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VikaD [51]
3 years ago
9

A ____ may be composed of a few individual objects or several complex groups of objects.

Computers and Technology
1 answer:
Diano4ka-milaya [45]3 years ago
4 0
A mixture can be composed of a few individual objects or several complex groups of objects.

Hope I helped! ( Smiles )
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You need to make use of the Brush tool for removing the unwanted part of an image.

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Yes, it is the brush tool or the lasso tool that you can make use of. and you will find brush tools in all sorts of image processing software may it be the paint or Photoshop. It is common to both the raster-based and vector-based image processing software. And you can easily find it inside the tool section, and then make use of it to remove the unwanted parts from an image.

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What is a data display in which the frequencies of different values are represented by small round points?
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Answer:

Answer to the following question is Dot plot

Explanation:

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Read 2 more answers
Produce an infinite collection of sets A1,A2,A3, . . . with the property that every Ai has an infinite number of elements, Ai ∩
atroni [7]

Answer:

Produce an infinite collection of sets A1,A2,A3, . . . with the property that every Ai has an infinite number of elements, Ai ∩ Aj = ∅ for all i = j, and [infinity] i=1 Ai = N.

Explanation:

Solution

For n ∈ N,

define  A_n = {2 ^n−1  ,(3)(2n−1 ),(5)(2^n−1 ),(7)(2^n−1 ), . . .}

I.e. A_n is all odd multiples of 2^n−1 . We must show that these sets satisfy the desired properties.

• (Infinite Number of Elements).

It is clear that the set A_n = {2 ^n−1 ,(3)(2^n−1 )(5)(2^n−1 ),(7)(2^n−1 ), . . .}  has infinitely many elements.

• (Disjoint).

Given A_n and A_m with n ≠ m, we can assume, without loss of generality, that n < m. Suppose  that there existed some x ∈ A_n ∩ A_m. Then by definition of these sets, there exists some odd numbers k  and l such that x = 2^n−1 . k = 2^m−1  . l.

However since n < m, we have that n ≤ m − 1, and therefore we  can write 2^m−1 = (2^n )(2 i ) with i ≥ 0. Hence we have 2^n−1 . k = 2^n. 2 ^i. l  

Dividing both sides by 2^n−1 yields  k = (2)(2^i ) .l, which contradicts the assumption that k is odd. Therefore A_n ∩ A_m = ∅.

• (Union is N).

We want to show that  [infinity] i=1 A_n = N.

(⊆). Since each A_n is a subset of N, the union of these sets is a subset of N as well.

(⊇).Given any x ∈ N, we can write x = 2^n−1 . k for some n ∈ N where k is odd. Then x ∈ A_n, as  desired.

5 0
3 years ago
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