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Art [367]
3 years ago
15

Let U = {all integers}.

Mathematics
2 answers:
elixir [45]3 years ago
7 0
All integer numbers are either even or odd. So, the complement to the set B (to the set of all even integers) is the set of all odd integers, that is D.

The integer number of the form 2x is always even, because it is always divided by 2, so the set C is the empty set.

Numbers 1, 3, 5, 7 are odd, therefore they belong to the set D.
valentina_108 [34]3 years ago
6 0

Answer:

D, C, D

Step-by-step explanation:

The complement of the set B is the set that contains all the elements that aren't in B. A number is either even or odd (it can't be both at the same time and it can't be neither of both). If an element is not in B, that means is not even. So, it's odd and it belongs to D. Therefore, D is the complement of B.

Let's notice that 4 belongs to U and is greater than 3, so it belongs to A. A is not empty. It also belongs to B because is even. So, B isn't empty. And 5 belongs to D because is an odd integer, D isn't empty as well.  

Let's see what happens with C: every number multiply by 2 is even, so 2x can't be an odd integer. Therefore, C is an empty set.

1, 3, 5 and 7 are all odd integers, so E is a subset of D.  

A can't contain it because 1 < 3.

B can't contain it because 1, 3, 5 and 7 aren't even.

C can't contain it because we said is empty.

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Which value is in the solution set of |x|+9=-2
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Answer:

The equation has no solutions

Step-by-step explanation:

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|x|=-11

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Attached is the representation of this function y= |x|

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4 0
3 years ago
Find the selling price. Round to the nearest cent.
Mrac [35]
10% of 13 is 1.3 so 13+1.3= 14.3
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3 years ago
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Determine the length of the longest metal rod which could be stored in a rectangle box 20 cm by 50 cm by 30 cm.
navik [9.2K]

Answer:

61.64 cm

Step-by-step explanation:

The longest metal rod which could be stored in a rectangular box will be equal to its diagonal:

\therefore \: length \: of \:metal \: rod  \\  =  \sqrt{ {l}^{2} +  {b}^{2}   +  {h}^{2} }  \\  =  \sqrt{ {20}^{2} +  {50}^{2}  +  {30}^{2}  }  \\  =  \sqrt{400 + 2500 + 900}  \\  =  \sqrt{3800}  \\  = 61.64414 \\  = 61.64 \: cm

5 0
3 years ago
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YOOO MARKIN BRAINIEST !!
Brilliant_brown [7]
Y ≥ 2x-5
y-intercept is -5, from there go 2 up 1 over ( slope is rise over run)
greater than or equal so the line is connected
now find the shaded area by plugging in (0,0)
0 ≥2(0)-5
0 ≥-5 is correct so shade the area that to the left of the line, (the whole area that including (0,0))

y<-3x
y intercept is 0 so start from there and go down 3 right 1 (or go up 3 left 1)
broken like cause no or equal sign
the (0,0) is on the line so use (1,1) to find the answer
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you can see the section where both shaded area cross, thats the answer so erase every area you shaded that isn’t the answer so
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3 years ago
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Whats the answer to 1/5x=121 x=?
weqwewe [10]
You have an algebraic expression in which you are solving for x.  When you are solving for a variable, you need to isolate it all alone.  Since you are multiplying by 1/5, you will have to undo it by multiplying by its reciprocal.  In this case you are multiplying both sides by 5/1.

5/1 *1/5x = 121*5/1
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