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The last date for BBA admission at DAITM Kolkata has not been officially announced yet. Students are advised to regularly check the official website of the institute for the latest updates and notifications regarding admission deadlines.

Normal to the required plane is perpendicular to the normals of the given planes.
n = n?  × n?  = (2i + j - k) × (i - j - k) = -2i + j - 3k.
Equation of the plane is -2 (x+1) + 1 (y-0) - 3 (z+2) = 0
-2x - 2 + y - 3z - 6 = 0
-2x + y - 3z - 8 = 0
2x - y + 3z + 8 = 0
Comparing with ax + by + cz + 8 = 0, we get a=2, b=-1, c=3.
a+b+c = 2-1+3 = 4.

Cr? O? ²? + 6Fe²? + 14H? → 2Cr³? + 6Fe³? + 7H? O
n-factor for Cr? O? ²? = 6, for Fe²? = 1
M.E Cr? O? ²? = M.E Fe²?
M? V? n? = M? V? n?
0.02 × 15 × 6 = M? × 10 × 1
M? = (0.02 × 15 × 6)/10 = 0.18 M = 18 × 10? ² M

 Yes, admissions are currently open for the BBA programme at DAITM Kolkata for the 2025 academic year. Interested candidates must have passed Class 12 and hold a valid MAKAUT CET score. It is recommended to apply early and regularly check the university's official website for updates on admission and exam deadlines.

|a × b|² + |a . b|² = |a|²|b|²
8² + (a . b)² = 2² * 5²
64 + (a . b)² = 100
(a . b)² = 36
a . b = 6 (since angle seems acute from options, but could be -6).

No, candidates are not allowed to choose more than one game for the Andhra Pradesh PECET skill test. They must select only one game during application process otherwise will be disqualified.

P (CO? ) = K? X (CO? )
X (CO? ) = P (CO? )/K? = 0.835 / (1.67 × 10³) = 0.5 × 10? ³
X (CO? ) = n (CO? )/ (n (CO? ) + n (H? O) ≈ n (CO? )/n (H? O) (since n (CO? ) << n (H? O)
n (H? O) in 0.9L = 900g/18gmol? ¹ = 50 mol
n (CO? ) = X (CO? ) × n (H? O) = 0.5 × 10? ³ × 50 = 25 × 10? ³ moles = 25 mmol

NIT Allahabad offers admission to MSc specialisations on the basis of IIT JAM exam results, followed by CCMN counselling. NIT Allahabad CCMN cutoff has been released up to round 4. The cutoff was released category-wise for the All India category candidates. Among the General AI category, the closing ranks varied between 739 and 1691. The most competitive specialisation among the General AI category candidates turned out to be M.Sc. in Mathematics and Scientific Computing. 

sinθ + cosθ = 1/2
16 (sin (2θ) + cos (4θ) + sin (6θ)
= 16 [2sin (4θ)cos (2θ) + cos (4θ)]
= 16 [4sin (2θ)cos² (2θ) + 2cos² (2θ) - 1] . (i)
Now, sinθ + cosθ = 1/2, squaring on both sides, we get
1 + sin (2θ) = 1/4
sin (2θ) = -3/4
cos² (2θ) = 1 - sin² (2θ) = 1 - 9/16 = 7/16
From equation (i)
16 [4 (-3/4) (7/16) + 2 (7/16) - 1]
16 [-21/16 + 14/16 - 16/16] = 16 [-23/16] = -23

C? H? + 13/2 O? → 4CO? + 5H? O
1 mole C? H? (58 g) produces 5 mole H? O (90 g)
∴ 90 g H? O obtained from 58 g C? H?
∴ 72g H? O obtained from (58/90) × 72g = 46.4 g
= 464 × 10? ¹g

Yes, Adarsh Group of Institutions offer Master of Buisness Administration through AIMIT. The course duration is two years that spans across four semesters. Selection into the MBA programme is based on KMAT scorecard. Selection of candidates is also subject to the fulfillment of the elgibility criteria set by the university.

No, the Candidate has to choose either High Jump or Long Jump. Candidate has to mention option in the Andhra Pradesh PECET application. Changing of option at the time of test is not permitted.

ΔH? = ΔH? + ΔH?
= 2.8 + 98.2 = 101 kJ/mole

S? : |z - 3 - 2i|² = 8
|z - (3 + 2i)| = 2√2
(x - 3)² + (y - 2)² = (2√2)²
S? : Re (z) ≥ 5
x ≥ 5
S? : |z - z? | ≥ 8
|2iy| ≥ 8
2|y| ≥ 8
|y| ≥ 4
y ≥ 4 or y ≤ -4
From the graph of the circle (S? ) and the regions (S? and S? ), we can see that there is one point of intersection at (5, 4).
∴ n (S? ∩ S? ∩ S? ) = 1

The tuition fee structure followed at Amrita School of Pharmacy for the BPharma courses is a per annum model. The total tuition fees at Amrita School of Pharmacy is INR 5 Lacs - INR 7 lakh. The duration for Amrita School of Pharmacy BPharma course is 4 years making each instalment of tuition fee INR 1.25 Lacs - INR 1.75 lakh. Candidates can check the tuition fee on the official website.

x² - |x| - 12 = 0
Case 1: x ≥ 0, |x| = x
x² - x - 12 = 0
(x-4) (x+3) = 0, x=4 (x=-3 is rejected)
Case 2: x < 0, |x| = -x
x² + x - 12 = 0
(x+4) (x-3) = 0, x=-4 (x=3 is rejected)
Two real solutions: 4 and -4.

Number of sigma bonds are 10.
[Structure showing 10 sigma bonds in the molecule]