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Arturiano [62]
3 years ago
15

Each chair in an auditorium is 1.8 feet wide. If there are 24 chairs in each row, about how long is a row?

Mathematics
2 answers:
kaheart [24]3 years ago
5 0

Answer:

48 feet

Step-by-step explanation:

1.8 feet * 24 chairs = 43.2 feet in a row (with no spaces between chairs)

48 is the closest given answer

gladu [14]3 years ago
4 0
The answer is 48 feet
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It would be: 11/18 - 1/6

= 11 - 3 / 18

= 8/18

= 4/9

In short, Your Answer would be Option D

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3 years ago
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Which statement correctly names the congruent triangles and justifies the reason for congruence?
Marat540 [252]

Answer:

  C. Triangle BAC is congruent to triangle FDE by AAS

Step-by-step explanation:

BAC names the vertices in the order longest-side, shortest-side. That same order is FDE in the other triangle, eliminating choiced B and D. The triangles are not right triangles, eliminating choice A.

The only viable answer choice is C.

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3 years ago
The constraint for the potential plant site 2 can be written as:
sdas [7]

Answer:

c. x21 + x22 + x23 less or equal than 2800 y2

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4 0
3 years ago
The total stopping distance T for a vehicle is T = 2.5x + 0.5x 2 , where T is in feet and x is the speed in miles per hour. Appr
Kobotan [32]

Answer:

%  change in stopping distance = 7.34 %

Step-by-step explanation:

The stooping distance is given by

T = 2.5 x + 0.5 x^{2}

We will approximate this distance  using the relation

f (x + dx) = f (x)+ f' (x)dx

dx = 26 - 25 = 1

T' =  2.5 + x

Therefore

f(x)+f'(x)dx = 2.5x+ 0.5x^{2} + 2.5 +x

This is the stopping distance at x = 25

Put x = 25 in above equation

2.5 × (25) + 0.5× 25^{2} + 2.5 + 25 = 402.5 ft

Stopping distance at x = 25

T(25) = 2.5 × (25) + 0.5 × 25^{2}

T(25) = 375 ft

Therefore approximate change in stopping distance = 402.5 - 375 = 27.5 ft

%  change in stopping distance = \frac{27.5}{375} × 100

%  change in stopping distance = 7.34 %

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3 years ago
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