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lilavasa [31]
3 years ago
12

Hello is it anyone up here that can help me with my question please

Mathematics
1 answer:
jok3333 [9.3K]3 years ago
7 0
Simple interest is basically the cost of borrowing money over a period of time.  So if you have borrowed $110.00 at 5% for two years, you will multiply the 5% by the two years (presuming that it is 5% annual percentage rate (APR).  So, You will multiply the 110 by 10% (or .1) to get $11 dollars of simple interest.
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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
Help a little eeeeeeeeeee
ozzi
0.8 or 4/5
Both are correct, but in different form.

Your welcome.
7 0
3 years ago
How many cards must be drawn​ (without replacement) from a standard deck of 52 to guarantee that ten of the cards will be of the
marta [7]
37 cards. 9 cards of every suit + 1 more card to guarantee the 10 cards by pigeonhole principle.
4 0
3 years ago
What is the rational number equivalent to 1 point 28 with a bar over 28? 1 and 6 over 97 1 and 28 over 99 1 and 8 over 33 1 and
katrin2010 [14]
In algebra, the conversion of a more complex number to a simpler mixed number is rather simple. For the repeated numbers, which are designated by the bar above them, we simply have to put over 9 for each number repeated and 0 for the non-repeated numbers.

In this item, we have 1.28 with bar above 28 signifying that the number can also be written as 1.28282828.... The mixed number is then equal to,
<em>1 and 28/99 or 1 and 28 over 99</em>

Therefore, the answer to this item is the second choice. 
7 0
3 years ago
Read 2 more answers
How do I graphy y = 2/3x - 1​
Kobotan [32]

Answer:

Look Below, if I'm wrong then ignore this answer.

Step-by-step explanation:

If we were to use the formula y=mx+b, 2/3 is the slope and (0,-1) is the y intercept, so the x intercept must be about (1.5,0), so the graph is probably

7 0
3 years ago
Read 2 more answers
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