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Andreyy89
3 years ago
13

At a recent marathon, spectators lined the street near the starting line to cheer for the runners. The crowd lined up 5 feet dee

p on both sides of the street for the first mile. You estimate that 14 people can comfortably fit in a square that measures 5 feet by 5 feet. Using this information, how many people cheered for the runners at the start of the race? 1mile=5280 feet
Mathematics
1 answer:
masha68 [24]3 years ago
5 0

Answer:

29568 people cheered for the runners at the start of the race

Step-by-step explanation:

From the question, the crowd lined up 5 feet deep on both sides of the street for the first mile.

This lined up crowd could be related to a rectangle that is 1 mile long and 5 feet wide.

First, Convert 1 mile to feet

1 mile = 5280 feet

Hence, the length of the rectangle is 5280 feet and the width is 5 feet.

Now, we will determine how many 5 feet by 5 feet square we can get from the 5280 feet by 5 feet rectangle. To do that, we will divide 5280 feet by 5 feet

5280 feet ÷ 5 feet = 1056

Hence, from the rectangle, we can get 1056 5 feet by 5 feet square.

From the question, you estimate that 14 people can comfortably fit in a square that measures 5 feet by 5 feet,

∴ 14 × 1056 people will comfortably fit in the crowed lined up 5 feet deep on one side of the street for the first mile.

14 × 1056 = 14784 people

This is the amount of people that will comfortably fit in the crowed lined up 5 feet deep on one side of the street for the first mile.

Since the crowd lined up on both sides of the street, then

2 × 14784 people will comfortably fit in the crowed lined up 5 feet deep on both sides of the street for the first mile

2 × 14784 = 29568 people

Hence, 29568 people cheered for the runners at the start of the race.

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3 years ago
Without multiplying all the terms, verify these equalities. a) 10!= 6!7! b) 10!=7!5!3! c) 16!= 14!5!2! d) 9!=7!3!3!2!
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Answer:

a)10! = 6×5×4×3×2 ×7!

10! = 6! 7! (Verified)

b)10! = 3! × 5! × 7! = 7!5!3! (Verified)

c)16! = 2! × 5! × 14!

16! = 14!5!2! (Verified)

d)9! = 3! × 3! ×2! × 7!

9! = 7!3!3!2! (Verified)

Step-by-step explanation:

a) 10!= 6!7!

10! = 10×9×8×7!

Reducing to their lowest factors

10! = (2×5)×(3×3)×(2×2×2) × 7!

Rearranging the factors, we have;

10! = (2×3)×(5)×(2×2)×(3)×2 × 7!

10! = 6×5×4×3×2 ×7!

10! = 6! 7! (Verified)

b) 10!=7!5!3!

10! = 10×9×8×7!

10! = (2×5)×(3×3)×(2×2×2) × 7!

Rearranging the factors, we have;

10! = (2×3)×(5)×(2×2)×(3)×2× 7!

10! = (3×2×1) × (5×4×3×2×1) × 7!

10! = 3! × 5! × 7! = 7!5!3! (Verified)

c) 16!= 14!5!2!

16! = 16×15×14!

16! = (2×2×2×2) × (5×3) × 14!

Rearranging the factors.

16! = (2) × (5) × (2×2) × (3) × (2) ×14!

16! = (2!) ×(5×4×3×2) × 14!

16! = 2! × 5! × 14!

16! = 14!5!2! (Verified)

d) 9!=7!3!3!2

9! = 9×8×7!

9! = (3×3) × (2×2×2) × 7!

Rearranging the factors

9! = (3×2) ×(3×2) ×(2) × 7!

9! = 3! × 3! ×2! × 7!

9! = 7!3!3!2! (Verified)

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