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andreev551 [17]
3 years ago
6

Gilbert and Brian drove 604 miles in 11.6 hours. Gilbert drove the first part of the trip and averaged 45 miles per hour. Brian

drove the remainder of the trip and averaged 65 miles per hour. For what length of the time did Gilbert drive?
Mathematics
1 answer:
AVprozaik [17]3 years ago
3 0

The length of hours Gilbert drove is 7.5 hours.

The length of hours Brian drove is 4.1 hours.

<u>Step-by-step explanation</u>:

  • Gilbert averaged = 45 miles per hour
  • Brain averaged = 65 miles per hour
  • Total hours = 11.6 hours
  • Total distance = 604 miles

Let 'x' be the hours Gilbert drove.

Let 'y' be the hours Brian drove.

x+y = 11.6  -----(1)

45x+65y = 604 -----(2)

Multiply (1) by 45 and subtract (2) from (1)

 45x+45y = 522

-(<u>45x+65y = 604</u>)

 <u>      -20y = -82 </u>

y = 82/20

y = 4.1 hours

The length of hours Brian drove = 4.1 hours

Substitute y=4.1 in (1)

x+4.1 = 11.6

x = 7.5 hours

The length of hours Gilbert drove = 7.5 hours

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A.How far is the library from the park?<br> B.How far is the park from the football field?
Scorpion4ik [409]

Answer:

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The last answer ⇒ 2√5 miles , 2√35 miles

Step-by-step explanation:

* Lets revise the rules in the right angle triangle when we draw the

 perpendicular from the right angle to the hypotenuse

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# Angle B is a right angle

# The hypotenuse is AC

# BD ⊥ AC

∴ (AB)² = AD × AC

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∴ (BD)² = AD × CD

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* Lets use one of these rules to solve the problem

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- Vertex D represents the position of the library

- AB represents the distance from the home to the park

- CB represents the distance from the football field to the park

- DB represents the distance from the library to the park

- AC represents the distance from the home to the football field

- AD represents the distance from the home to the library

- CD represents the distance from the football field to the library

* Now lets solve the problem

a.

∵ The distance from the home to the library is 4 miles

∴ AD = 4

∵ The distance from the football field to the library is 10 miles

∴ CD = 10

∵ The distance from the library to the park represented by BD

- Lets use the rule (BD)² = AD × CD

∴ (BD)² = 4 × 10 = 40 ⇒ take √ for both sides

∴ BD = √40 = 2√5 miles

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b.

∵ The distance from the park to the football field represented by BC

* Lets use the rule (BC)² = CD × AC

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II: A=l*w
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II': A=l*w=
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substitute l in III with I to remove l:
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substitute A and B from II' and III' into IV:
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