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Komok [63]
2 years ago
14

how many integers from 100 to 999, inclusive, don't include the numbers 2,4,6,8? PLS HELP I WILL MARK BRAINLIEST

Mathematics
1 answer:
UNO [17]2 years ago
7 0

Answer:

243

Step-by-step explanation:

999-100= 899 integers

200s 400s 600s 800s excluded so 400 numbers excluded

899 - 400 = 499

20s 40s 60s 80s which are 40 integers x 4 = 160

499 - 160 = 339

2s 4s 6s 8s which are 4 integers x 6 x 4 ( 4 from each 100 ) = 96

339 - 96 = 243 integers

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It is fine that you did not include the measure of angle XYZ in your posting.

This question is testing your knowledge of the four types of transformations.
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3) Rotations - an item is rotated, usually around the origin (the point (0,0) is the center of most rotations, especially in high school math).
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It the first three, the image after the transformation is congruent to the pre-image.  It has the same size and shape.  It is simply flipped, rotated, slid...

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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:

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  • 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)

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∼ is Reflexive by definition

∼ is Symmetric:

Let a,b ∈ R and suppose a∼b

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b∼a, ∀a,b ∈ R

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∼ 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

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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)

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