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New answer posted

a year ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

V = 1 3 π r 2 h

V = π 8 h 3 t a n 2 α

d v d t = π h 2 t a n 2 α . d h d t

CSA = π r l

d ( C S A ) d t = 1 5 1 6 π 2 h . d h d t

1 5 8 * 2 3 = 5 m 2 / h r s

New answer posted

a year ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

t n = 2 ( 2 3 + 4 3 + 6 3 + . . . . + ( 2 n ) 3 ) − ( 1 3 − 2 3 + 3 3 + . . . . . ( 2 n ) 3 ) n ( 4 n + 3 )

2 * 2 3 * ( n * ( n + 1 ) 2 ) 2 − ( 2 n * ( 2 n + 1 ) 2 ) 2 n ( 4 n + 3 )

tn = n

n o w     ∑ n = 1 1 5 T n = 1 5 . 1 6 2

New answer posted

a year ago

0 Follower 17 Views

V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

New answer posted

a year ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

  x 2 − 1 0 x + 9 ≤ 0

  ( x − 1 ) ( x − 9 ) ≤ 0              

A = {1, 2, 3, ……, 9}

for set B,   f ( x ) ≤ ( x − 3 ) 2 + 1

total number of such function = 2 * 1 * 1 * 1 * 2 * 3 * 4 * 5 * 6 = 2 * 6! = 1440

New answer posted

a year ago

0 Follower 27 Views

V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

New answer posted

a year ago

0 Follower 6 Views

P
Payal Gupta

Contributor-Level 10

take z = x + iy

z2+z¯=0

⇒x2−y2+x+i  2xy−yi=0

⇒x2−y2+x=0  and  y (2x−1)=0

if y = 0 x = 0, 1

i f     x = 1 2 ⇒ y = ± 3 2

Σ ( R e ( z ) + l m ( z ) ) = ( 0 − 1 + 1 2 + 1 2 ) + ( 0 + 0 + 3 2 − 3 2 ) = 0

New answer posted

a year ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

  (x+1)2λ+5= (y+1)2λ+54=1

length of latus rectum = 2b2a=2 (λ+54)5+λ=4

λ+5=8⇒λ=59

Major axis = 2λ+5=16

New answer posted

a year ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

Let A =  [abcdef9hi]

Now ATA

trace will be a2+b2+c2+d2+e2+f2+92+h2=6

total ways = 9C6.26.1.1.1 = 5376

New answer posted

a year ago

0 Follower 5 Views

P
Payal Gupta

Contributor-Level 10

|f (x)|≤800⇒2n2−n−1≤800

⇒2n2−n−801≤0

∑x∈Sf (x)=∑ (2x2−x−1)

=2 (192+182+........12+02+12+.....+202)

= 10620

New answer posted

a year ago

0 Follower 11 Views

P
Payal Gupta

Contributor-Level 10

y5 – 9xy + 2x = 0

differentiate 5y4 – 9x dydx 9y + 2 = 0

dydx=9y−25y4−9x

For horizontal tangent dydx=0⇒y=29 which does not satisfy the equation so no horizontal

For vertical tangent 5y4−9x=0

m = 0, N = 2

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