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Nitella [24]
3 years ago
10

A person will pay a car dealer a total of $14,196 to be paid in 60 equal monthly installments. What is her monthly bill ?

Mathematics
1 answer:
Snezhnost [94]3 years ago
8 0

Answer:

$236.60

Step-by-step explanation:

You have to divide $14,196 by 60 months to get $236.60.

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What is the value of y in the equation 3(3y – 12) = 0? (5 points)
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Set equal to 0

3=0, nope

3y-12=0
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5p-4p-8=-2+3
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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)\}

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It's parallel. Definitely true

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Write the following series in sigma notation.<br> 7 + 16 + 25 +34 +43 +52 + 61
Solnce55 [7]

The series 7 + 16 + 25 +34 +43 +52 + 61 is an illusration of arithmetic series

The sigma notation of the series is: \sum\limits^7_{n=1} {9n - 2}

<h3>How to write the series in sigma notation?</h3>

The series is given as:

7 + 16 + 25 +34 +43 +52 + 61

The above series is an arithmetic series, with the following parameters

  • First term, a = 7
  • Common difference, d = 9
  • Number of terms, n = 7

Start by calculating the nth term using:

a(n) = a + (n - 1) * d

This gives

a(n) = 7 + (n - 1) * 9

Evaluate the product

a(n) = 7  - 9 + 9n

Evaluate the difference

a(n) = 9n - 2

So, the sigma notation is:

\sum\limits^7_{n=1} {9n - 2}

Read more about arithmetic series at:

brainly.com/question/6561461

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