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butalik [34]
3 years ago
11

Use the graph below to find f(1)

Mathematics
1 answer:
Vladimir [108]3 years ago
7 0

Answer:

I dont understand sorry

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Empat sebutan pertama suatu jajang aritmetik ialah 3x-1,4x,5x+1,4.cari
wariber [46]
Here, A.P. is 3x-1, 4x, 5x+1, 4

i) Common difference (d)
 (5x-1) - 4x = x - 1
It would be same as 4 - (5x-1) = 4-5x+1 = 5-5x
Now, x - 1 = 5 - 5x
5x + x = 5 + 1
6x = 6
x = 6/6
x = 1

So, Common difference would be 1

ii) Eight term will be: a + (n-1)d = 2 + (8-1)2 = 2+7(2) = 2+14 = 16

So, eight term would be 16


Hope this helps!
3 0
3 years ago
What is the length of the hypotenuse of a right triangle with legs of length 6 and 4 ?
liberstina [14]

Answer:

6²+4²=h²

36+16=h²

\sqrt{52}  = h \:  \:  \:  \:  \:  \\ hyphotenuse \: is \: 2 \sqrt{13  \:} \\ or \: 7.2111

6 0
4 years ago
Which two values of x are roots of the polynomial below? x2 + 3x - 5
lesya692 [45]

Answer:

Step-by-step explanation:

The roots are the values of x for which x² + 3x - 5 = 0.

Quadratic formula:

x = [-3±√(3²-4(1)(-5))]/(2·1) = [-3±√29]/2

3 0
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)\}

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