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Yes, Ambalika Institute of Management and Technology offers direct admission. The institute also relies on entrance exams for many programs. Direct admissions are usually offered for courses like BBA, BCA, and BCom, but for BTech and MBA, entrance exams and subsequent counseling are required. The institute accepts national entrance exams such as CUET, JEE Main,  CAT, MAT, and others.

RV Institute of Technology and Management Bangalore offers seats to candidates based on their entrance exam scores. Candidates can get admission under Government Quota or Management Quota. The institute selects candidates for BE programme based on their KCET or COMEDK score. For MCA, aspirants must have a valid score in KMAT or Karnataka PGCET. Once the aspirants have passed the entrance exam, they need to participate in KEA Counselling for seat allotment. Once selected, aspirants have to pay the course fee to confirm their seat.

19. Given, x-coordinate of R = 4

Let R divides line segment joining points P(2, –3, 4) and Q(8, 0,10) internally in the ratio k : 1. Then coordinate of R is

$\left(\frac{k(8)+1(2)}{k+1},\frac{k(0)+1(-3)}{k+1},\frac{k(10)+1(4)}{k+1}\right)$

= $\left(\frac{8k+2}{k+1},\frac{-3}{k+1},\frac{10k+4}{k+1}\right)$

Then,

$\frac{8k+2}{k + 1}$ = 4

=> $8k+2=4\left(k+1\right)$

=> $8k+2=4k+4$

=> $8k-4k=4-2$

=> $4k=2$

=> $k=\frac{2}{4}$

=> $k=\frac{1}{2}$

Hence, $y=$ $\frac{-3}{k + 1}$

= $\frac{-3}{\frac{1}{2} + 1}$

= $-3÷\frac{1+2}{2}$

= $-\mathrm{3\times } \frac{2}{3}$

= $-2$

And,

z = $\frac{10k + 4}{k + 1}$

= $\frac{\left(10\frac{1}{2}\right) + 4}{\frac{1}{2} + 1}$

= $\left(5+4\right)÷\frac{1+2}{2}$

= $9\frac{2}{3}$

= 6

Therefore, coordinates of R is (4, –2, 6).

18. Let Q be the point on y-axis which are at a distance  from point P. As Q is on y-axis it has the coordinates of form (0, y, 0).

=> ${\left(0-3\right)}^2+{\left(y+2\right)}^2+{\left(0-5\right)}^2$  = 25 x 2

=> 9 +  $y^2+2^2+2.y.2+25=50$

=> $y^2+4y+34+4-50=0$

=> $y^2+4y-12=0$

=> $y^2+6y-2y-12=0$

=> $y\left(y+6\right)-2\left(y+6\right)=0$

=> $\left(y+6\right)\left(y-2\right)=0$

=> $y=-6, y=2$

So the coordinates Q are (0, 2, 0) and (0, –6, 0).

Yes, RVITM Bangalore accepts COMEDK scores for admission to BE programme. However, candidates can also get selected for BE based on their score in KCET. COMEDK is a national-level entrance exam conducted once in a year in Computer-based mode. The exam is usually conducted in May every year. 

In 2024, COMEDK cutoff for RVITM Bangalore BE was 14477 for CSE and 22853 for ECE. The ranks are not very competitive. Hence, aspirants can get admission with a decent rank in COMEDK.

17. We know that, the centroid of a triangle with vertices (x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃) is

$\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3},\frac{z_1+z_2+z_3}{3}\right)$

So, $\left(\frac{2a+(-4)+8}{3},\frac{2+3b+14}{3},\frac{6+(-10)+2c}{3}\right)$  = (0, 0, 0)

Equating the coordinates we get,

$\frac{2a+\left(-4\right)+8}{3}$ = 0

=> $2a-4+8=0$

=> $2a+4=0$

=> $2a=-4$

=> $a=\frac{-4}{2}$

=> $a=-2$

And,

$\frac{2+3b+14}{3}=0$

=> $2+3b+14=0$

=> $3b+16=0$

=> $3b=-16$

=> $b=\frac{-16}{3}$

And,

$\frac{6+\left(-10\right)+2c}{3}=0$

=> $6-10+2c=0$

=> $-4+2c=0$

=> $2c=4$

=> $c=\frac{4}{2}$

=> $c=2$

16. In a triangle ABC, the medians are the line segment that joins a vertex to the mid-point of the side that is opposite to that vertex. So, AE, BF and CG are the three medians.

15. Let D(x, y, z) be the fourth vertex of the parallelogram ABCD.

In a parallelogram, the diagonal AC and BD bisects each other at point say O.

=> $\left(\frac{3-1}{2},\frac{-1+1}{2},\frac{2+2}{2}\right)=\left(\frac{1+x}{2},\frac{2+y}{2},\frac{-4+z}{2}\right)$

=> (1, 0, 2) = $\left(\frac{1+x}{2},\frac{2+y}{2},\frac{-4+z}{2}\right)$

Equating the coordinates we get,

$\frac{1+x}{2}$ = 1

=> $1+x=2$

=> $x=1$

And

$\frac{2 + y}{2}$ = 0

=> $y=-2$

And

$\frac{- 4 + z }{2}$ = 2

=> $-4+z=4$

=> $z=4+4$

=> $z=8$

So, coordinates of fourth vertex is (1, –2, 8)

AIMT has an affordable fee structure. The institute offers UG and PG programmes across various streams such as Engineering, Management, etc. AIMT fee ranges between INR 27,000- INR 3.6 L. The fee structure includes tution fee, caution fee, application fee, and others. 

Integral university is an best choice for BPT course but to take admission you should have cleared the eligibility criteria to appear on the admission process

10+2 with PCM with more than 50%

You can reach out their website and can apply online and submit your documents and later they'll shortlist students based on marks

Entrance test will be held and if you pass that test you would be eligible to be an student in Integral university 

If you have further queries feel free to connect :)

Yes, Karnavati University BBA (Hons) admissions 2025 are currently open. As of now, the university is accepting online registrations and applications for the academic year 2025-26. Interested candidates can visit the university's official website to apply for admission. Before applying, they must ensure that they are eligible as per the course-specific eligibility criteria specified by the university. Further, candidates seeking admission based on their CUET UG/ UGAT scores must separately apply for the entrance exams.

4.39 Given, k₂ = 4k₁, T₁ = 293K and T₂ = 313K

We know that from the Arrhenius equation, we obtain

On solving, we get,

E_a = 58263.33 J mol^-1 or 58.26 kJ mol^-1

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4.37 From Arrhenius equation, we obtain

Also, k₁ = 4.5 × 10³ s ^-1

T₁ = 273 + 10 = 283 K

k₂ = 1.5 × 10⁴ s ^-1

E_a = 60 kJ mol ^-1 = 6.0 × 10⁴ J mol ^-1

Then,

→ 0.5229 = 3133.627 × (T₂-283)/ (283 × T₂)

→ 0.0472T₂ = T₂-283 T₂ = 297K or T₂ = 24⁰ C