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Dima020 [189]
2 years ago
11

Please help me with this question!

Mathematics
2 answers:
pogonyaev2 years ago
6 0

Answer:

OK,okoko,OK,OK,OK,OK,ok

Kamila [148]2 years ago
5 0

Answer:

         g4g5igi5gm4gngm lil nine

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At a recent marathon, spectators lined the street near the starting line to cheer for the runners. The crowd lined up 5 feet dee
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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
PLS HELP WILL MARK YOU BRIANLIEST
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BC=16-12 = 4

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Cross multiplication

4x=60

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2(2x+1)= (2 x 2x) + (2 x 1) = 4x + 2
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