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mω²acosθ = mgsinθ
ω = √ (gtanθ/a)
y = 4cx²

tanθ = dy/dx = 8xC
(tanθ)? , b = 8aC
ω = √ (g×8aC/a) = 2√ (2gC)

ρ? = 98 × 10?
ρ? = 2.65 × 10?
ρc = 1.724 × 10?
ρT = 5.65 × 10?

1. Let t? denotes r+1th term of (? x? +? x? )¹?
   t? = ¹? C? (? x? )¹? (? x? )? = ¹? C? ¹? x? ¹?
   If t? is independent of x
   90-15r=0? r=6
   This differs from the solution.
   Let's follow the solution's powers.
   (10-r)/9 - r/6 = 0? r=4
   maximum value of t? is 10K (given)
   ? ¹? C? is maximum
   By AM? GM (for positive numbers)
   (? ³/2+? ³/2+? ²/2+? ²/2)/4? (? /16)¹/?
   ? ? ? 16
   So, 10K = ¹? C?16
   ? K=336

ΔP = dsinθ = dθ
dy/D = (10? ³ × 1.270mm)/1m = 1.27µm

International students planning to prolong their stay in NJ are required to make on average around USD 108,992 (INR 95.44 L) per year, assuming the plan is to live alone. 

For families, the minimum annual salary requirement is around USD 282,714 (INR 2.47 Cr).

Source: SmartAsset via northjersey.com

∫? ² |x-1|-x|dx
Let f (x)=|x-1|-x|
= {|1-2x|, x≤1; 1, x≥1}
A = 1/2+1=3/2
∫? ¹/² (1-2x)dx+∫? /? ¹ (2x-1)+∫? ²1dx
= [x-x²]? ¹/²+ [x]? ² = 3/21dx
= [x-x²]? ¹/²+ [x]? ² = 3/2

Allahabad University offers PG courses via the CUET PG exam. Allahabad University has released a cutoff ranging from 54 to 246 for students belonging to the General AI quota. Wherein, M.Tech. in Materials Science and Technology was the easiest specialisation with a cutoff score of 54 and MA in Painting, the toughest to secure with a cutoff of 246. Other categories will have different cutoff scores accordingly.

Circle x²+y²-2x-4y+4=0
⇒ (x-1)²+ (y-2)²=1
Centre: (1,2) radius=1
line 3x+4y-k=0 intersects the circle at two distinct points.
⇒ distance of centre from the line < radius
⇒ |3×1+4×2-k|/√ (3²+4²) < 1
⇒ |11-k|<5
⇒ 6⇒ k∈ {7,8,9, .,15} since k∈I
Number of K is 9.

M = ∫ ρdV
M = ∫? (k/r) 4πr²dr
M = 4πkR?²/2 = 2πkR?²
F? = GMm/R?² = 2ω?²R

In the Andhra Pradesh EdCET exam, the Part C (Methodology/Subject) section carries the maximum marks, with 100 questions for 100 marks. This section tests the candidate's knowledge in a specific subject like Mathematics, Physical Sciences, Biological Sciences, Social Studies, or English, which they choose before the exam

lim (x→1) (x+x²+.+x? -n)/ (x-1) = 820
⇒ lim (x→1) (x-1)/ (x-1)+ (x²-1)/ (x-1)+.+ (x? -1)/ (x-1) = 820
⇒ 1+2+.+n = 820
⇒ n (n+1)=2×820
⇒ n (n+1)=40×41
Since n∈N, so n=40

Kindly consider the following Image

|a|=|b|=|c|=1
|a-b|²+|a-c|²=8
⇒|a|²+|b|²-2a.b+|a|²+|c|²-2a.c=8
⇒4-2 (a.b+a.c)=8
⇒a.b+a.c=-2
|a+2b|²+|a+2c|²
=|a|²+4|b|²+4a.b+|a|²+4|c|²+4a.c
=10+4 (a.b+a.c)
=10-8=2

FR > mgcosθR
F > mgcosθ
F > mg √ (R²- (R-a)²)/R ⇒ Mg√ (1- (R-a)²/R²)

MOTHER
1? E
2? H
3? M
4? O
5? R
6? T
So position of word MOTHER in dictionary
2*5!+2*4!+3*3!+2!+1
=240+48+18+2+1
=309

You should aim to practice at least 5 to 7 years of Andhra Pradesh EdCET previous year papers to be well-prepared for the exam. Focusing on papers from recent years is crucial, as they reflect the current trends and patterns of the exam. Target the last 3-5 years: At a minimum, you should solve the last 3 years' worth of papers to understand the latest exam trends and question types.

Zero error = 0 + 7 × 0.1 = 0.070
Vernier reading = (3.1 + 4 × 0.01) – 0.07 = 3.07

The New Jersey Institute of Technology fees for international students range between INR 1.6 L and INR 21.15 L per year. While the estimated cost of living for Indian students is INR 23.5 L. 

Thus, Indian students planning to study at NJIT can expect to spend in the range of INR 25 L-INR 45 L per year, depending on their course and the standard of living they opt to pursue. 

Hence, NJIT can be considered moderately costly.

Ambalika Institute of Professional Studies admits students based on entrance exam scores. The institute offers three Diploma courses, namely, Diploma in Electrical Engineering, Diploma in Computer Science and Diploma in Mechanical Engineering Automobile. There are two ways to enrol for admission. The first route is through JEECUP, and the second one is through JEE exam scores. 

Kindly consider the following figure

|x|<1, |y|<1, x? y
(x+y) + (x²+xy+y²) + (x³+x²y+xy²+y³) + .
By multiplying and dividing by x-y:
(x²-y²)+ (x³-y³)+ (x? -y? )+.)/ (x-y)
= (x²+x³+x? +.)- (y²+y³+y? +.)/ (x-y)
= (x²/ (1-x)- (y²/ (1-y)/ (x-y)
= (x²-y²-xy (x-y)/ (1-x) (1-y) (x-y)
= (x+y-xy)/ (1-x) (1-y)