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Nezavi [6.7K]
3 years ago
11

Rewrite each expression using each base only once.

Mathematics
1 answer:
mezya [45]3 years ago
8 0
Rewrite each expression using each base only once. 

 (-6)^12 * (-6)^3 * (-6)^2 
 (-6)^(12+3+2)
 (-6)^(17)
 Answer: 
 (-6)^(17)


 2^2 * 2^7 * 2^0 
 (2)^(2+7+0)
 (2)^(9)
 Answer: 
 (2)^(9)

 Simplify each expresion.

 5c^4 * c^6 
 5*c^(4+6)
 5*c^(10)
 Answer:
 5*c^10
 (-2.4n^4)(2n^-1)
 (-2.4*2)(n^4)(n^-1)
 (-2.4*2)(n^(4+(-1))
 (-4.8)(n^(4-1))
 (-4.8)(n^(3))
 Answer: 
 (-4.8)(n^(3))

 (4c^4)(ac^3)(-3a^5c)
 ((4)*(-3))*(c^(4+3+1))*(a^(1+5))
 (-12)*(c^(8))*(a^(6))
 Answer: 
 (-12)*(c^8)*(a^6)

 a^6b^3 * a^2b^-2
 (a^(6+2))*(b^(3+(-2)))
 (a^(8))*(b^(3-2))
 (a^(8))*(b^(1))
 (a^8)*(b)
 Answer: 
 (a^8)*(b)

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3 years ago
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
Please help me and please write the answer out​
Vitek1552 [10]
15 degrees is the answer
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LaSanda bought 20 shares of stock for $175. She sold the stock for a total profit of $25. What was the selling price of each sha
podryga [215]
So if she bought 20 share of stock for $175 that means that she paid $8.75 for each stock so if she made $25 you add then you divide

175÷20 = $8.75
175+$25=$200 she sold them for
$200 ÷ 20= $10
4 0
3 years ago
Tim has some video games. Mark has twice as
BabaBlast [244]

7 or more games

2x = 20+

so if he had 7 games, mark would have double (14 games)z 14+7=21. that's more that 20 games.

*Aggressively drops microphone*

4 0
3 years ago
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