Write the specified set using the roster method givenU={1,2,3,…,20},E={2,4,6,…,20},F={5,10,15,20}U=\left\lbrace{1,2,3,\ldots,20}\right\rbrace,E={\left\lbrace2,4,6,\ldots,20\right\rbrace},F={\left\lbrace5,10,15,20\right\rbrace}.F∩UF∩U

F∩U={5,10,15,20}F\cap U=\left\lbrace5,10,15,20\right\rbrace

Write the specified set using the roster method given U = { 1 , 2 , 3 , … , 20 } , E = { 2 , 4 , 6 , … , 20 } , F = { 5 , 10 , 15 , 20 } U=\left\lbrace{1,2,3,\ldots,20}\right\rbrace,E={\left\lbrace2,4,6,\ldots,20\right\rbrace},F={\left\lbrace5,10,15,20\right\rbrace} . F ∩ U F∩U

Welcome back. For this problem, we've been given the universal set U, as well as sets E and F, and we want to find F intersect U. We can find this by checking through the elements of F and seeing which elements are also in you. Maybe we can see where this is going, but just in case, we are going to work through the whole problem. Now, we already have F written so that we can see all of the elements, but U is written with ellipses, so we can't actually see all of the elements in you. So, a good first step is going to be to rewrite you, so that we can see all of the elements. U contains 123, and so on and so forth up to 20. So this is going to be the set of whole numbers up to 20. So you is going to include 123456789, 10, 117, 1213, 1415, 1617. 1819, and 20. Now that we have you listed out, we can check through each of the elements of F to see which ones are also in you. Now 5 is also in you, so it will be in our intersection. Similarly, 10 is also in you, so it will be in the intersection, and 15 is also in you, so it will be in the intersection, and 20 is also in you, so it will be in the intersection. And we can say that F intersect U is a set containing 5, 1015, and 20, which is actually just Now, why did this happen? Well, as a reminder, F lives inside of the universal set. So actually, all of the elements of F have to be inside of you, the universal set. And because every element of F is also in you, every element of F will be inside of the intersection. That means that whenever we take any set and intersect it with its universal set, we will always get that set back, because everything inside of one of our sets is going to also live inside the universal set and will be in the intersection. So F intersect U is going to be set F. Great work.

Write the specified set using the roster method givenU={1,2,3,…,20},E={2,4,6,…,20},F={5,10,15,20}U=\left\lbrace{1,2,3,\ldots,20}\right\rbrace,E={\left\lbrace2,4,6,\ldots,20\right\rbrace},F={\left\lbrace5,10,15,20\right\rbrace}.F∩UF∩U

F∩U={5,10,15,20}F\cap U=\left\lbrace5,10,15,20\right\rbrace

F∩U={10,20}F\cap U=\left\lbrace10,20\right\rbrace

F∩U={1,2,3,…,20}F\cap U=\left\lbrace1,2,3,\ldots,20\right\rbrace

F∩U={2,4,6,8,10,12,14,16,18,20}F\cap U=\left\lbrace2,4,6,8,10,12,14,16,18,20\right\rbrace

2. Sets

Intersections and Unions

Liberal Arts Math

2. Sets

Intersections and Unions

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Multiple Choice

Write the specified set using the roster method given
$U = {1, 2, 3, \dots, 20}, E = {2, 4, 6, \dots, 20}, F = {5, 10, 15, 20}$ .
$F \cap U$

$F \cap U = {10, 20}$

$F \cap U = {5, 10, 15, 20}$

$F \cap U = {1, 2, 3, \dots, 20}$

$F \cap U = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}$

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Struggling with Liberal Arts Math?

Write the specified set using the roster method givenU={1,2,3,…,20},E={2,4,6,…,20},F={5,10,15,20}U=\(\left\]\lbrace{1,2,3,\ldots,20}\[\right\]\rbrace\),E={\(\left\[\lbrace\)2,4,6,\(\ldots\),20\(\right\]\rbrace\)},F={\(\left\[\lbrace\)5,10,15,20\(\right\]\rbrace\)}.F∩UF∩U

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Write the specified set using the roster method given
$U = {1, 2, 3, \dots, 20}, E = {2, 4, 6, \dots, 20}, F = {5, 10, 15, 20}$ .
$F \cap U$