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Pavel [41]
4 years ago
15

Anyone, I need help... Just answer the 6 (c)....and also proper working.☺️

Mathematics
1 answer:
Ainat [17]4 years ago
5 0

Answer:

(i) The area of the rabbit cage when the width is 5.2 m is 81.5 m²

(ii) The area of the rabbit cage if Wilson has 40 meters of wire mesh is 75 m²

Step-by-step explanation:

(i) The given relation of the area, A to the width P of the rabbit cage is A = 3·p²

The graph of the function between the values of 0 and 6 inclusive is found as follows;

A,              3·p²

0,               0

1,                1

2,               12

3,               27

4,               48

5,               75

6,               108

Please find attached the graph of A to 3·p²

From the graph, we have when the the width, p, of the rabbit cage = 5.2, the area, A ≈ 81.5 m²

The area of the rabbit cage when the width is 5.2 m = 81.5 m²

(ii) Also from the graph given that the total wire mess with Wilson = 40 meters, we have;

The formula for the perimeter of the cage = The formula for the perimeter of a rectangle = 2×length + 2×width

The formula for the perimeter of the cage = 2×3×p + 2× p = 8·p

Where the total length of the wire mesh available = 40 meters for the cage

The 40 meters of wire mesh will be used round the perimeter of the cage

∴ 40 m. = 8·p

p = 40/8 = 5 m.

At p = 5 m. the area is given as A = 75 m².

Therefore, the area of the rabbit cage if Wilson has 40 meters of wire mesh = 75 m².

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It’s a new semester!students are grouped into three clubs, which each has 10,4 and 5 students.In how many ways can teacher sleep
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Since there are three clubs which each has 10, 4 and 5 students, and we require the number of ways a teacher can select 2 students so that they are from different clubs.

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<h3 /><h3>Number of ways of selecting from the first two clubs</h3>

Since we have two slots, to select the first person from the first club, we ¹⁰C₁. For the second student from the club of 4, we have ⁴C₁. Also, there are 2 ways of selecting the two students.

So, there are ¹⁰C₁ × ⁴C₁/2!

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<h3 /><h3>Number of ways of selecting from the next two clubs</h3>

For the next two clubs of 4 and 5 students, for the first slot, we have ⁴C₁. For the second student, we have ⁵C₁. Also, there are 2 ways of selecting the two students.

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<h3 /><h3>Number of ways of selecting from the last two clubs</h3>

For the first and last club of 10 and 5 students, for the first slot, we have ¹⁰C₁. For the second student, we have ⁵C₁.  Also, there are 2 ways of selecting the two students.

So, there are ¹⁰P₁ × ⁵P₁/2!

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= 25 ways from the first and last club.

<h3 /><h3>Total number of ways of selecting two students from the 3 clubs.</h3>

So, the total number of ways to select 2 students from the 3 clubs is 20 + 10 + 25 = 55 ways.

So, the total number of ways to select 2 students from the 3 clubs is 55 ways.

Learn more about combinations here:

brainly.com/question/25990169

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