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djyliett [7]
3 years ago
13

What does Pi = Vxbbbbnnnnnnnnnnnbvvvcccccccccvggvvvvffffff

Mathematics
2 answers:
777dan777 [17]3 years ago
8 0

Answer: 3.14

Step-by-step explanation : 3.14159265359

Liono4ka [1.6K]3 years ago
4 0

Pi is used in formulas of circles it is a never ending decimal.

3.14 or 22/7 are common values used for PI

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In the ASA Congruence Postulate, the side between the angles is called the ___ side.
rusak2 [61]
No English plees enpanio
6 0
3 years ago
I just learned this today and it’s a little confusing , is there anyone that could explain how to do these problems for me ?
evablogger [386]

Step-by-step explanation:

I'll do the first problem as an example.

∠P and ∠H both have one mark.  That means they're congruent.

∠T and ∠G both have two marks.  So they're congruent.

∠W and ∠D both have three marks.  So they're congruent.

So we can write a congruence statement:

ΔPTW ≅ ΔHGD

We can write more congruence statements by rearranging the letter, provided that corresponding pairs have the same position (P is in the same place as H, etc.).  For example:

ΔWPT ≅ ΔDHG

ΔTWP ≅ ΔGDH

3 0
3 years ago
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.

4 0
3 years ago
Your teacher asked you to represent 3 more than 8 multiplied by a number t as an expression. You wrote 3 + 8t, and your friend w
Leni [432]

Both answers are not the same. You are correct and your friend is wrong. The situation your teacher gave could have only one other interpretation that would be acceptable and that would be 8t+3 because it is the same number just in a different order.

3 0
3 years ago
37:35-(32:24-10).2+√289=
Mkey [24]

Answer:

to simplify this it is 1295/1804

Step-by-step explanation:

3 0
2 years ago
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