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New Question

a month ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

t a n π 8 = 2 h 8 0 . . . . . . . . . ( i )

t a n θ = h 8 0 . . . . . . . . . . ( i i )

t a n π 8 = 2 t a n θ

t a n 2 θ = 3 2 2 4

New Question

a month ago

0 Follower 3 Views

N
Nishtha Shukla

Guide-Level 15

Jaipuria Institute of Management, Jaipur offers admission based on entrance exam scores of the aspirants. Those who meet the eligibility criteria have to first apply for the preferred programme. Aspirants must have a valid score in CAT, MAT, XAT or CMAT for PGDM programme. Admission to FPM is based on score of the candidates in the aptitude test conducted by the institute. Further, aspirants also have to pass a personal interview and other selection rounds for admission.

New Question

a month ago

0 Follower 2 Views

New Question

a month ago

0 Follower 3 Views

N
Nishtha Shukla

Guide-Level 15

Jaipuria Institute of Management, Jaipur considers entrance exam scores for admission to PGDM and FPM programmes. The institute accepts scores in CAT, MAT, XAT or CMAT for PGDM programme. Jaipuria Jaipur conducts an aptitude test to shortlist candidates for FPM programme. Hence, entrance exam is required for admission at Jaipuria Institute of Management, Jaipur.

New Question

a month ago

0 Follower 3 Views

N
Nishtha Shukla

Guide-Level 15

Jaipuria Institute of Management, Jaipur provides stipends to FPM scholars during the course. Aspirants need to meet the below conditions to avail the scholarship:

  • Aspirants must have obtained his or her master's from one of the top NIRF ranked institutions in 'Management' Category or 'Overall' Category.
  • Candidates should be an alumnus/ almuna of Jaipuria Institute of Management

New Question

a month ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

P : ( x + 3 y z 6 ) = λ ( 6 x + 5 y z 7 ) = 0

passes ( 2 , 3 , 1 2 )

( 2 + 9 1 2 6 ) = λ ( 1 2 + 1 5 1 2 7 ) = 0

λ = 1

| 1 3 a | 2 d 2 = ( 1 3 ) 2 ( 9 3 ) ( 1 3 ) 2 = 9 3

New Question

a month ago

0 Follower 1 View

V
Virajita Kumari

Contributor-Level 7

A raw GRE score is the total number of correct answers the a test taker get in different sections. The raw score is then scaled using statistical techniques to account for slight test difficulty and differences. This results in your final GRE score for Verbal or Quantitative sections.

New Question

a month ago

0 Follower 3 Views

N
Nishtha Shukla

Guide-Level 15

Jaipuria Institute of Management Jaipur provides scholarships to eligible and deserving candidates pursuing PGDM from the institute. The scholarships up to INR 3.5 lakh based on CAT/ CMAT/ MAT/ XAT scores to PGDM students. Aspirants must have documents to prove their eligibility to get the scholarship.

New Question

a month ago

0 Follower 3 Views

N
Nishtha Shukla

Guide-Level 15

Jaipuria Institute of Management, Jaipur application forms are available online. Candidates meeting the eligibility criteria can follow the below steps to apply for admission:

Step 1: Visit the official website of the Jaipuria Institute of Management.

Step 2: Click on the “Apply Now” tab.

Step 3: Register witht the required details to generate login credentials.

Step 4: Login to fill out the application form and upload the documents.

Step 5: Pay an application fee of INR 1,000 for PGDM and INR 1,050 for FPM and submit the form.

New Question

a month ago

0 Follower 2 Views

A
Aashi Tiwari

Contributor-Level 7

No, GRE score calculators only give you an overview of your score. There is nothing to do with official GRE scoring process with calculators. The ETS uses adaptive algorithms and equating methods that are not publicly released. On the other hand, GRE calculators give you a reliable estimate, this will help you to be prepare for upcoming goals.

New Question

a month ago

0 Follower 5 Views

N
Nishtha Shukla

Guide-Level 15

CAT is one of the accepting exams required for admission to PGDM programme. However, it is not compulsory. Aspirants with a valid score in any other accepted entrance exams such as MAT, XAT or CMAT can also get admission in PGDM. Selected aspirants also have to pass the case-based group discussion and a personal interview round for final admission.

New Question

a month ago

0 Follower 4 Views

N
Nishtha Shukla

Guide-Level 15

Jaipuria Institute of Management, Jaipur offers admission to both PGDM and FPM programmes based on entrance exam scores of the candidates. The institute considers scores in CAT, MAT, XAT or CMAT for PGDM programme. Admission to FPM is based on aptitude test conducted by the institute. Further, selected aspirants have to pass GDPI round for PGDM and an interview round for admission to FPM. Selected candidates have to confirm their seat by paying the course fee.

New Question

a month ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

( 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  

New Question

a month ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

Let vector along L is x

  x = | i j k 1 2 1 3 5 2 |  

 = i ^ j ^ k ^  

Area of  Δ P Q R = 1 2 | P Q × P R |  

  = 1 2 | i ^ j ^ k ^ 1 4 1 5 3 4 3 1 1 3 |  

= 4 3 3 8  

New Question

a month ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

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  

 

New Question

a month ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

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 Question

a month ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

f ( x ) = x 7 + 5 x 3 + 3 x + 1

f ' ( x ) = 7 x 6 + 1 5 x 4 + 3 > 0 x R

f ( x ) is increasing

 for x - , f (x)

x = 0, f (x) = 1

f ( x ) = 0 has only one real root.

 

New Question

a month ago

0 Follower 3 Views

New Question

a month ago

0 Follower 8 Views

R
Raj Pandey

Contributor-Level 9

[ x x 2 y 2 + e y / x ] x d y d x = x + [ x x 2 y 2 + e y / x ] y

e y / x [ x d y y d x ] = x d x + x x 2 y 2 ( y d x x d y )

e y / x d ( y / x ) = d x x d ( y / x ) 1 ( y / x ) 2

Integrating

e y / x = l n x s i n 1 ( y x ) + c

Passes (1, 0)

⇒ 1 = c

α = 1 2 e x p ( e 1 + π 6 )

New Question

a month ago

0 Follower 3 Views

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