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If $\mathrm{A}=\{3,6,9,12,15,18,21\}, \mathrm{B}=\{4,8,12,16,20\}$,$\mathrm{C}=\{2,4,6,8,10,12,14,16\}, \mathrm{D}=\{5,10,15,20\} ;$ find(i) $\mathrm{A}-\mathrm{B}$(ii) $A-C$(iii) $A-D$(iv) $\quad B-A$(v) $\mathrm{C}-\mathrm{A}$(vi) $\mathrm{D}-\mathrm{A}$(vii) $\quad \mathrm{B}-\mathrm{C}$(viii) $\mathrm{B}-\mathrm{D}$(ix) $\quad C-B$(x) $\mathrm{D}-\mathrm{B}$(xi) $\quad C-D$(xii) $\mathrm{D}-\mathrm{C}$

Algebra

Chapter 1

Sets

Section 4

Finite and Infinite Sets

Equations and Inequalities

Missouri State University

McMaster University

Baylor University

University of Michigan - Ann Arbor

Lectures

01:32

In mathematics, the absolu…

01:11

02:21

MULTIPLE CHOICE Which equa…

00:50

If $\mathbf{u}=\langle a, …

01:28

Let $A=\left[\begin{array}…

01:46

00:26

Find $A B$ and $B A$ in ea…

01:45

01:13

01:12

00:25

01:30

00:31

01:59

00:49

01:01

01:10

02:05

given said E. Equal to 34 most six woman nine 12 15 home 18 former 21 and that be equal to four comma eight comma 12 coma 16 coma 20 sets equal to common foe. Coma six cuomo a moment 10 moment 12 comma 14 for 16 said the equal to five home and and 15 moment 20. Now here we have to find E minus speak the set of elements which is belongs to A. But not B. Three comma six common nine number 15 from 18 21. So it is over answer. Four seconds E. Negative C. Equal to the sets of elements which belongs to A. But not see three common nine 15 call 18 21. So it is over answer now we saw arthur E minus B. The set of elements which belongs to A. But not two D. So here 34 months six roman nine coma 12 from 18 come on 21. So it is over answer now we saw poor fool be negative minus A. Equal to the third of elements which belongs to be but not E. So here 48 I was 16. Common 20 now we sold part of it see negative E. The set of elements which belongs to see but not A. So here to cuomo fu come on eight. Come on 10 cuomo 14 almost 16. No we sold for six B 98 equal and you can say D minus E. The set of elements which belongs to D. But not to A. So here we get right, coma then coma 20 seven B. Negative T. Here the set of elements which belongs to be but not to see. So here we get 20 and well A. B minus be equal to the set of element which belongs to be not to be four koma. Eight, coma 12 to 16. Now we saw part nine see negative B the sex of elements which belongs to see not to be. So here we get to coma 6, 10 come A 14 now we sold part 10 be negative B. Equal to the set of elements which belongs to be not but not to be. So here we get five home or 10 15. Now it's all part 11 C minus D. Equal to set of elements which belongs to see but not today. Two more for comma six, comma eight, comma 12 come on 14, coma 16. No, it's all part would be minus C. The set of elements which belongs to be but not to see. So it is A pointing on.

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