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Harlamova29_29 [7]
2 years ago
13

Please help with the correct answer:

Mathematics
1 answer:
Mrac [35]2 years ago
7 0

Step-by-step explanation:

Quick Tip - Plug the pure number inside the parenthesis into all x of the function given.

f(2) = 2^2 - 3×2 + 2 = 4 - 6 + 2 = 0

f(5) = 5^2 - 3×5 + 2 = 25 - 15 + 2 = 12

g(3) = 3^2 + 2×3 - 5 = 10.

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Workers who drive to work in a city have a one in three chance of experiencing traffic delays.
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Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:
professor190 [17]

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

  • 1R4 and 4R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 4R1 and 1R2, then 4 must be related with 2.

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

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

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

Observe that

  • 1R2 and 2R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 2R1 and 1R4, then 2 must be related with 4.
  • 4R1 and 1R2, then 4 must be related with 2.
  • 2R4 and 4R2, then 2 must be related with itself

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)\}

5 0
3 years ago
Bobby is 28cm shorter than Daniel. Roy is 19 cm taller than Daniel. Find the difference in height between Roy and Bobby.
Tcecarenko [31]

Answer:

There is a 47cm difference between the two....

So you just added 28cm to 19cm to get 47cm....

Because 19-(-28) is 19 + 28.

Hope this helps :D

Step-by-step explanation:

6 0
3 years ago
Adding mixed numbers. Can someone help me correct my answer?
Olin [163]

Answer:

27 13/60

Step-by-step explanation:

What you have tried to do is not an impossible way to do it. What you could do is set your fractions and whole numbers like this.

( 7 + 1/10) * (3 + 5/6) I don't know if you know what FOIL is, but that is the way to proceed.

  • First: (meaning the first in each factor ).   3 * 7 =                                    21
  • Outside: (meaning the two terms near outside the brackets) 7 * 5/6 = 5 5/6
  • Inside: The two terms closest to the * sign      1 / 10  *3 =                         3/10
  • Last: The two end terms    (1/10 * 5/6 = 5/60                                              5/60

Now you can add. 21 + 5 is the whole number amount = 26

Add the fractions

  • 5/6 + 3/10 + 5/60 The lowest common multiple is 60
  • 50/60 + 18/60 +5/60 = 73 / 60 = 1 and 13/60

Add that to 26 and you get 27 13/60

3 0
3 years ago
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