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5 months agoContributor-Level 10
No, the admissions for BTech courses at GITAM have been closed in the year 2025. The BTech admissions at GITAM require candidates to clear the JEE Mains entrance examination with an apropriate rank. The JEE Mains 2025 examination has been concluded. Candidates who appeared for the JEE Mains exam had to apply till July to gain admission into the BTech courses of GITAM.
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5 months agoContributor-Level 10
Dr DY Patil Arts, Commerce, and Science College Placement is known for its excellent placement record. Check out the tabulated data given below to know more about Dr DY Patil Arts, Commerce, and Science College placements 2024:
Course | Total Students (2024) | Students Placed (2024) | Median Salary (2024) |
|---|---|---|---|
UG 3-year | 1,983 | 510 | INR 4.5 LPA |
PG 2-year | 573 | 340 | INR 6 LPA |
NOTE: The above data is obtained from the NIRF report 2025
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5 months agoContributor-Level 10
P will be the centroid of triangle ABC.
The centroid P is (x? +x? +x? )/3, (y? +y? +y? )/3).
The coordinates of P are given as (17/6, 8/3).
The coordinates of Q are not given, but a calculation is shown.
PQ = √ (24/6)² + (9/3)²) = √ (4² + 3²) = √ (16+9) = √25 = 5.
This implies the coordinates of Q are such that the difference in coordinates with P leads to this result. For example if P= (x? , y? ) and Q= (x? , y? ), then x? -x? =4 and y? -y? =3.
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5 months agoContributor-Level 10
The maximum cutoff rank required in 2025 for admission to B.Tech. in CSE was 48300 at IIIT Bhopal for the students belonging to the General AI quota. Therefore, a cutoff rank of up to 45000 is considered to be a safe rank for CSE admission at this institute. Please note that the required cutoff rank varies every year for different categories. The round-wise maximum required cutoff ranks for CSE admission are tabulated below for OBC category students in 2025.
- Round 1: 10027
- Round 2: 10975
- Round 3: 11046
- Round 4: 11046
- Round 5: 11046
- Round 6: 11401
- Round 7: 45228
- Round 8: 46788
- Round 9: 48300
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5 months agoContributor-Level 10
Let t = 3^ (x/2). As x→2, t→3^ (2/2) = 3.
The limit becomes lim (t→3) [ (t² + 27/t²) - 12 ] / [ (t - 3²/t) ].
lim (t→3) [ (t? - 12t² + 27)/t² ] / [ (t² - 9)/t ].
lim (t→3) [ (t²-9) (t²-3) / t² ] * [ t / (t²-9) ].
lim (t→3) [ (t²-3) / t ].
Substituting t=3: (3²-3)/3 = (9-3)/3 = 6.
(The provided solution arrives at 36, let's re-check the problem statement)
The denominator is t - 9/t, not t - 3²/t.
lim (t→3) [ (t²-9) (t²-3) / t² ] * [ t / (t-3) (t+3)/t ]
This leads to the same cancellation. Let's re-examine the image's steps.
lim (t-3
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5 months agoContributor-Level 10
At CGC University, Mohali, students are offered several undergraduate, postgraduate, and doctoral courses. One must satisfy the admission criteria of the desired course to be eligible for admission. Students can check out the list below to know about the programmes offered by the university:
- BTech
- MTech
- BBA
- BCA
- PhD
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5 months agoContributor-Level 10
The TBSE 12th supplementary result is generally declared after two to three weeks on completion of compartment exams. The students can expect the board to release the TBSE HS supplementary result 2026 in September 2026. Students are suggested to keep their roll number ready before checking the result.
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5 months agoContributor-Level 10
Var (1, 2, ., n) = (Σn²/n) - (Σn/n)² = 10.
(n (n+1) (2n+1)/6n) - (n (n+1)/2n)² = 10.
(n+1) (2n+1)/6 - (n+1)/2)² = 10.
(n+1)/12 * [2 (2n+1) - 3 (n+1)] = 10.
(n+1)/12 * (4n+2 - 3n-3) = 10.
(n+1) (n-1)/12 = 10.
n² - 1 = 120 ⇒ n² = 121 ⇒ n = 11.
Var (2, 4, ., 2m) = Var (2* (1, 2, ., m) = 2² * Var (1, 2, ., m) = 16.
4 * Var (1, 2, ., m) = 16.
Var (1, 2, ., m) = 4.
Using the formula from above: (m²-1)/12 = 4.
m² - 1 = 48 ⇒ m² = 49 ⇒ m = 7.
m + n = 7 + 11 = 18.
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5 months agoContributor-Level 10
Yes, all candidates who are interested in securing admissions into the BTech courses at GITAM must appear for the JEE Mains entrance examination. The JEE Mains exam is conducted in two sessions in one year. Candidates who wish to register for the JEE Mains entrance exam can do so by visiting the official portal for the exam and following the instructions of the announcement circular.
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5 months agoContributor-Level 10
No, the TBSE class 12 compartment mark sheet is as same as the regular one. It will simply display your updated marks after clearing the subject. The board does not usually mention separately that it is a supplementary exam. This means you can use it for college admissions without any issues. Both schools and universities accept it as a valid result.
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5 months agoContributor-Level 9
f (x) + f (x + 1) = 2 (1)
replace x with x + 1: f (x + 1) + f (x + 2) = 2 (2)
(2) - (1) ⇒ f (x + 2) = f (x)
∴ f (x) is periodic with period 2.
I? = ∫? f (x)dx = 4 ∫? ² f (x)dx.
I? = ∫? ³ f (x)dx = ∫? f (u-1)du. Let u = x+1.
I? = ∫? f (x-1)dx = 2 ∫? ² f (x-1)dx.
From (1), f (x-1) + f (x) = 2.
I? + 2I? = 4∫? ² f (x)dx + 2 (2∫? ² f (x-1)dx) = 4∫? ² f (x)dx + 4∫? ² (2 - f (x)dx
= 4∫? ² (f (x) + 2 - f (x)dx = 4∫? ² 2 dx = 4 [2x] from 0 to 2 = 16.
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5 months agoContributor-Level 10
IIIT Kalyani cutoff 2025 varies for each round and category. For the General AI category, the cutoff score range for M.Tech in VLSI and Embedded Systems varied from 354 to 358 as per the last and first rounds' results. Different categories will have a variation in their GATE cutoff 2025, as in, for the OBC AI category, the cutoff score fell between 342 and 330. Please find below the cutoff score for the first and last rounds of different categories accordingly.
| Category | First Round Cutoff 2025 | Last Round Cutoff 2025 |
|---|---|---|
| OBC | 330 | 342 |
| General | 358 | 354 |
| SC | 261 | 354 |
| ST | 264 | 264 |
| EWS | 320 | 323 |
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5 months agoContributor-Level 9
y (x) = ∫? (2t² - 15t + 10)dt
dy/dx = 2x² - 15x + 10.
For tangent at (a, b), slope is m = dx/dy = 1 / (dy/dx) = 1 / (2a² - 15a + 10).
Given slope is -1/3.
2a² - 15a + 10 = -3
2a² - 15a + 13 = 0 (The provided solution has 2a²-15a+7=0, suggesting a different problem or a typo)
Following the image: 2a² - 15a + 7 = 0
(2a - 1) (a - 7) = 0
a = 1/2 or a = 7.
a = 1/2 Rejected as a > 1. So a = 7.
b = ∫? (2t² - 15t + 10)dt = [2t³/3 - 15t²/2 + 10t] from 0 to 7.
6b = [4t³ - 45t² + 60t] from 0 to 7 = 4 (7)³ - 45 (7)² + 60 (7) = 1372 - 2205 + 420 = -413.
|a + 6b| = |7 - 413| = |-406|
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5 months agoContributor-Level 10
You can check the TBSE 12th supplementary result on the official TBSE website by entering your roll number. The board provide a separate link to check TBSE HS compartment results on its websites at tbresults.tripura.gov.in and tbse.tripura.gov.in. Students should enter their roll number correctly to check the result. After the result appears, download and take a printout of the same.
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5 months agoContributor-Level 10
For IIT Madras summer internships, a candidate must fulfill the following criteria:
- Working Professionals are not eligible to apply for internships
- Students must pursue the IITM BSc Degree without other commitments (on-campus degree or full-time job)
- Students will have to fulfill training requirements before they apply for internships.
- Students should have completed 4 courses + 2 course projects at the Diploma level.
- Students will be allowed to apply once they complete at least one Diploma course.
- Non-refundable Registration fees of INR 1,800 (incl GST) must be submitted
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5 months agoContributor-Level 10
There are various different ways to apply for a visa to Sweden. See if there is any external service provider. You can go to the Swedish embassy website to see if there is any external service provider that has an agreement with the embassy. You can also directly go to the Swedish embassy and apply for a visa or use their e-visa services if provided for your country by Sweden.
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5 months agoContributor-Level 10
If your visa application for Sweden is rejected, then the embassy will inform you of the reason for the rejection of your application. If you are not satisfied with the decision, you can submit a fresh application for a visa after a cooling off period of 3 weeks from the day you received the decision.
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5 months agoContributor-Level 9
Yes, LNMIIT Jaipur admits BTech courses through JEE Main merit. Candidates are also required to meet the 60% eligibility criteria in Class 12 with PCM as subjects. Shortlisted candidates are given consideration for seat allocation during counselling. Selection is determined by merit and the availability of seats.
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