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goblinko [34]
3 years ago
11

Understand mean and mad quiz -level f

Mathematics
1 answer:
love history [14]3 years ago
3 0

Answer:

What exactly do you need help with?

Step-by-step explanation:

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How many ways can David pick four of the first twelve positive integers such that no two of the numbers he picks are consecutive
vekshin1

Answer:

70 times

Step-by-step explanation:

5 consecutive even integers: 2n, 2n+2, 2n+4, 2n+6, 2n+8 for any integer n The sum of the two smallest of five consecutive even integers is 50 less than the sum of the other three integers: 2n + 2n+2 = 2n+4 + 2n+6 + 2n+8 - 504n + 2 = 6n -3234 = 2nn = 17 The smallest integer in our sequence is 2n = 2*17 = 34 Check:2n + 2n+2 = 2n+4 + 2n+6 + 2n+8 - 5034 + 36 = 38 + 40 + 42 - 5070 = 120 - 5070 = 70

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What are some ways I can have a good wardrobe without breaking the bank?
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Fast fashion even though you promoting child labor if you were that poor go ahead
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How many tens do you need to get too 805
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The answer is 80.5 ... 805 / 5 
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For x, y ∈ R we write x ∼ y if x − y is an integer. a) Show that ∼ is an equivalence relation on R. b) Show that the set [0, 1)
vodomira [7]

Answer:

A. It is an equivalence relation on R

B. In fact, the set [0,1) is a set of representatives

Step-by-step explanation:

A. The definition of an equivalence relation demands 3 things:

  • The relation being reflexive (∀a∈R, a∼a)
  • The relation being symmetric (∀a,b∈R, a∼b⇒b∼a)
  • The relation being transitive (∀a,b,c∈R, a∼b^b∼c⇒a∼c)

And the relation ∼ fills every condition.

∼ is Reflexive:

Let a ∈ R

it´s known that a-a=0 and because 0 is an integer

a∼a, ∀a ∈ R.

∼ is Reflexive by definition

∼ is Symmetric:

Let a,b ∈ R and suppose a∼b

a∼b ⇒ a-b=k, k ∈ Z

b-a=-k, -k ∈ Z

b∼a, ∀a,b ∈ R

∼ is Symmetric by definition

∼ is Transitive:

Let a,b,c ∈ R and suppose a∼b and b∼c

a-b=k and b-c=l, with k,l ∈ Z

(a-b)+(b-c)=k+l

a-c=k+l with k+l ∈ Z

a∼c, ∀a,b,c ∈ R

∼ is Transitive by definition

We´ve shown that ∼ is an equivalence relation on R.

B. Now we have to show that there´s a bijection from [0,1) to the set of all equivalence classes (C) in the relation ∼.

Let F: [0,1) ⇒ C a function that goes as follows: F(x)=[x] where [x] is the class of x.

Now we have to prove that this function F is injective (∀x,y∈[0,1), F(x)=F(y) ⇒ x=y) and surjective (∀b∈C, Exist x such that F(x)=b):

F is injective:

let x,y ∈ [0,1) and suppose F(x)=F(y)

[x]=[y]

x ∈ [y]

x-y=k, k ∈ Z

x=k+y

because x,y ∈ [0,1), then k must be 0. If it isn´t, then x ∉ [0,1) and then we would have a contradiction

x=y, ∀x,y ∈ [0,1)

F is injective by definition

F is surjective:

Let b ∈ R, let´s find x such as x ∈ [0,1) and F(x)=[b]

Let c=║b║, in other words the whole part of b (c ∈ Z)

Set r as b-c (let r be the decimal part of b)

r=b-c and r ∈ [0,1)

Let´s show that r∼b

r=b-c ⇒ c=b-r and because c ∈ Z

r∼b

[r]=[b]

F(r)=[b]

∼ is surjective

Then F maps [0,1) into C, i.e [0,1) is a set of representatives for the set of the equivalence classes.

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The equation of a circle in general form is ​ x2+y2+20x+12y+15=0 ​ . What is the equation of the circle in standard form?
mestny [16]

Answer:

(x)²+(y)²=0

Step-by-step explanation:

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