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stira [4]
3 years ago
6

All 231 students in the math club went on a field trip in 15 vehicles. Some students rode in vans which holds 7 students each an

d some students ride in buses which holds 25 students each. How many vans and buses were used?
Mathematics
1 answer:
Rina8888 [55]3 years ago
7 0

They used 8 vans and 7 buses

Step-by-step explanation:

The given is:

  • All 231 students in the math club went on a field trip in 15 vehicles
  • Some students rode in vans which holds 7 students each and some students ride in buses which holds 25 students each

We need to find how many vans and buses were used

Assume that the number of vans is x and the number of buses is y

∵ There were x vans in the trip

∵ There were y buses in the trip

∵ They went in 15 vehicles

- Add x and y and equate the sum by 15

∴ x + y = 15 ⇒ (1)

∵ Each van holds 7 students

∵ Each bus holds 25 students

∵ 231 students went to the trip

- Multiply x by 7 and y by 25, then add the products and equate

  the sum by 231

∴ 7x + 25y = 231 ⇒ (2)

Now we have a system of equation to solve it

Multiply equation (1) by -7 to eliminate x

∵ -7x - 7y = -105 ⇒ (3)

- Add equations (2) and (3)

∴ 18y = 126

- Divide both sides by 18

∴ y = 7

- Substitute the value of y in equation (1) to find x

∵ x + 7 = 15

- Subtract 7 from both sides

∴ x = 8

They used 8 vans and 7 buses

Learn more:

You can learn more about the system of equations in brainly.com/question/6075514

#LearnwithBrainly

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yulyashka [42]

Answer : The specific heat of the metal is, 2178.67J/kg^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of metal = ?

c_2 = specific heat of ice = 2000J/kg^oC

m_1 = mass of metal = 1.00 kg

m_2 = mass of ice = 1.00 kg

T_f = final temperature of mixture = -8.88^oC

T_1 = initial temperature of metal = 5.00^oC

T_2 = initial temperature of ice = -24.0^oC

Now put all the given values in the above formula, we get:

(1.00kg)\times c_1\times (-8.88-5.00)^oC=-[(1.00kg)\times 2000J/kg^oC\times (-8.88-(-24.0))^oC]

c_1=2178.67J/kg^oC

Therefore, the specific heat of the metal is, 2178.67J/kg^oC

4 0
4 years ago
four-fifths of the senior class went to prom. if 324 seniors went to prom, how many are there in the senior class
romanna [79]

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Step-by-step explanation:

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4 0
2 years ago
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