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10 months ago

0 Follower 8 Views

V
Vipra Bisht

Contributor-Level 10

The University of Cincinnati has a huge alumni network that includes several famous people. Also, Cincinnati University has over 435,000 alumni who are working all over the world. A few leading working sectors where the alumni work are Information Technology, Engineering, Sales, Marketing, etc. Some popular personalities that are part of Cincinnati University alumni include:

Earl Hamner Jr.
Urban Meyer
Randy Harrison
Kathleen Battle
Rich Franklin
Joseph Strauss
Michael Graves

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10 months ago

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10 months ago

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P
Payal Gupta

Contributor-Level 10

25. Kindly go through the solution

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10 months ago

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Nishtha Rawat

Contributor-Level 7

After pursuing a University of East London program, graduates secure employment at the university campus and in the following companies:

Top Employers
NHSSainsbury's
AccentureNHS England
AmazonDeloitte
UCLInternational Business Times UK
PwCQueen Mary University of London
MicrosoftBusiness Advice Centre at the University of East London (BAC)

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10 months ago

0 Follower 6 Views

M
Manashjyoti Sharma

Contributor-Level 10

Cincinnati University rich alumni network include 350,000 members all over the world. The University of Cincinnati alumni network include mentors, leaders, etc. Some notable alumni at the university are Kathleen Battle, Randy Harrison, etc. 

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10 months ago

0 Follower 2 Views

H
Himanshi Pandey

Contributor-Level 10

No, there is no provision for direct admission to the PGDM course of FIIB Delhi. PGDM admission at the institute is based on the applicants' eligibility, entrance exam scores, and performance in the selection rounds. Candidates willing to take admission to the programme must have scored at least 50% in their graduation. Further, they are evaluated on the following criteria:

  • CAT/ GMAT/ XAT/ CMAT/ MAT/ ATMA score
  • Academic background

  • Work experience (if any)

  • Extracurricular achievements

  • Performance in Written Ability Test (WAT) and PI

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10 months ago

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Vishal Baghel

Contributor-Level 10

The equation of any plane through the intersection of the planes,

3x  y +2z ­4=0and x + y + z 2=0, is

(3x  y +2z 4)+α (x + y + z 2)=0,where,αR..........(1)

The plane passes through the point (2,2,1). Therefore, this point will satisfy equation (1).

(3×22+2×14)+α(2+2+12)=02+3α=0α=23

Substituting α=23 in equation (1), we obtain

(3xy+2z4)23(x+y+z2)=03(3xy+2z4)2(x+y+z2)=0(9x3y+6z12)2(x+y+z2)=07x5y+4z8=0

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10 months ago

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10 months ago

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Vishal Baghel

Contributor-Level 10

The equation of the plane ZOX is

y = 0

Any plane parallel to it is of the form,  y = a

Since the y-intercept of the plane is 3,

∴ a = 3

Thus, the equation of the required plane is y = 3

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10 months ago

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P
Payal Gupta

Contributor-Level 10

24. Kindly go through the solution

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10 months ago

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Vishal Baghel

Contributor-Level 10

2x+yz=5 ......... (1)

Dividing both sides of equation (1) by 5, we obtain

25x+y5z5=1x52+y5+z5=1.......... (2)

It is known that the equation of a plane in intercept form is xa+yb+zc=1 , where a,  b,  c are the intercepts cut off by the plane at x,  y, and z axes respectively.

Therefore, for the given equation,

a=52, b=5andc=5

Thus, the intercepts cut off by the plane are 52 , 5and5.

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10 months ago

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H
Himanshu Singh

Contributor-Level 10

Currently, NIILM University has not released the final application deadline for the BPharm programme 2025. Prospective students are advised to keep an eye on the official website or reach out to the admissions office for timely updates. Staying informed will help ensure that important dates, including the BPharm application deadline, are not missed.

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10 months ago

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Nikita Chauhan

Contributor-Level 9

Yes, if the degree is from a recognised institution, many recruiters accept it for Interior Design jobs. Some factors that affect the online degree are its curriculum, college rating, affiliation, practical exposure, etc.

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10 months ago

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I
Ishita Chauhan

Contributor-Level 10

The admit card for JET 2025 will be released on the official website on june 24, 2025. The exam date for Rajasthan JET is scheduled for June 24, 2025. 

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10 months ago

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A
alok kumar singh

Contributor-Level 10

41. Kindly consider the following

 

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10 months ago

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Vishal Baghel

Contributor-Level 10

We know that through three collinear points A,B,C i.e., through a straight line, we can pass an infinite number of planes.

(a) The given points are A(1,1,1),B(6,4,5),andC(4,2,3).

|111645423|=(1210)(1820)(12+16)

=2+24=0

Since A,B,C are collinear points, there will be infinite number of planes passing through the given points.

(b) The given points are A(1,1,0),B(1,2,1),andC(2,2,1).

|110121221|=(22)(2+2)=80

Therefore, a plane will pass through the points A, B, and C.

It is known that the equation of the plane through the points,  (x1, y1, z1),(x2, y2, z2)&(x3, y3, z3) , is

|xx1yy1zz1x2x1y2y1z2z1x3x1y3y1z3z1|=0|x1y1z011311|=0(2)(x1)3(y1)+3z=02x3y+3z+2+3=02x3y+3z=52x+3y3z=5

This is the Cartesian equation of the required plane.

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10 months ago

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P
Payal Gupta

Contributor-Level 10

23. (i) sin 75°= sin (45°+30°)

Using sin (x + y)= sin x cos y + cos x sin y we can write

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10 months ago

0 Follower 8 Views

I
Ishita Chauhan

Contributor-Level 10

The Rajasthan JET 2025 application process began on April 30, 2025, as per the official notification released by the exam conducting authority. The deadline to submit the Rajasthan JET 2025 Application form is May 28, 2025, without late fee, and May 31, 2025, with late fee. 

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10 months ago

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A
alok kumar singh

Contributor-Level 10

40. Kindly go through the solution

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10 months ago

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V
Vishal Baghel

Contributor-Level 10

(a) The position vector of point (1,0,2) is  a=i^2k^

The normal vector N perpendicular to the plane is  N=i^+j^k^

The vector equation of the plane is given by,  (ra).N=0

[r(i^2k^)].(i^+j^k^)=0.........(1)

r is the position vector of any point (x, y, z) in the plane.

r=xi^+yj^+zk^

Therefore, equation (1) becomes

[(xi^+yj^+zk^)(i^2k^)].(i^+j^k^)=0[(x1)i^+yj^+(z+2)k^].(i^+j^k^)=0(x1)+y(z+2)=0x+yz3=0x+yz=3

This is the Cartesian equation of the required plane.

(b) The position vector of the point (1,4,6) is  a=i^+4j^+6k^

The normal vector  N perpendicular to the plane is  N=i^2j^+k^

The vector equation of the plane is given by,  (ra).N=0

[r(i^+4j^+6k^)].(i^2j^+k^)=0.........(1)

r is the position vector of any point P(x, y, z) in the plane.

r=xi^+yj^+zk^

Therefore, equation (1) becomes

[(xi^+yj^+zk^)(i^+4j^+6k^)].(i^2j^+k^)=0[(x1)i^+(y4)j^+(z6)k^].(i^2j^+k^)=0(x1)+2(y4)+(z6)=0x2y+z+1=0

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