)[/tex]: [tex]\( z = 8900 \times (2)^2 = 8900 \times 4 = 35600 \)[/tex]
- For [tex]\( d = 2.5 \)[/tex]: [tex]\( z = 8900 \times (2.5)^2 = 8900 \times 6.25 = 55625 \)[/tex]
- For [tex]\( d = 3 \)[/tex]: [tex]\( z = 8900 \times (3)^2 = 8900 \times 9 = 80100 \)[/tex]
3. Plot these points on a graph and draw the curve [tex]\( z = 8900d^2 \)[/tex]. The curve will be a parabola opening upwards, starting from the origin.
### Part (c): Comparison of Diameters for Different Breaking Strengths
Suppose we have two ropes with diameters [tex]\( d_1 \)[/tex] and [tex]\( d_2 \)[/tex], and the breaking strength of the second rope is four times that of the first rope. Let's determine whether [tex]\( d_2 = 4d_1 \)[/tex]:
1. Let [tex]\( z_1 \)[/tex] be the breaking strength of the rope with diameter [tex]\( d_1 \)[/tex]:
[tex]\[ z_1 = 8900 \cdot d_1^2 \][/tex]
2. Let [tex]\( z_2 \)[/tex] be the breaking strength of the rope with diameter [tex]\( d_2 \)[/tex], and given [tex]\( z_2 = 4z_1 \)[/tex]:
[tex]\[ z_2 = 4z_1 = 4 \cdot 8900 \cdot d_1^2 = 35600 \cdot d_1^2 \][/tex]
3. The breaking strength of the rope with diameter [tex]\( d_2 \)[/tex] is also given by:
[tex]\[ z_2 = 8900 \cdot d_2^2 \][/tex]
4. Equate the two expressions for [tex]\( z_2 \)[/tex]:
[tex]\[ 8900 \cdot d_2^2 = 35600 \cdot d_1^2 \][/tex]
5. Solve for [tex]\( d_2 \)[/tex]:
[tex]\[ d_2^2 = 4 \cdot d_1^2 \][/tex]
[tex]\[ d_2 = 2d_1 \][/tex]
Therefore, if one rope has four times the breaking strength of another, the diameter of the stronger rope is twice the diameter of the other rope, not four times.
### Conclusion:
- Domain of [tex]\( z = 8900d^2 \)[/tex]: [tex]\( (0, \infty) \)[/tex]
- Range of [tex]\( z = 8900d^2 \)[/tex]: [tex]\( [0, \infty) \)[/tex]
- The stronger rope has a diameter twice the size of the weaker rope, not four times the size.
