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आप जो सवाल पूछ रहे हैं वह अस्पष्ट या अधूरा है।
आप अध्याय नाम या पुस्तक नाम जैसे अधिक विवरण जोड़ सकते हैं।
विशिष्ट प्रश्न पूछें जो स्पष्ट और संक्षिप्त हों।
उत्तर के लिए ठीक से पूछे गए प्रश्न पूछें।

 The Central Board of Secondary Education (CBSE) on Wednesday released announced introduction of ‘Applied Mathematics (241)’ as an academic elective subject at senior secondary level. The course can be opted for the class XI students from this Academic year (2020-21) as an Academic elective.
The students who have passed Basic Mathematics (241) in class X are now allowed to offer the new academic elective Applied Mathematics (241) at Senior Secondary Level.
Accordingly, the students who have passed Basic Mathematics (241) as well as Standard Mathematics (041) in Class X of CBSE exam are eligible for this course. However, it is once again clarified that students who have passed Mathematics-Basic (241) at Secondary level are not eligible to opt for Mathematics (041) at Sr. Secondary Level.
The detailed modalities for opting this subject shall be made available at the time of registration for this course at class XI.
It may be noted that those students who have earlier offered Applied Mathematics as a Skill Elective will offer Applied Mathematics as an Academic Elective in the current year 2020-21 for class XII. They have to follow the syllabus of Applied Mathematics offered as an Academic subject. Applied Mathematics Course will not be available from the current Academic Session 2020-21 as Skill subject.
 

A set is a well-defined collection of objects, whose elements are fixed and cannot vary. It means set doesn’t change from person to person. Like for example, the set of natural numbers up to 7 will remain the same as {1,2,3,4,5,6,7}. Still, if we say the set of best players in a football team, then the name of footballers could vary every time we ask about the best players, as each person has its own choice to consider the best player. Similarly, if we speak about the set of rivers in India, the elements of the set will remain the same. So, this is a real-life example of a set. In mathematics, we represent the sets in curly brackets { }.

So the important point about circular arrangement is as follows:

1. If the clockwise and counter clockwise orders CAN be distinguished then total number of circular permutation of n elements taken all together = (n-1)!
2. If the clockwise and counter clockwise orders CANNOT be distinguished then total number of circular permutation of n elements taken all together = (n-1)! / 2

Example of indistinguishable case is, if you consider 5 diamonds and you want to make a necklace. In this case 5 diamonds can be arranged in a circle in (5-1)! = 24 ways. But in case of forming a necklace the clockwise and counter clockwise arrangements cannot be distinguished. So the total circular permutation in this case = (5-1)! / 2 = 4! / 2 = 12 ways.

P= x

A=2x

T= 6years

Step-by-step explanation:

SI= A-P

=2x- x

= x

SI=P*R*T/100

=x*r*6/100

R = 100/6

P = x

A = 4x

SI = 3x

SI = PRT/100

= x*100*T/ 6*100

= 18 years

18 YEARS

(i) There are 11 letters in the word 'MATHEMATICS' . Out of these letters M occurs twice, A occurs twice, T occurs twice and the rest are all different.
Hence, the total number of arrangements of the given letters
=11!/ (2!)×(2!)×(2!)=4989600. 
(ii) The given word contains 4 vowels AEAI as one letter, we have to arrange 8 letters MATHMTCS + AEAI, out of which M occurs twice, T occurs twice and the rest are all different.
So, the number of all such arrangements =8!/ (2!)×(2!)=10080 
Now, out of 4 vowels, A occurs twice and the rest are all distinct.
So, the number of arrangements of these vowels =4!/2!=12 
Hence, the number of arrangement in which 4 vowels are together =(10080×12)=120960.

whenever you have log on both sides you can remove log from both sides
now
on removing log from both sides..

log(2x+3)=log7
Let us simplify the expression,
2x+3=7
2x=7−3
2x=4
x=4/2
=2

n(A × B) = 5(4) = 20

n[(A × B) ∩ (B × A)] = 9

Step by step explanation:

Solution :

Given:

n(A) = 5 and n(B) = 4

Thus, we have:

n(A × B) = 5(4) = 20

A and B are two sets having 3 elements in common.

Now,

Let:

A = (a, a, a, b, c) and B = (a, a, a, d)

Thus, we have:

(A × B) = {(a, a), (a, a), (a, a), (a, d), (a, a), (a, a), (a, a), (a, d), (a, a), (a, a), (a, a), (a, d), (b, a),

(b, a), (b, a), (b, d), (c, a), (c, a), (c, a), (c, d)}

(B × A) = {(a, a), (a, a), (a, a), (a, b), (a, c), (a, a), (a, a), (a, a), (a, b), (a, c), (a, a), (a, a), (a, a),

(a, b), (a, c), (d, a), (d, a), (d, a), (d, b), (d, c)}

[(A × B) ∩ (B × A)] = {(a, a), (a, a), (a, a), (a, a), (a, a), (a, a), (a, a), (a, a), (a, a)}

∴ n[(A × B) ∩ (B × A)] = 9