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RUDIKE [14]
3 years ago
7

Find the angle(s) that are Alternate Interior to angle 5.

Mathematics
1 answer:
gizmo_the_mogwai [7]3 years ago
7 0
B no problem lol so ok
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Simplify without using the absolute value sign<br><br> | x- ( - 18) | if x &gt; - 18
slega [8]
The expression would be > 0.

Subtracting a negative is the same as adding a positive; therefore -18--18 = -18+18 = 0.  Since we know that x>-18, 0 is below anything we could get from this.
7 0
4 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
The sum of 5 times a number is 7
IRISSAK [1]

Answer:

5*x=7

The answer is made by division

x=7/5

x=1.4

Step-by-step explanation:

7 0
3 years ago
Multiply.<br> (d+7)(d-7)<br> Simplify your answer.
romanna [79]

Answer:

\huge\boxed{(d+7)(d-7)=d^2-49}

Step-by-step explanation:

\bold{METHOD\ 1:}\\\\(d+7)(d-7)\qquad|\text{use}\ (a+b)(a-b)=a^2-b^2\\\\=d^2-7^2=d^2-49\\\\\bold{METHOD\ 2:}\\\\(d+7)(d-7)\qquad|\text{use FOIL}\ (a+b)(c+d)=ac+ad+bc+bd\\\\=(d)(d)+(d)(-7)+(7)(d)+(7)(-7)\\\\=d^2-7d+7d-49\qquad|\text{combine like terms}\\\\=d^2+(-7d+7d)-49\\\\=d^2-49

7 0
3 years ago
The following data points are the yearly salaries (in thousands of dollars) of the 4 high school cheerleading
GarryVolchara [31]

Answer:

6

Step-by-step explanation:

5 0
3 years ago
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