Day of the Week:

January has 31 days. It means that every date in February will be 3 days later than the same date in January(28 is 4 weeks exactly). The below table is calculated in such a way. Remember this table which will help you to calculate.

January	0
February	3
March	3
April	6
May	1
June	4
July	6
August	2
September	5
October	0
November	3
December	5

Step1: Ask for the Date. Ex: 23^rd June 1986
Step2: Number of the month on the list, June is 4.
Step3: Take the date of the month, that is 23
Step4: Take the last 2 digits of the year, that is 86.
Step5: Find out the number of leap years. Divide the last 2 digits of the year by 4, 86 divide by 4 is 21.
Step6: Now add all the 4 numbers: 4 + 23 + 86 + 21 = 134.
Step7: Divide 134 by 7 = 19 remainder 1.
The reminder tells you the day.

Sunday	0
Monday	1
Tuesday	2
Wednesday	3
Thursday	4
Friday	5
Saturday	6

Answer: Monday

Set Theory formulas

Set Theory is a branch of mathematics which deals with the study of sets or the collection of similar objects.
Set theory is one of the most fundamental branch of mathematics.

Set Theory Formulas:

Notations used in set theory formulas:

$n(A)$ – Cardinal number of set A.

$n_o(A)$ – cardinality of set A.

$\bar{A} = A^c$ – Complement of set A.

$U$ – universal set

$A \subset B$ – Set A is proper subset of subset of set B.

$A \subseteq B$ – Set A is subset of set B.

$\phi$ – Null set.

$a \in A$ – element “a” belongs to set A.

$A \cup B$ – Union of set A and set B.

$A \cap B$ – Intersection of set A and set B.

Formulas:

For a group of two sets:

1. If A and B are overlapping set, $n(A \cup B) = n(A) + n(B) - n(A \cap B)$

2. If A and B are disjoint set, $n(A \cup B) = n(A) + n(B)$

3. $n(A) = n(A \cup B) +n(A \cap B) - n(B)$

4. $n(A \cap B) = n(A) +n(B) - n(A \cup B)$

5. $n(B) = n(A \cup B) +n(A \cap B) - n(A)$

6. $n(U) = n(A) + n(B) - n(A \cap B) + n((A \cup B)^c)$

7. $n((A \cup B)^c) = n(U) + n(A \cap B) - n(A) - n(B)$

8. $n(A \cup B) = n(A - B) + n(B - A) + n(A \cap B)$

9. $n(A - B) = n(A \cup B) - n(B)$

10. $n(A - B) = n(A) - n(A \cap B)$

11. $n(A^c) = n(U) - n(A)$

For a group of three sets:
1. $n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(A \cap C) - n(B \cap C) + n(A \cap B \cap C$

2. $n(A \cup B \cup C) = n(U) - n((A \cup B \cup C)^c)$

3. $n(A \cap B \cap C) = n(A \cup B \cup C)+n(A \cap B)+n(A \cap C)+n(B \cap C) - n(A) - n(B) - n(C)$

4. $n_0(A) = n(A)-n(A \cap B)-n(A \cap C)+n(A \cap B \cap C)$

5. $n_0(B) = n(B)-n(A \cap B)-n(B \cap C)+n(A \cap B \cap C)$

6. $n_0(C) = n(C)-n(A \cap C)-n(B \cap C)+n(A \cap B \cap C)$

7. $n(A \cap B only) = n(A \cap B)-n(A \cap B \cap C)$

8. $n(B \cap C only) = n(B \cap C)-n(A \cap B \cap C)$

9. $n(A \cap C only) = n(A \cap C)-n(A \cap B \cap C)$

Class 10th - Construction

Proof of Pythagoras Theorem

Sphere and its Surface Area

Day of the Week:

Set Theory formulas

Set Theory formulas

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