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Lilit [14]
2 years ago
7

If roni and allie work together to mow the field, what part of the field would roni mow?

Mathematics
1 answer:
marissa [1.9K]2 years ago
6 0

Based on the Roni using a lawn mower and Allie pushing one, Roni would be able to mow 0.71 of the field.

<h3>How much of the field would Roni mow?</h3>

Assume that the part of the field that Roni would mow is x.

The relevant equation would be:

x/30 + x/75 = 1

5x/150 + 2x/150 = 1

7x / 150 = 1

x = 21.43

In 30 minutes, Roni would have mowed:

= 21.43 / 30

= 0.71

Full question is:

Roni and Allie are mowing the grass at the soccer field. Roni has a riding lawn mower and can mow the field in 30 minutes. Allie is pushing a lawn mower and can mow the field in 75 minutes.

Find out more on the rate of work at brainly.com/question/4976375.

#SPJ1

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Answer:

The most correct option is;

(B) 958.2 ft.²

Step-by-step explanation:

From the question, the dimension of each square = 3 ft.²

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Based on the above dimensions, the dimension of the small semicircle is found by counting the number of square sides ti subtends as follows;

The dimension of the diameter of the small semicircle = 10·√3

Radius of the small semicircle = Diameter/2 = 10·√3/2 = 5·√3

Area of the small semicircle = (π·r²)/2 = (π×(5·√3)²)/2 = 117.81 ft.²

Similarly;

The dimension of the diameter of the large semicircle = 10·√3 + 2 × 6 × √3

∴ The dimension of the diameter of the large semicircle = 22·√3

Radius of the large semicircle = Diameter/2 = 22·√3/2 = 11·√3

Area of the large semicircle = (π·r²)/2 = (π×(11·√3)²)/2 = 570.2 ft.²

Area of rectangle = 11·√3 × 17·√3 = 561

Area, A of large semicircle cutting into the rectangle is found as follows;

A_{(segment \, of \, semicircle)} = \frac{1}{4} \times (\theta - sin\theta) \times r^2

Where:

\theta = 2\times tan^{-1}( \frac{The \, number \, of  \, vertical  \, squrare  \, sides  \ cut  \,  by  \  the  \  large  \,  semicircle}{The \, number \, of  \, horizontal \, squrare  \, sides  \ cut  \,  by  \  the  \  large  \,  semicircle} )

\therefore \theta = 2\times tan^{-1}( \frac{10\cdot \sqrt{3} }{5\cdot \sqrt{3}} ) = 2.214

Hence;

A_{(segment \, of \, semicircle)} = \frac{1}{4} \times (2.214 - sin2.214) \times (11\cdot\sqrt{3} )^2 = 128.3 \, ft^2

Therefore; t

The area covered by the pavers = 561 - 128.3 + 570.2 - 117.81 = 885.19 ft²

Therefor, the most correct option is (B) 958.2 ft.².

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Step-by-step explanation:

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(hope this helps can i plz have brainlist :D hehe)

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