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New answer posted

10 months ago

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Payal Gupta

Contributor-Level 10

 dydx+2xx1y=1 (x1)2

IF =e2xx1dx

e2x (x1)2

ye2x (x1)2= {e2x (x1)2 (x1)2dx+C

y = e2x2 (x1)2+C (x1)2

y (2) = 1+e42e4, C=12

y (3) = eα+1βeα=e6+18e6

New answer posted

10 months ago

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Payal Gupta

Contributor-Level 10

Data contradiction.

a* (b*c)= (ac)b (ab)c

New answer posted

10 months ago

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Payal Gupta

Contributor-Level 10

9 n 8 n 1 = ( 1 + 8 ) n 8 n 1

= ( 1 + 8 n + n ? C 2 8 2 + n ? C 3 8 3 + . . . . ] 8 n 1

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Payal Gupta

Contributor-Level 10

pva ⇒ (rvp)

(pq) (rvp)

its negation as asked in question

(pq) (pr)

(ppr) (qrp)

(prp) [asppisfalse]

New answer posted

10 months ago

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Payal Gupta

Contributor-Level 10

Let base = b

tan60°=hb

tan30°=h? 20b

New question posted

10 months ago

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New answer posted

10 months ago

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Payal Gupta

Contributor-Level 10

Mean=3+12+7+a+ (43a)5=13

Variance = 32+122+72+a2+ (43a)25 (13)2

2a2a+15Naturalnumber

Let 2a2 – a + 1 = 5x

D = 1 – 4 (2) (1 – 5n)

= 40n – 7, which is not 4λor4λ+1from.

As each square form is 4λor4λ+1

New answer posted

10 months ago

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Payal Gupta

Contributor-Level 10

Total number of possible relation = 2n2=24=16

Favourable relations = ? , { (x, x)}, { (y, y)}

{ (x, x), (y, y)}

{ (x, x), (y, y), (x, y), (y, x)}

Probability = 516

New answer posted

10 months ago

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Payal Gupta

Contributor-Level 10

Mid point of BC is 12 (5i^+ (α2)j^+9k^)

AB¯=i^+ (α4)j^+k^

AC¯=i^+ (2α)j^+k^

For = 1,  AB¯ and AC¯ will be collinear. So for non collinearity

= 2

New answer posted

10 months ago

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Payal Gupta

Contributor-Level 10

 x11=y22=z12=2 (1+4+216)1+22+22

(x, y, z) = (3, 6, 5)

now point Q and line both lies in the plane.

So, equation of plane is

|xyz+136612|=0

2x – z = 1

option (B) satisfies.

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