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gulaghasi [49]
3 years ago
12

Tap picture and answer all

Mathematics
1 answer:
creativ13 [48]3 years ago
8 0
Check the attachment or picture for the completed worksheet.

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Consider U = {x|x is a positive integer greater than 1}.
Zinaida [17]
B) {x|x ∈ U and 2x is prime}
x is greater than 1.
the only even prime number is 2
2 times anything is even.
the minimum x value is 2. 2x is 4, therefore higher than the only even number
every value 2x will have a multiple of 2 so it will not be prime.
6 0
3 years ago
Read 2 more answers
ABC IS a right angled triangle at B and D is a point on BC. If AD=18cm , BD=9cm and CD = 4 cm, find AC
leva [86]
AB =  \sqrt{ AD^{2} - BD^{2} } =  \sqrt{243} = 9 \sqrt{3} 

AC = \sqrt{ AB^{2} + BC^{2} } =  \sqrt{412} = 2 \sqrt{103}
4 0
3 years ago
If a bike is $125 how much will it be when its discounted by 20%
weeeeeb [17]

Answer:

160 pages.

Step-by-step explanation:

That is 125 - 0.20 * 125

= 125 - 25

= 160 pages.

6 0
3 years ago
Read 2 more answers
What is the answer and work for these attached below
GrogVix [38]
-4(-5-b)=1/3(b+16) Multiply both sides by 3 to get rid of the fraction
-12(-5-b)=b+16  distribute the -12 to get rid of the parenthesis 
60+12b=b+16  get the b on the left side and non b values to the right side
11b=-44 solve for b
b=-44/11 simplify the fraction
b=-4

3/5(t+18)=-3(2-t)   multiply both sides by 5/3 to get rid of thefraction
t+18=-5(2-t)   distribute the -5 to get rid of the parenthises
t+18=-10+5t  get the t to the left side and non t values to the right
-4t=-28  solve for t
t=7
5 0
3 years ago
What is the number of subsets of S= {1, 2, 3…10} that contain five element include 3 or 4 but not both?
Ivenika [448]

Answer:

140

Step-by-step explanation:

To construct a subset of S with said property, we have two choices, include 3 in the subset or include four in the subset. These events are mutually exclusive because 3 and 4 can not both be elements of the subset.

First, let's count the number of subsets that contain the element 3.

Any of such subsets has five elements, but since 3 is already an element, we only have to select four elements to complete it. The four elements must be different from 3 and 4 (3 cannot be selected twice and the condition does not allow to select 4), so there are eight elements to select from. The number of ways of doing this is {}_8C_4=70.

Now, let's count the number of subsets that contain the element 4.

4 is already an element thus we have to select other four elements . The four elements must be different from 3 and 4 (4 cannot be selected twice and the condition does not allow to select 3), so there are eight elements to select from, so this can be done in {}_8C_4=70 ways.

We conclude that there are 70+70=140 required subsets of S.

7 0
3 years ago
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