Class 11th

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New answer posted

4 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

(x/m) = k (P)¹/?
log (x/m) = log k + (1/n) log P
Slope = 1/n = 2 So n = ½
Intercept ⇒ log k = 0.477 So k = Antilog (0.477) = 3
So (x/m) = k (P)¹/? = 3 [0.04]² = 48 * 10?

New answer posted

4 months ago

0 Follower 5 Views

R
Raj Pandey

Contributor-Level 9

In ΔCDF
sin 30° = z/1 [CD = 1 km (given)]
z = 1/2

cos 30° = y/1 ⇒ y = √3/2

Now in ΔABC
tan 45° = h/ (x+y)
⇒ h = x+y
⇒ x = h - √3/2

Now
In ΔBDE,
tan 60° = (h-z)/x
√3x = h - z
√3 (h - √3/2) = h - 1/2
√3h - 3/2 = h - 1/2
h (√3 - 1) = 1
h = 1/ (√3-1) km

New answer posted

4 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Na? SO? + H? SO? → SO? + Na? SO? + H? O
SO? + 2NaOH → Na? SO? + H? O
Na? SO? + SO? + H? O → 2NaHSO?

New answer posted

4 months ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

x? = Σf? x? / Σf? = (10 + 15x + 50) / (4+x)
= (60+15x)/ (4+x) = 15
σ² = 50 = Σf? x? ²/Σf? - (x? )²
50 = (50+225x+1250)/ (4+x) - (15)²
50 = (1300+225x)/ (4+x) - 225
⇒ 275 (4+x) = 1300 + 225x
⇒ 50x = 200 ⇒ x = 4

New answer posted

4 months ago

0 Follower 12 Views

V
Vishal Baghel

Contributor-Level 10

Let P (2cosθ, 2sinθ)
∴ Q (-2cosθ, -2sinθ)
Given line x+y-2=0
∴ α = |2cosθ + 2sinθ – 2| / √2
β = |-2cosθ - 2sinθ – 2| / √2
∴ αβ = √2 (cosθ + sinθ – 1) · √2 (cosθ + sinθ + 1)
= 2|cos²θ + sin²θ + 2sinθcosθ – 1| = 2|sin2θ|
Max |sin2θ| = 1
∴ maximum αβ = 2.

New answer posted

4 months ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

Ways of selecting correct questions =? C? = 15
Ways of doing them correct = 1
Ways of doing remaining 2 questions incorrect = 3² = 9
∴ No. Of ways = 15 * 1 * 9 = 135

New answer posted

4 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Given,
300 = 1 + (N – 1)d
⇒ (N − 1)d = 299
∴ (N, d) = (24,13) is the only possible pair
∴ a? = 1 + 19 (13) = 248 and, S? = (1+248)/2 * 20
= 2490

New answer posted

4 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Any point (x, y) on perpendicular bisector equidistant from p and q
∴ (x − 1)² + (y − 4)² = (x − k)² + (y − 3)²
At x = 0, y = -4
∴ 1 + 64 = k² + 49
k² = 16

New answer posted

4 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Let number of elements in T is R.
∴ 20R = 500 ⇒ R = 25
and 6R = 5N ⇒ N = 30

New answer posted

4 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Given : a/e = 4 and 1/4 = 1 - b²/a²
Solving : a = 2, b = √3
Parametric co - ordinates are
(2cosθ, √3sinθ) = (1, β)
∴ θ = 60°
∴ Equation of normal is axsecθ − bycosecθ = a² − b²
⇒ 4x - 2y = 1

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