Principal =
Required average
= (( 7 - 1 ) + ( 8 - 1 ) + ( 10 - 1 ) + ( 13 - 1 ) + (6 - 1) +( 10 - 1 ) )/ 6
= ( 7 + 8 +10 + 13 + 6 + 10 ) / 6 - ( 6 x 1 ) / 6
=9 - 1 = 8
Hexagon has 6 sides .
? n = 6
? Required number of diagonals = 10C2 - n
= 6C2 - 6
= 6! / [2!(6 - 2)!] - 6 = 6! / (2!4!) - 6
= (6 x 5 x 4!) / (2 x 4!) - 6
= 15 - 6
= 9
⟹ a = 5.
∴ 5(a - 3) = 5(5 - 3) = 52 = 25.
AB = 60 m, BC = 40 m and AC = 80 m
? s = (60 + 40 + 80 ) / 2 m = 90 m
(s-a) = 90 - 60 = 30 m,
(s-b) = 90 - 40 = 50 m and
(s-c) = 90 - 80 = 10 m
? Area of ? ABC =
?s(s-a)(s-b)(s-c)
= ?90 x 30 x 50 x 10 m2
= 300?15 m2
? Area of parallelogram ABCD = 2 x area of ? ABC
= 600?15 m2
(9.95)2 x (2.01)3 = 2 x (?)2
Here, 9.95 is Approximated to 10, 2.01 is approximate to 2 because calculating the square or cubes of whole number is much easier than to calculate the square or cube of decimal numbers
2 x (?) 2 = ( 10)2 x (2)3 = 100 x 8 = 800
? (?)2 = 400
? ? =20
Since, particular player is always chosen. It means that 11 - 1 = 10 players are selected out of the remaining 15 - 1 = 14 players.
? Required number of ways = 14C10
= 14 ! / (10! x 4!)
= (14 x 13 x 12 x 11) / (4 x 3 x 2 x 1)
= 7 x 13 x 11
= 91 x 11
= 1001
Thus, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.
∴ Number of ways of arranging these letters = | 8! | = 10080. |
(2!)(2!) |
Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.
Number of ways of arranging these letters = | 4! | = 12. |
2! |
∴ Required number of words = (10080 x 12) = 120960.
Total ways = 8C3 x 8C3
= (8 x 7 x 6)/(3 x 2) x (8 x 7 x 6)/(3 x 2)
= 56 x 56 = 3136
After fixing up one boy on the table, the remaining can be arranged in 4 ! ways, but boys and girls have to be alternate. There will be 5 places, one place each between two boys. These 5 place can be filled by 5 girls in 5 ! ways .
Hence, by the principle of multiplication, the required number of ways = 4 ! x 5 ! = 2880
In ||gm ABCD , we have
Since diagonals of parallelogram(||gm ) bisect each other,
? M will be the mid-point of each of the diagonal AC and BD
? In ?ABC AB2 + BC2 = 2(AM2 + MB2) ......... ( 1 ) [Appolonius Theorem]
In ?ADC AD2 + CD2 = 2(AM2 + DM2) ......... ( 2 )
= 2(AM2 + MB2) [? DM = BM]
Adding equations ( 1 ) and ( 2 )
AB2 + BC2 + CD2 + DA2 = 2(AM2 + MB2 + 2(AM2 + MB2 = 4AM2 + 4MB2
= (2AM)2 + (2MB)2= AC2+ BD2. { ? AM = MC , MB = MD }
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