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9 months ago

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9 months ago

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A
alok kumar singh

Contributor-Level 10

Last two digit must be in form

23, 4, 5163243252}3×4=12

Total number of required number = 12 + 18 = 30

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9 months ago

0 Follower 4 Views

A
Akansha Bhandari

Contributor-Level 7

No. Candidates DO NOT need to send “Confirmation Page” printed after completion and submission of online UPTAC application form to the University. However, the candidate should keep the printout of the Confirmation page for record. Candidates are also advised to note down the Application Form Number for any future correspondence.

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9 months ago

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V
Vishal Baghel

Contributor-Level 10

 02+3p61p [23, 43], 02p81p [6, 2]and01p21p [1, 1]

0<P (E1)+P (E2)+P (E3)101312p81p [23, 263] Taking intersection to all p [23, 1]

p1+p2=53

New Question

9 months ago

0 Follower 3 Views

A
Abhay Arora

Contributor-Level 10

Netaji Subhas University of Technology appoints experienced and trained faculty members. NSUT Delhi faculty focuses on theoretical and practical learning. The faculty is committed to helping the students in their career pathways, they engage the students with the latest activities innovative teaching and quality research. 

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9 months ago

0 Follower 4 Views

S
Saurabh Khanduri

Contributor-Level 10

Yes, internships are a part of BTech curriculum in IIIT Ranchi. Students are required to undergo mandatory internships throughout their BTech course and submit an internship report for gaining the maximum credits for those semesters. 

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9 months ago

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alok kumar singh

Contributor-Level 10

The normal vector to the plane is n1¯×n2¯=|i3k1a111a|= (1a)i^+j^+k^

equationofplaneis (1a) (x1)+ (y1)+z=0

(1 – a)x + y + z = 2 – a …… (i)

Now distance from (2, 1, 4) = 3

3=|2 (1a)+1+4 (2a) (1a)2+1+1|

a2+2a8=0a=4, 2 the largest value of a = 2.

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9 months ago

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U
Upasana Kumari

Contributor-Level 10

Netaji Subhas University of Technology admission process begins with an online application. Students applying through JAC Delhi refer to the application process mentioned below:

Follow the steps given below to submit the online application form:

Step 1: Appear for the entrance exam.

Step 2: Apply at DTU/JAC Counselling.

Step 3: JAC Delhi & University Counselling.

Step 4: Document Verification for admission.

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9 months ago

0 Follower 3 Views

M
Ms Shruti Gupta

Contributor-Level 10

SRM IST Vadapalani BTech cutoff for 2025 is mainly decided based on SRMJEEE, which is university's own entrance exam held at a national level. The exam is fully online and takes place in three diffrent phases. Students are ranked according to their performance in the exam, and the cutoff varies for each specialisation and campus. Cutoffs may be higher for popular branches like CSE or AI-ML and slightly lower for others.

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9 months ago

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V
Vishal Baghel

Contributor-Level 10

Given, mean = np = . and variance = npq = α3q=13andp=23

P (X=1)=np1qn1=4243n (23)1 (13)n1=4243n=6

P (X=4or5)=6C4 (23)4 (13)2+6C5 (23)5 (13)1=1627

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9 months ago

0 Follower 10 Views

P
PRIYANKA

Contributor-Level 10

According to some unofficial sources, the top employers for university graduates are as follows:

  • NHS
  • Tees, Esk and Wear Valleys
  • Amazon
  • NHS England
  • Middlesbrough College
  • Wood
  • FUJIFILMSBiotechnologies
  • Northumbria University
  • GSK 

New Question

9 months ago

0 Follower 8 Views

A
alok kumar singh

Contributor-Level 10

 CAandCB=φ

If C is formed only by {1, 2, 4, 5} total number of subsets of A = 27.

Total number of subsets of {1, 2, 4, 5} = 24

 Number of subsets where CBφ

= 27 – 24 = 112

New Question

9 months ago

0 Follower 10 Views

R
Rashmi Sharma

Contributor-Level 7

Atma Ram Sanatan Dharma (ARSD) College has a vibrant campus life with numerous active clubs, societies, and events. Some of the notable ones include the Cultural Society, Nimbus (English Debating Society), Enactus (entrepreneurship and social inclusion), Rangayan (Dramatics Society), and Sarang (Singing Society). The college also hosts events such as TIDE (the annual fest), Aarambh (fresher's party), and various departmental fests and seminars. 

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9 months ago

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M
Ms Shruti Gupta

Contributor-Level 10

Students should focus on topics based on their Class 12 CBSE syllabus. The entrance test includes questions from Physics, Chemistry, Mathematics or Biology (depending on stream), English, and Aptitude. The test format is multiple-choice and it don't have negative marking. Preparation should start early as the exam can be tricky and competitive. The application fee is INR 1200 per phase, and students can appear in more than one phase if needed to improve their score.

New Question

9 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

Given : a=(α,1,1)andb=(2,1,α)c=a×b=|i^j^k^α1121α|

=(α+1)i^+(α22)j^+(α2)k^ Projection of c on d=i^+2j^2k^=|c.d|d||=30{Given}

|α14+2α22α+41+4+4|=30

On solving α=132 (Rejected as > 0) and = 7

New Question

9 months ago

0 Follower 2 Views

A
Anupama Singh

Contributor-Level 10

Netaji Subhas University of Technology admission is based on merit and entrance. The institute accepts national entrance exams such as JEE Main, JAC Delhi,  CUET, MAT, NATA, CAT,  CMAT, UCEED, and others. The institute admission process begins with an online application. Interested candidates must fill out the online application for the respective course on the official website of the Netaji Subhas University of Technology Delhi. 

New Question

9 months ago

0 Follower 11 Views

A
alok kumar singh

Contributor-Level 10

 Area=202 (1|x21|)dx=2 [01 (1 (1x2))dx+12 (2x2)dx]=83 (21)

New Question

9 months ago

0 Follower 4 Views

New Question

9 months ago

0 Follower 5 Views

A
Akansha Thakur

Beginner-Level 5

The NEET scores have to be added to the Kerala NEET 2025 counselling form in online mode by visiting the official website of counselling. The step by step process is given below. 

·       Visit the official website of Kerala NEET counselling 2025

·       Click the link 'KEAM 2025-Candidate Portal' and provide KEAM 2025 Application number and   Password to enter the Home Page of the candidate.

·       Click the Menu 'NEET Result Submission' and provide NEET (UG) 2025 Roll Number, NEET (UG)-2025 Application Number and NEET (UG)- 2025

...more

New Question

9 months ago

0 Follower 12 Views

A
alok kumar singh

Contributor-Level 10

H 1 = v 2 R 1 t 1 = v 2 R 1 * 2 0

H2 = v 2 R 2 * t 2 = v 2 R 2 * 6 0

from question,

H1 = H2

v 2 R 1 * 2 0 = v 2 R 2 * 6 0

R 2 R 1 = 3

Now connected is parallel

Req = R 1 R 2 R 1 + R 2

H = ( v 2 R 1 + v 2 R 2 ) t '

H = v 2 [ R 1 + R 2 R 1 R 2 ] t '

Now, According to question

H = H1 + H2

v 2 ( R 1 + R 2 R 1 R 2 ) t ' = v 2 R 1 * 2 0

t ' [ 1 + R 2 R 1 R 2 ] = 2 0 R 1

t ' [ 1 + 3 ] = 2 0 * 3

t ' = 6 0 4 = 1 5 m i n

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