Class 11th

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New answer posted

2 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Average speed = (f (t? )-f (t? )/ (t? -t? ) = a (t? +t? )+b.
Instantaneous speed = f' (t)=2at+b.
2at+b=a (t? +t? ) ⇒ t= (t? +t? )/2.

New answer posted

2 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

α, β are roots of x²+5√2x+10=0. α+β=-5√2, αβ=10.
P? + 5√2P? + 10P? = 0.
So P? +5√2P? =-10P? P? +5√2P? =-10P?
Expression = P? (-10P? )/P? (-10P? ) = 1.

New answer posted

2 months ago

0 Follower 16 Views

R
Raj Pandey

Contributor-Level 9

α+110+54+30+β=584 ⇒ α+β=390.
Median=45. L=40, N=584, C=α+110+54=α+164, f=30, h=10.
45 = 40 + [ (292- (α+164)/30]*10 = 40 + (128-α)/3.
5 = (128-α)/3 ⇒ 15=128-α ⇒ α=113.
β = 390-113=277.
|α-β|=164.

New answer posted

2 months ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

Ratio = ²? C? / (¹? C? + ¹? C? ) = ²? C? / ²? C? = 1.

New answer posted

2 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

I = m (l/√2)²*2 + m (√2l)² = 3ml²
L=Iω = 3ml²ω

New answer posted

2 months ago

0 Follower 1 View

A
Aadit Singh Uppal

Contributor-Level 10

Here are the important applications of alkanes in our everyday lives:

  • Fuel in Vehicles
  • Heating and Cooking
  • Lubrication
  • Power Generation
  • Wax and Solvents

New answer posted

2 months ago

0 Follower 25 Views

A
alok kumar singh

Contributor-Level 10

m (AP)=tan60°=√3. y-2√3=√3 (x-2). At x=1, y=√3. A= (1, √3).
m (AB)=tan120°=-√3. y-√3=-√3 (x-1) ⇒ √3x+y=2√3. (3, -√3) satisfies

 

New answer posted

2 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

(x+1)/ (x²/³-x¹/³+1) - (x-1)/ (x-x¹/²) = (x¹/³+1) - (1+x? ¹/²)
= (x¹/³ - x? ¹/²)¹?
general term = ¹? C? (x¹/³)¹? (-x? ¹/²)? = ¹? C? x^ (20-5r)/6)
For independent of x, 20-5r=0 ⇒ r=4
∴ Coefficient = ¹? C? = 210

New answer posted

2 months ago

0 Follower 9 Views

V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

New answer posted

2 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

| Class | No. of students | Number of possible cases |
| :-: | :-: | :- |
| 10 | 5 | (I) 2 | (II) 3 | (III) 2 |
| 11 | 6 | (I) 2 | (II) 2 | (III) 3 |
| 12 | 8 | (I) 6 | (II) 5 | (III) 5 |
Total cases =? C? *? C? *? C? +? C? *? C? *? C? +? C? *? C? *? C?
= 23,800 = 100K
∴ K = 238

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