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Anastasy [175]
3 years ago
9

A car went 60 km in 5/6 of an hour while a second car went 54 km in 2/3 h. Which car was faster? How many times faster?

Mathematics
2 answers:
saveliy_v [14]3 years ago
7 0
This is the relationship between distance, velocity and time
v = d/t

To find which car was faster, we need to find the velocity
First car
d = 60 km
t = 5/6 of an hour

Calculate the velocity of the first car
v = d/t
v = 60 ÷ 5/6
v = 60 × 6/5
v = 72
The velocity of the first car is 72 km/hour

Second car
d = 54 km
t = 2/3 of an hour

Calculate the velocity of the second car
v = d/t
v = 54 ÷ 2/3
v = 54 × 3/2
v = 81
The velocity of the second car is 81 km/hour

81 is bigger than 72. The second car was more faster

81 - 72 = 9 km/hour
The second car was 9 km/hour faster than the first car
Margarita [4]3 years ago
3 0
This is the relationship between distance, velocity and timev = d/t
To find which car was faster, we need to find the velocityFirst card = 60 kmt = 5/6 of an hour
Calculate the velocity of the first carv = d/tv = 60 ÷ 5/6v = 60 × 6/5v = 72The velocity of the first car is 72 km/hour
Second card = 54 kmt = 2/3 of an hour
Calculate the velocity of the second carv = d/tv = 54 ÷ 2/3v = 54 × 3/2v = 81The velocity of the second car is 81 km/hour
81 is bigger than 72. The second car was more faster
81 - 72 = 9 km/hourThe second car was 9 km/hour faster than the first car
 Read more on Brainly - brainly.com/sf/question/8668399
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The algebraic solution to the situation is as follows:

x + y = 750

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The number of adult and student are 590 and 160 respectively.

<h3>How to find and solve an algebraic equation?</h3>

One adult ticket for a school play costs 10 dollars and one student ticket costs 6 dollars.

One evening,750 people attended the play and the total receipts were 6860.

Therefore,

let

x = number of adult ticket

y = number of student tickets

Therefore,

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4y = 640

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y = 160

Hence,

x  = 750 - y

x = 750 - 160

x = 590

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