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New answer posted

4 months ago

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P
Payal Gupta

Contributor-Level 10

A = {1, 2, 3, ….50}

R1 = (2, 1), (2, 2), (2, 4)…. (2, 32)

(3, 1) (3, 3) (3, 9) (3, 27)

(13, 1) (13, 13)   etc.

R2 = { (2, 1), (2, 2), (3, 1), (3, 3), (5, 1), (5, 5), (7, 1), (7, 7), (1, 1), (11, 11)…}

R 1 R 2 contains only 9 elements.

New answer posted

4 months ago

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P
Payal Gupta

Contributor-Level 10

Consider the following image

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4 months ago

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P
Payal Gupta

Contributor-Level 10

tanπ8=2h80......... (i)

tanθ=h80.......... (ii)

tanπ8=2tanθ

tan2θ=3224

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4 months ago

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Payal Gupta

Contributor-Level 10

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New answer posted

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P
Payal Gupta

Contributor-Level 10

 P: (x+3yz6)=λ (6x+5yz7)=0

Passes  (2, 3, 12)

(2+9126)=λ (12+15127)=0

λ=1

|13a|2d2= (13)2 (93) (13)2=93

New answer posted

4 months ago

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Payal Gupta

Contributor-Level 10

( 5 x + 8 y + 1 3 z 2 9 ) + λ ( 8 x 7 y + z 2 0 ) = 0  

P1 passing (2, 1, 3)

(10 + 8 + 39 – 29) + λ ( 1 6 7 + 3 2 0 ) = 0  

2 8 8 λ = 0 λ = 7 / 2               

2X – Y + Z – 6 = 0      ….(i)

For P2 passes (0, 1, 2)

( 8 + 2 6 2 9 ) + λ ( 7 + 2 2 0 ) = 0              

5 2 5 λ = 0 λ = 1 5               

P 2 : x + y + 2 z 5 = 0 . . . . . . . . ( i i )               

Acute angle between the planes

c o s θ = 2 1 + 2 4 + 1 + 1 1 + 1 + 4 = 1 2               

θ = π 3            

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4 months ago

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Payal Gupta

Contributor-Level 10

Let vector along L is x

x=|ijk121352|

i^j^k^

Area of ΔPQR=12|PQ*PR|

=12|i^j^k^1415343113|

= 4 3 3 8

New answer posted

4 months ago

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P
Payal Gupta

Contributor-Level 10

x 2 + y 2 2 x 4 y = 0

 Centre (1, 2) r = 5  

Equation of OQ is x . 0 + y . 0 – (x + 0) – 2 (y + 0) = 0

⇒ x + 2y = 0       ……… (i)

Equation of PQ is  x ( 1 + 5 ) + 2 y ( | x + 1 + 5 ) 2 ( y + 2 ) = 0

Solving (i) & (ii), Q ( 5 + 1 , 5 + 1 2 )  

= 1 2 | 6 + 2 5 + 4 5 = 4 2 | = 6 5 + 1 0 4 = 3 5 + 5 2

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Payal Gupta

Contributor-Level 10

a 2 ( e 2 1 ) = b 2  

e = 5 2 b 2 = 3 a 2 2                

Length of latus rectum 2 b 2 a = 6 2  

3 a = 6 2 a = 2 2                

b = 2 3                

y = 2x + c is tangent to hyperbola

c 2 = a 2 m 2 b 2 = 2 0        

New answer posted

4 months ago

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P
Payal Gupta

Contributor-Level 10

x (1x2)dydx+ (3x2yy4x3)=0

x (1x2)dydx+ (3x21)y=4x3

y=2x1 (x3x)y (3)=18

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