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sergeinik [125]
3 years ago
6

Please help right away

Mathematics
2 answers:
Nezavi [6.7K]3 years ago
6 0

Answer:

5,985 fans

Step-by-step explanation:

First you will need to get the floor area of the store

The dimensions are:

150' by 100'

This is 150 ft by 100 ft

Area would be:

Side x side (Assuming that the floor area is a quadrilateral)

150 ft x 100 ft = 15000 ft²

Next we solve for the area of the stars only:

6' by 6'

Side x side

6 ft x 6 ft = 36 ft²

So here we subtract the stars only area from the total floor area:

15,000 ft² - 36 ft² = 14,964 ft²

This is the floor area that the fans can occupy.

Next since the fire marshal said 1 person must occupy an area no less than 2.5ft², we divide the floor area for fans by the requirement.

<u>14,964 ft² </u>  = 5,985.6 fans

   2.5ft²

Since we cannot have a decimal for people, we need to round it down. If it goes beyond 5,985.6 that means we do not meet the minimum requirement of 2.5 ft² per person. So we need to round it down, to the nearest whole number, the answer would be:

<u>5,985 fans. </u>

avanturin [10]3 years ago
5 0

Answer:

D) 5985

Step-by-step explanation:

Total area: 150×100 = 15000

Area for stars: 6×6 = 36

Total area - area for stars = area for fans

15000-36= 14964

14964ft² for fans, each takes up 2.5 ft²

14964 ÷2.5 = 5985.6 fans

Since we can't have 0.6 of a person, we round down to 5985 fans

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The coordinates of the vertices of the image of the trapezoid are given as;

A'(x, y) = (- 4, 1), B'(x, y) = (- 2, 1), C'(x, y) = (- 5 / 2, 5 / 2) , D'(x, y) = (- 7 / 2, 5 / 2).

<h3>How to find the image of a trapezoid by dilation?</h3>

In this question we have a representation of a trapezoid, whose image has to be generated by a kind of rigid transformation known as dilation, whose equation is described :

P'(x, y) = O(x, y) + k · [P(x, y) - O(x, y)]

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And P(x, y) - Coordinates of the original point, P'(x, y) - Coordinates of the resulting point.

Since k = 1 / 2, A(x, y) = (- 5, - 2), B(x, y) = (- 1, - 2), C(x, y) = (- 2, 1), D(x, y) = (- 4, 1), O(x, y) = (- 3, 4),

Therefore, the coordinates of the vertices of the image are:

Point A'

A'(x, y) = O(x, y) + k · [A(x, y) - O(x, y)]

A'(x, y) = (- 3, 4) + (1 / 2) [(- 5, - 2) - (- 3, 4)]

A'(x, y) = (- 3, 4) + (1 / 2)  (- 2, - 6)

A'(x, y) = (- 3, 4) + (- 1, - 3)

A'(x, y) = (- 4, 1)

Point B';

B'(x, y) = O(x, y) + k [B(x, y) - O(x, y)]

B'(x, y) = (- 3, 4) + (1 / 2) [(- 1, - 2) - (- 3, 4)]

B'(x, y) = (- 3, 4) + (1 / 2)  (2, - 6)

B'(x, y) = (- 3, 4) + (1, - 3)

B'(x, y) = (- 2, 1)

Point C';

C'(x, y) = O(x, y) + k · [C(x, y) - O(x, y)]

C'(x, y) = (- 3, 4) + (1 / 2)  [(- 2, 1) - (- 3, 4)]

C'(x, y) = (- 3, 4) + (1 / 2) (1, - 3)

C'(x, y) = (- 3, 4) + (1 / 2, - 3 / 2)

C'(x, y) = (- 5 / 2, 5 / 2)

Point D'

D'(x, y) = O(x, y) + k  [D(x, y) - O(x, y)]

D'(x, y) = (- 3, 4) + (1 / 2) [(- 4, 1) - (- 3, 4)]

D'(x, y) = (- 3, 4) + (1 / 2) (- 1, - 3)

D'(x, y) = (- 3, 4) + (- 1 / 2, - 3 / 2)

D'(x, y) = (- 7 / 2, 5 / 2)

To learn more on dilations:

brainly.com/question/13176891

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