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Sagot :
Sure, let's tackle the two parts of the question step by step.
4.1.3 Mpho's Claim:
Mpho claims that if the surface area of a sphere increases, then its radius will decrease. To evaluate this claim, we need to consider the formula for the surface area of a sphere.
The formula for the surface area ([tex]\(S\)[/tex]) of a sphere is given by:
[tex]\[ S = 4 \pi r^2 \][/tex]
where [tex]\(r\)[/tex] is the radius of the sphere.
Now, let's analyze the relationship:
1. If the surface area, [tex]\(S\)[/tex], increases, this implies that [tex]\(4 \pi r^2\)[/tex] must also increase.
2. To maintain the equation, [tex]\(r^2\)[/tex] needs to increase since [tex]\(4 \pi\)[/tex] is a constant.
3. Hence, for [tex]\(r^2\)[/tex] to increase, the radius [tex]\(r\)[/tex] itself must increase.
Therefore, if the surface area of a sphere increases, its radius will also increase. Consequently, Mpho's claim is incorrect.
4.2 Calculating the Common Interval of Ordinates:
Given the ordinates of an irregular metal plate in millimeters are:
[tex]\[ 120, 125, 135, 140, 145, 160, 155, 150, 145, 135, 118 \][/tex]
and the area of the plate is:
[tex]\[ 21135 \, \text{mm}^2 \][/tex]
We are to calculate the common interval of the ordinates.
Let's use the Trapezoidal Rule for an irregular shape. When applying the trapezoidal rule, the formula for the area ([tex]\(A\)[/tex]) with uniform intervals ([tex]\(h\)[/tex]) is given by:
[tex]\[ A = \frac{h}{2} \left[ y_0 + 2(y_1 + y_2 + \cdots + y_{n-1}) + y_n \right] \][/tex]
Where:
- [tex]\(y_0, y_1, \ldots, y_n\)[/tex] are the ordinates.
- [tex]\(h\)[/tex] is the common interval.
- [tex]\(n\)[/tex] is the number of intervals.
In this case, the sum of the ordinates is:
[tex]\[ y_0 + y_1 + y_2 + \cdots + y_{n-1} + y_n \][/tex]
Breaking it down:
- [tex]\(y_0 = 120\)[/tex]
- [tex]\(y_n = 118\)[/tex]
- sum of middle ordinates: [tex]\(125 + 135 + 140 + 145 + 160 + 155 + 150 + 145 + 135 = 1290\)[/tex]
- sum_ordinates including 2x middle ordinates: [tex]\(120 + 118 + 2 \times 1290\)[/tex]
Thus:
[tex]\[ y_0 + y_n + 2 \times \sum_{i=1}^{n-1} y_i = 120 + 118 + 2 \times 1290 = 2818 \][/tex]
Given the total area [tex]\(A = 21135 \, \text{mm}^2\)[/tex], we solve for [tex]\(h\)[/tex]:
[tex]\[ 21135 = \frac{h}{2} \times 2818 \][/tex]
[tex]\[ 21135 = 1409h \][/tex]
[tex]\[ h = \frac{21135}{1409} \][/tex]
[tex]\[ h = 15.0 \, \text{mm} \][/tex]
Given 11 ordinates creates 10 intervals, the number of intervals is indeed:
[tex]\[ 10 \][/tex]
Thus, the common interval [tex]\(h\)[/tex] is:
[tex]\[ 15.0 \, \text{mm} \][/tex]
In summary:
- Mpho's claim is incorrect because an increase in the surface area of a sphere will lead to an increase in its radius.
- The common interval of the ordinates given an area of [tex]\(21135 \, \text{mm}^2\)[/tex] is [tex]\(15.0 \, \text{mm} \)[/tex].
4.1.3 Mpho's Claim:
Mpho claims that if the surface area of a sphere increases, then its radius will decrease. To evaluate this claim, we need to consider the formula for the surface area of a sphere.
The formula for the surface area ([tex]\(S\)[/tex]) of a sphere is given by:
[tex]\[ S = 4 \pi r^2 \][/tex]
where [tex]\(r\)[/tex] is the radius of the sphere.
Now, let's analyze the relationship:
1. If the surface area, [tex]\(S\)[/tex], increases, this implies that [tex]\(4 \pi r^2\)[/tex] must also increase.
2. To maintain the equation, [tex]\(r^2\)[/tex] needs to increase since [tex]\(4 \pi\)[/tex] is a constant.
3. Hence, for [tex]\(r^2\)[/tex] to increase, the radius [tex]\(r\)[/tex] itself must increase.
Therefore, if the surface area of a sphere increases, its radius will also increase. Consequently, Mpho's claim is incorrect.
4.2 Calculating the Common Interval of Ordinates:
Given the ordinates of an irregular metal plate in millimeters are:
[tex]\[ 120, 125, 135, 140, 145, 160, 155, 150, 145, 135, 118 \][/tex]
and the area of the plate is:
[tex]\[ 21135 \, \text{mm}^2 \][/tex]
We are to calculate the common interval of the ordinates.
Let's use the Trapezoidal Rule for an irregular shape. When applying the trapezoidal rule, the formula for the area ([tex]\(A\)[/tex]) with uniform intervals ([tex]\(h\)[/tex]) is given by:
[tex]\[ A = \frac{h}{2} \left[ y_0 + 2(y_1 + y_2 + \cdots + y_{n-1}) + y_n \right] \][/tex]
Where:
- [tex]\(y_0, y_1, \ldots, y_n\)[/tex] are the ordinates.
- [tex]\(h\)[/tex] is the common interval.
- [tex]\(n\)[/tex] is the number of intervals.
In this case, the sum of the ordinates is:
[tex]\[ y_0 + y_1 + y_2 + \cdots + y_{n-1} + y_n \][/tex]
Breaking it down:
- [tex]\(y_0 = 120\)[/tex]
- [tex]\(y_n = 118\)[/tex]
- sum of middle ordinates: [tex]\(125 + 135 + 140 + 145 + 160 + 155 + 150 + 145 + 135 = 1290\)[/tex]
- sum_ordinates including 2x middle ordinates: [tex]\(120 + 118 + 2 \times 1290\)[/tex]
Thus:
[tex]\[ y_0 + y_n + 2 \times \sum_{i=1}^{n-1} y_i = 120 + 118 + 2 \times 1290 = 2818 \][/tex]
Given the total area [tex]\(A = 21135 \, \text{mm}^2\)[/tex], we solve for [tex]\(h\)[/tex]:
[tex]\[ 21135 = \frac{h}{2} \times 2818 \][/tex]
[tex]\[ 21135 = 1409h \][/tex]
[tex]\[ h = \frac{21135}{1409} \][/tex]
[tex]\[ h = 15.0 \, \text{mm} \][/tex]
Given 11 ordinates creates 10 intervals, the number of intervals is indeed:
[tex]\[ 10 \][/tex]
Thus, the common interval [tex]\(h\)[/tex] is:
[tex]\[ 15.0 \, \text{mm} \][/tex]
In summary:
- Mpho's claim is incorrect because an increase in the surface area of a sphere will lead to an increase in its radius.
- The common interval of the ordinates given an area of [tex]\(21135 \, \text{mm}^2\)[/tex] is [tex]\(15.0 \, \text{mm} \)[/tex].
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