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New answer posted
a year agoContributor-Level 9
Let P be (x?,y?)
Equation of normal at P is x/2x? - y/y? = 1/2
It passes through (-1/3√2, 0) ⇒ -1/(6√2x?) = 1/2 ⇒ x? = -1/(3√2)
So y? = 2√2/3 (as P lies in 1st Quadrant)
So β = y?/x? = (2√2/3)/(-1/3√2) = -4. (The solution gives a positive value, likely an error in the problem or my interpretation)
New answer posted
a year agoContributor-Level 10
If A ⊆ B and B ⊆ D then A ⊆ C
Contrapositive is
If A ⊄ C, then A ⊄ B or B ⊄ D
New answer posted
a year agoContributor-Level 9
We know, ?C? is max at middle term
a = ¹?C? = ¹?C? = ¹?C?
b = ²?C_q = ²?C?
c = ²¹C? = ²¹C? = ²¹C?
a/¹?C? = b/(²?C?) = c/(²¹C?) = 20/10, 21/11
New answer posted
a year agoContributor-Level 9
f(3)=f(4) ⇒ α=12
f'(x) = (x²-12)/(x(x²+12))
∴ f'(c)=0 ⇒ c=√12
∴ f''(c) = 1/12
New answer posted
a year agoContributor-Level 10
(1 + x)¹? + x(1+x)? + x²(1+x)? + . . + x¹?
= (1-x)¹? [1-(x/(1+x))¹¹]/[1-x/(1+x)]
⇒ (1+x)¹¹ - x¹¹
Coefficient of x? is ¹¹C? = 330
New answer posted
a year agoContributor-Level 10
2(cos²θ/sin²θ) - 5/sinθ + 4 = 0
(2sinθ - 1)(sinθ - 2) = 0
sinθ = 1/2 only
∴ θ = π/6, 5π/6
↓↓
θ?, θ?
∫(π/6 to 5π/6) cos²3θdθ = ∫(π/6 to 5π/6) (1+cos6θ)/2 dθ = π/3
New answer posted
a year agoContributor-Level 10
11.00
Let probability of hitting the target = p ⇒ p=1/2
Let n be the minimum number of bombs
According to given condition
1 - (?C?P?(1-P)? + ?C?P¹(1-P)?¹) ≥ 99/100
⇒ 2? ≥ (n+1)100
n=10 ⇒ 2¹? ≥ 1100 Reject
n=11 ⇒ 2¹¹ ≥ 1200 Select
New answer posted
a year agoContributor-Level 9
If each observation is multiplied with p and then q is subtracted
New mean x? = px? - q ⇒ 10 = p(20)-q
and new standard deviations σ? = |p|σ? ⇒ 1 = |p|(2) ⇒ |p|=1/2 ⇒ p=±1/2
If p=1/2, then q=0. If p=-1/2, q=-20.
New answer posted
a year agoContributor-Level 10
6.00
b·a = c·a
|a+b-c|² = |a|²+|b|²+|c|²+2(a·b - b·c - a·c)
= 4+16+16+2(a·b - 0 - a·b) = 36
⇒ |a+b-c| = 6
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