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Juli2301 [7.4K]
3 years ago
15

The floor of a conference hall can be covered completly with tiles.Its length is 36ft. longer than its width.The area of the flo

or is less than 2040squ
are ft.How would you represent the with of the floor?How about its lenght?What mathematical sentence would represent the given situation?What are the possible dimensions of the floor?How about the possible areas of the floor?Would it be realistic for the floor to have an area of 144sq.ft?Explain your answer.
Mathematics
1 answer:
Masteriza [31]3 years ago
5 0

Answer:

Step-by-step explanation:

Given

Area of a floor = 2040ft²

If Its length is 36ft. longer than its width, then L = W+36

Area = LW

L is the length

W is the width

A <  W(W+36)

2040 < W²+36W

0<W²+36W-2040

W²+36W-2040 >0

W = -36±√36²-4(-2040)/2

W =  -36±√1296+8160)/2

W =  -36±√9456/2

W =  -36±97.25/2

W = -36+97.25/2

W = 61.24/2

W > 30.62ft

Since L = W+36

L > 30.62+36

L > 66.62ft

The width of the floor is 30.62ft

The length is 66.62ft

The mathematical sentence would represent the given situation is the width is greater than 30.62ft while the length must be greater than 66.62ft

The possible dimension of the floor is 31ft by 67 ft

The possible areas is 31*67 = 2077ft²

It won't be realistic to get an area of 144sqft because the initial area is greater than 144. Hence the feasible area will be greater than 2040ft²

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Answer:

Option c) y=3\frac{17}{25} is correct

The value of y is 3\frac{17}{25}

Step-by-step explanation:

Given equation is  

y-\frac{12}{25}=3\frac{1}{5}

To solve the given equation for y

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y-\frac{12}{25}=\frac{16}{5}  (converting mixed fraction to normal fraction )

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Option c) y=3\frac{17}{25} is correct

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Answer:

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Step-by-step explanation:

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In this formula n = number of trials

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In this sum n = 5

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Now putting these values in the formula

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