Ask & Answer: India's Largest Education Community

(6!/ (5!1!) * 2! + (6!/ (4!2!) * 2! + (6!/ (3!)²2!) * 2! = 62

Kindly consider the following Image 

The last date to register for GATE 2026 with late fees was October 13. For submission with late fees, the application form window  remained active from October 7 to October 13. The registration fee with late fees is as follows -

• For female/SC/ST and PwD candidates, the fee is INR 1500
• For all other candidates, including foreign nationals, the fee is INR 2500

First element in 20^th group is equal to ‘20’.

(20, 21, ….upto 39 times)

$\therefore Sum=\frac{39}{2}\left(2\times 20+\left(39-1\right)\times 1\right)$        

(k-3)/ (h-2) × (k-0)/ (h-0) = -1
⇒ k (2k – 3) = -2 (h – 2)h
⇒ 2h² + 2k² – 4h – 3k = 0
2x² + 2y² – 4x – 3y = 0
(0,0)

$\stackrel{\stackrel{+4}{{\mathrm{N}\mathrm{O}}_2}}{2}+{\mathrm{H}}_2\mathrm{O}\to \stackrel{+3}{{\mathrm{H}\mathrm{N}\mathrm{O}}_2}+\stackrel{+5}{\mathrm{H}\mathrm{N}\mathrm{O}}{}_3$

$\stackrel{0}{{\mathrm{C}\mathrm{l}}_2}+{\mathrm{H}}_2\mathrm{O}\to \stackrel{-1}{\mathrm{H}\mathrm{C}\mathrm{l}}+\mathrm{H}\mathrm{O}\mathrm{C}\mathrm{l}$

$2\stackrel{+4}{{\mathrm{C}\mathrm{l}\mathrm{O}}_2}+{\mathrm{H}}_2\mathrm{O}\to \stackrel{+3}{{\mathrm{H}\mathrm{C}\mathrm{l}\mathrm{O}}_2}+\stackrel{+5}{{\mathrm{H}\mathrm{C}\mathrm{l}\mathrm{O}}_3}$

$\stackrel{+6}{{\mathrm{C}\mathrm{l}}_2}{\mathrm{O}}_6+{\mathrm{H}}_2\mathrm{O}\to \stackrel{+5}{{\mathrm{H}\mathrm{C}\mathrm{l}\mathrm{O}}_3}+\stackrel{+{+}^{+}}{{\mathrm{C}\mathrm{l}\mathrm{O}}_4}$

(5³)^740 * 5² = 5² (125)^740 = 5² (126 – 1)^740
So remainder = 4.

C.V. = (σ/x? ) × 100 ⇒ σ = (C.V. × x? ) / 100
∴ σ? = (50×30)/100 = 15 and σ? = (60×25)/100 = 15 ⇒ σ? - σ? = 0

${\mathrm{C}\mathrm{r}\mathrm{O}}_4^{2-}+2{\mathrm{H}}^{+}+2{\mathrm{H}}_2{\mathrm{O}}_2?\underset{\text{Chromium Pentoxide}}{{\mathrm{C}\mathrm{r}\mathrm{O}}_5+3{\mathrm{H}}_2\mathrm{O}}$

Equation of tangent is (x/a) (1/2) + (y/b) (√3/2) = 1
Equation of auxiliary circle is x² + y² = a²
Homogenising (ii) with (i) and making coefficient of x² + coefficient of y² = 0
⇒ (3a²/4b²) - (7/4) = 0 ⇒ e = 2/√7

The authority will announce the UPESEAT 2026 exam date online.

$N_1{V}_1=N_2{V}_2$

$0.01*V_1=0.01*20*3$

$V_1=60\mathrm{m}\mathrm{L}$

f' (x) = cosx + sinx − k ≤ 0∀x ∈ R
k ≥ √2

Molarity $=\frac{\text{moles of}\text{solute}}{\text{volume of solution in}L}=\frac{1*1.15}{1342}*1000=0.86\mathrm{M}$

Candidates can download JK SET answer key PDF using the following steps:

1. Visit the official website 
2. Click on the 'Display/ Challenge Answer Key' link
3. Accept the declaration
4. Enter the roll number and mobile number
5. Download the key PDF

UPES, University of Petroleum and Energy Studies, is a private university.

The JKSET answer key 2025 will be released soon on the official website- jujkset.in. The JKSET exam answer key PDF download link will be shared here once released - https://www.shiksha.com/exams/jk-set-exam-answer-key

The number of >C = 0 groups in tripeptide is 0 .

Kindly consider the following figure

log? (abc) = 6 ⇒ abc = 6?
a = b/r & c = br
⇒ b = 36 and a = 36/r
⇒ r = 2,3,4,6,9,12,18
Also b - a = 36 (1 - 1/r) is a perfect cube.
∴ r = 4 ⇒ a + b + c = 9 + 36 + 144 = 189