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a month ago

0 Follower 11 Views

A
Advik

Beginner-Level 5

Chandigarh University's BBA FinTech program offers a forward looking curriculum designed to match the evolving financial technology industry. Developed with KPMG, it introduces students to areas like cryptocurrency, regulatory technology (RegTech), digital payments, and risk analytics.  The program encourages practical learning through workshops, live instructor-led sessions, and projects where students design fintech solutions or simulate financial scenarios. Students also gain exposure to cloud computing tools, data visualization software, and AI applications in finance.

The course balances core business management subjects like

...more

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a month ago

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a month ago

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a month ago

1 Follower 12 Views

L
Lakshay Garg

Beginner-Level 1

You can't do BBA from CUET with PCB in 12 th as CUET requires the same subjects as in 12th for BBA you need math's English and general test

 

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a month ago

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K
Kanishka Gambhir

Contributor-Level 10

Yes, English language requirement is one of the important requirements for admission in MBA in Finland. Universities in Finland accepts IELTS exam scores as a proof of proficiency in the English language and students can refer to the table below for minimum score details:

University

Minimum IELTS Score Requirements

University of Helsinki

A minimum overall score of 6.5

(with a minimum of 6.0 in the writing section)

Aalto University
University of Turku
Lappeenranta University of Technology

A minimum overall score of 6.5

University of Eastern Finland

A minimum overall score of 6.5

(with a minimum of 5.5 in the writing section)

New Question

a month ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

z1 + z2 = 5

z 1 3 + z 2 3 = 2 0 + 1 5 i            

  z 1 3 + z 2 3 = ( z 1 + z 2 ) 3 3 z 1 z 2 ( z 1 + z 2 )           

z 1 3 + z 2 3 = 1 2 5 3 z 1 z 2 ( 5 )            

 ⇒ 20 + 15i = 125 – 15z1z2

3z1z2 = 25 – 4 – 3i

3z1z2 = 21– 3i

z1z2 = 7 – i

(z1 + z2)2 = 25

z 1 2 + z 2 2 = 2 5 2 7 ( 7 i )     

= 11 + 2i

  ( z ? 1 2 + z 2 2 ) 2         = 121 − 4 + 44i

  z 1 4 + z 2 4 + 2 ( 7 i ) 2 = 1 1 7 + 4 4 i

  z 1 4 + z 2 4 = 117 + 44i − 2(49 −1−14i )

= 21 + 72i

  | Z 1 4 + Z 2 4 | = 7 5

New Question

a month ago

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A
alok kumar singh

Contributor-Level 10

f is increasing function

x < 5x < 7x

f (x) < f (5x) < f (7x)

  f ( x ) f ( x ) < f ( 5 x ) f ( x ) < f ( 7 x ) f ( x )           

l i m x f ( x ) f ( x ) < l i m x f ( 5 x ) f ( x ) < l i m x f ( 7 x ) f ( x )             

-> 1 < l i m x f ( 5 x ) f ( x ) < 1 l i m x f ( 5 x ) f ( x ) = 1

l i m x ( f ( 5 x ) f ( x ) 1 ) = 0            

New Question

a month ago

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A
alok kumar singh

Contributor-Level 10

Let   A = [ x 1 y 1 z 1 x 2 y 2 z 2 x 3 y 3 z 3 ]

Given  A = [ 1 0 1 ] = [ 2 0 2 ]   .(1)


[ x 1 + z 1 x 2 + z 2 x 3 + z 3 ] = [ 2 0 2 ]

∴    x1 + z1 = 2                … (2)

x2 + z2 = 0               … (3)

x3 + z3 = 0                … (4)

Given   A = [ 1 0 1 ] = [ 4 0 4 ]

[ x 1 + z 1 x 2 + z 2 x 3 + z 3 ] = [ 4 0 4 ]

⇒   – x1 + z1 = −4             … (5)

–x2 + z2 = 0              

...more

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a month ago

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A
alok kumar singh

Contributor-Level 10

Area =   | 1 4 [ ( 4 x ) 2 ( x 4 ) 2 3 ] | d x

Area =   = | ( 1 6 6 4 3 ) ( 4 1 3 + 2 7 9 ) |

= | ( 1 6 6 4 3 ) ( 4 1 3 + 2 7 9 ) |

= | 1 6 6 4 3 4 + 1 3 + 3 |

= | 1 5 2 | = 6

New Question

a month ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

a + d, a + 7d and a + 43d are 1st, 2nd, 3rd term of G.P.

a + 7 d a + d = a + 4 3 d a + 7 d              

(a + 7d)2 = (a + d) (a + 43d)

a2 + 49d2 + 14d = a2 + 44ad + 43d3

6d2 = 30ad

d2 = 5d

d = 0, 5

a = 1, d = 5

  S 2 0 = 2 0 2 [ 2 + ( 1 9 ) 5 ]          

= 10 [95 + 2]

= 970

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a month ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

  a + b + 6 8 + 4 4 + 4 0 + 6 0 6 = 5 5

212 + a + b = 330

a + b = 118

x i 2 n ( x ¯ ) 2 = 1 9 4          

a 2 + b 2 + ( 6 8 ) 2 + ( 4 4 ) 2 + ( 4 0 ) 2 + ( 6 0 ) 2 6 = ( 5 5 ) 2 = 1 9 4

= 3219

11760 + a2 + b2 = 19314

a2 + b2 = 19314 – 11760

= 7554

(a + b)2 –2ab = 7554

From here b = 41.795

a + b = 118

a + b + 2b = 118 + 83.59

= 201.59

New Question

a month ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Let probability of tail is   1 3

Probability of getting head = 2 3  

Probability of getting 2 heads and 1 tail

= ( 2 3 × 2 3 × 1 3 ) × 3

= 4 2 7 × 3

= 4 9                  

                   

                   

New Question

a month ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

a = sin−1 (sin5) = 5 − 2π

and b = cos−1 (cos5) = 2π − 5

∴    a2 + b2 = (5 − 2π)2 + (2π − 5)2

= 8π2 − 40π + 50

New Question

a month ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Kindly go through the image 

 

New Question

a month ago

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A
alok kumar singh

Contributor-Level 10

This is an example of chromyl chloride test

K2Cr2O7 + 4KCl + 6H2SO4? 6KHSO4 + 2CrO2Cl2 + 3H2O

Oxidation state of Cr is +6.

New Question

a month ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

The balanced reaction is as follows

2 M n O 4 + 5 C 2 O 4 2 + + 1 6 H + 2 M n 2 + + 1 0 C O 2 + 8 H 2 O            

2 mole  M n O 4 react with 16 mole H+

1 mole  M n O 4 will react with 8 mole H+

New Question

a month ago

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A
alok kumar singh

Contributor-Level 10

A, D, E, K vitamins are fat soluble vitamins, are stored in liver and adipose tissue.

While vitamin B and vitamin C are water soluble and must be supplied regularly in diet (not stored) (except vitamin B12)  

New Question

a month ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

w = n R T l n V 2 V 1

| w | = 2 . 3 0 3 n R T l o g V 2 V 1

| w | = 1 × 2 . 3 0 3 × 8 . 3 1 4 × 3 0 0 l o g 1 0 0 1 0        

|w| = 5744 J

|w| = 5.744 kJ » 6 kJ

New Question

a month ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

t 9 0 = 2 . 3 0 3 k l o g ( 1 0 0 1 0 0 9 0 )

= 2 . 3 0 3 × 3 6 2 . 3 0 3 × l o g 2 × l o g 1 0 = 3 6 0 . 3 = 1 2 0

New Question

a month ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

A(g) ->B(g) + 1 2 (g)

Initial moles             n                         0             0

Eqb. moles               n(1 – a)   n α 2           na            

total moles =   n ( 1 + α 2 )

Eqb. pressure   ( 1 α ) p 1 + α 2 α p 1 + α 2 ( α 2 ) p 1 + α 2

K p = α p ( 1 + α 2 ) × [ α p ( 2 + α ) ] 1 2 ( 1 + α ) p 1 + α 2

K p = α 3 2 p 1 2 ( 2 + α ) 1 2 ( 1 α )        

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