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New answer posted

a year ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

[ a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 ] a 1 , b 1 , c 1 t { − 1 , 0 , 1 }

a 1 + a 2 + a 3 + . . . . ? 9     t i m e s = 5

Total = 414

New answer posted

a year ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

λ = 1 + μ

2 λ = 3 λ μ _ λ = 2 → S ( 2 , 4 , 6 )

μ = 1

n → = | i ^ j ^ k ^ 1 2 3 1 1 5 | = ( 7 , − 2 , − 1 )

7x – 2y – z + d = 0

(2, 4, 6) Þ d = -14 + 8 + 6

d = 0

7x – 2y – z = 0

7 – 2 = 5

New answer posted

a year ago

0 Follower 5 Views

R
Raj Pandey

Contributor-Level 9

g (1) =?

= f ( f ( f ( 1 ) ) ) + f ( f ( 1 ) )

f ( 1 ) = ( 2 ( 1 − 1 2 ) ( 2 + 1 ) ) 1 5 0 = 3 1 5 0

3 1 5 0 + 1 3 1 5 0 > 1 1 5 0

3 > 1

3 1 5 0 > 2   2 > 3 1 5 0 > 1

3 < 2 5 0   [ 3 1 5 0 + 1 ] = 1 + 1 = 2

 

New answer posted

a year ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

α n = 1 9 n − 1 2 n

3 1 α 9 − α 1 0 5 7 . α 2 = 3 1 ( 1 9 9 − 1 2 9 ) − ( 1 9 1 0 − 1 2 1 0 ) 5 7 ( 1 9 8 − 1 2 8 )

= 1 9 9 ( 3 1 − 1 9 ) − 1 2 9 ( 3 1 − 1 2 ) 5 7 ( 1 9 8 − 1 2 8 )

= 1 9 9 . 1 2 − 1 2 9 . 1 9 5 7 ( 1 9 8 − 1 2 8 ) = 4

New question posted

a year ago

0 Follower 4 Views

New answer posted

a year ago

0 Follower 16 Views

R
Raj Pandey

Contributor-Level 9

P ( x 1 , y 1 )

Q ( x 2 , y 2 )

x 1 + x 2 = r 2 , x 1 x 2 = P 2

y 1 + y 2 = s , y 1 , y 2 = − 9

( x − x 1 ) ( x − x 2 ) + ( y − y 1 ) ( y − y 2 ) = 0

2 ( x 2 + y 2 ) − r x − 2 s y + p − 2 q = 0

r = 11, s = 7, p – 2q = -22

New answer posted

a year ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

| 2 ( a → * b → ) | 2 + 4 ( a → . b → ) 2

= 4 | a → | 2 | b → | 2

=4 * 16 * 9 = 576

New answer posted

a year ago

0 Follower 14 Views

R
Raj Pandey

Contributor-Level 9

x 1 + x 2 = 6 , 1 3 → 6 + 6 = 1 2

x 1 + x 2 = 4 , 1 1 , 1 8 → 4 + 8 + 1 = 1 3 x 1 + x 2 = 2 , 9 , 1 6 → 2 + 9 + 3 = 1 4 x 1 + x 2 = 7 , 1 4 → 7 + 5 = 1 2 x 1 + x 2 = 5 , 1 2 → 5 + 7 = 1 2

= 63

New answer posted

a year ago

0 Follower 61 Views

R
Raj Pandey

Contributor-Level 9

∑ k = 1 1 0 ( 2 k − 1 ) . k . 1 0 C k

= 2 ∑ k 2     1 0 C k − ∑ k . 1 0 C k

x = 1 ⇒ ∑ k = 0 1 0 k . 1 0 C k = 1 0 . 2 9

x = 1 ⇒ ∑ k 2     1 0 C k = 9 0 . 2 8 + 1 0 . 2 9 = 2 8 ( 9 0 + 2 0 1 ) = 2 8 . 1 1 0

S = 29 . 110 – 10.29 = 29 . 100

S = 29 . 100

2 1 1 . 2 5 2 1 − 1

New answer posted

a year ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

  2 − 4 = − 1 2  

y – 4 = 2 (x – 3)

y = 2x – 2

x2 + (2x – 2)2 = 25

5 x 2 − 8 x − 2 1 = 0  

z ( − 7 5 , − 2 4 5 )  

             

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