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goblinko [34]
2 years ago
8

Each side of a square is increasing at a rate of 5 cm/s. At what rate (in cm2/s) is the area of the square increasing when the a

rea of the square is 49 cm2
Mathematics
2 answers:
pickupchik [31]2 years ago
5 0

Answer:

25 cm^2/sec

Step-by-step explanation:

I my understanding of the question.  It doesn't matter what the area of the square is currently is, for example if the area is 25 cm^2, 36 cm^2 or 49cm^2, the rate will ALWAYS increase by rate squared.

      rate(a)  x  rate(b)  =  rate(area)           in this case rate(a) = rate(b)

        (5 cm^2/s) (5 cm^2/s)  = 25 cm^2/s

nikklg [1K]2 years ago
4 0

Answer:

knkkl

Step-by-step explanation:

bjkjjjkhjhjhjkhjhj

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Ainat [17]

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

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∴ 40 m. = 8·p

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

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