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photoshop1234 [79]
2 years ago
5

During batting practice, two pop flies are hit from the same location, 2 s apart. The paths are modeled by the equations h = -16

t^2 + 56t and h = -16t^2 + 156t - 248, where t is the time that has passed since the first ball was hit. Explain how to find the height at which the balls meet. Then find the height to the nearest tenth.
Mathematics
1 answer:
Schach [20]2 years ago
4 0

Answer:

2.5 seconds since the first ball was hit.

Step-by-step explanation:

The equations that model the heights are:

  • h = -16t² + 56t
  • h = -16t² + 156t - 248

When the balls meet, both heights are equal, therefore:

-16t² + 56t = -16t² + 156t - 248

-16t² + 56t + 16t² - 156t + 248 = 0

- 100t + 248 = 0

248 = 100t

248/100 = t

t = 2.48 ≈ 2.5

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Answer:

Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is \{1,2,3,4\}.

a) We need to find a relation R reflexive and transitive that contain the relation R1=\{(1, 2), (1, 4), (3, 3), (4, 1)\}

Then, we need:

1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,

2. Observe that

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Therefore \{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\} is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

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and the analysis for be transitive is the same that we did in a).

Observe that

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  • 4R1 and 1R4, then 4 must be related with itself.
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  • 4R1 and 1R2, then 4 must be related with 2.
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Therefore, the smallest relation containing R1 that is symmetric and transitive is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

c) We need a new relation reflexive, symmetric and transitive containing R1.

For be reflexive

  • 1 must be related with 1,
  • 2 must be related with 2,
  • 3 must be related with 3,
  • 4 must be related with 4

For be symmetric

  • since 1R2, 2 must be related with 1,
  • since 1R4, 4 must be related with 1.

For be transitive

  • Since 4R1 and 1R2, 4 must be related with 2,
  • since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

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Answer:

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