Class 12th

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New answer posted

7 months ago

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V
Vishal Baghel

Contributor-Level 10

Given vectors a? and b? such that |a? | = |b? | and a? ⋅ b? = 0 (they are orthogonal).
The problem implies |a? |=|b? |=1.
Let c? = a? + b? + a? x b?
To find the magnitude of c? , we calculate |c? |²:
|c? |² = c? ⋅ c? = (a? + b? + a? x b? ) ⋅ (a? + b? + a? x b? ).
This expands to |a? |² + |b? |² + |a? x b? |² because all other dot products are zero (e.g., a? ⋅ b? = 0, a? ⋅ (a? x b? ) = 0).
|a? x b? |² = (|a? |b? |sin (90°)² = |a? |²|b? |².
So, |c? |² = |a? |² + |b? |² + |a? |²|b? |² = 1² + 1² + 1²*1² = 3.
∴ |c? | = √3.
To find the angle θ between c? and a? , we compute their dot product:
c? ⋅ a? = (a? + b? + a

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New answer posted

7 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Given functions f (x) = (x-2)/ (x-3) and g (x) = 2x-3.
First, find the inverse functions f? ¹ (x) and g? ¹ (x).
For f? ¹ (x): y = (x-2)/ (x-3) ⇒ y (x-3) = x-2 ⇒ xy - 3y = x-2 ⇒ xy-x = 3y-2 ⇒ x (y-1) = 3y-2 ⇒ x = (3y-2)/ (y-1). So, f? ¹ (y) = (3y-2)/ (y-1).
For g? ¹ (x): y = 2x-3 ⇒ y+3 = 2x ⇒ x = (y+3)/2. So, g? ¹ (y) = (y+3)/2.
We are given f? ¹ (x) + g? ¹ (x) = 13/2.
(3x-2)/ (x-1) + (x+3)/2 = 13/2.
Multiply by 2 (x-1): 2 (3x-2) + (x+3) (x-1) = 13 (x-1).
6x - 4 + x² + 2x - 3 = 13x - 13.
x² + 8x - 7 = 13x - 13.
x² - 5x + 6 = 0.
(x-2) (x-3) = 0.
The possible values of x are 2 and 3. Note that x=3 is not in the domain of t

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New answer posted

7 months ago

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R
Raj Pandey

Contributor-Level 9

Match the following:
Antifertility drug → Norethindrone
Antibiotic → Salvarsan
Tranquilizer → Meprobamate
Artificial Sweetener → Alitame

New answer posted

7 months ago

0 Follower 31 Views

V
Vishal Baghel

Contributor-Level 10

A relation R is defined as ARB if PAP? ¹ = B for a non-singular matrix P.

·       Reflexive: ARA requires PAP? ¹ = A. This holds if P is the identity matrix I, as IAI? ¹ = A. Assuming P can be I, the relation is reflexive.

·       Symmetric: We need to show that if ARB, then BRA.
ARB ⇒ PAP? ¹ = B.
To get the reverse, we need to express A in terms of B.
From PAP? ¹ = B, pre-multiply by P? ¹ and post-multiply by P:
P? ¹ (PAP? ¹)P = P? ¹BP ⇒ A = P? ¹BP. This shows BRA where the matrix is P? ¹. Thus, the relation is symmetric.

·       Trans

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New answer posted

7 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

Non- biodegradable wastes are generated by thermal power plants which produce fly ash. Bio-degradable detergents leads to eutrophication by decreasing oxygen level in water.

New answer posted

7 months ago

0 Follower 1 View

R
Raj Pandey

Contributor-Level 9

Covalent solids have high melting point due to strong covalent bonds, also they are insulator in both solid as well as in molten state.

New answer posted

7 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

In basic medium H? O? can behaves as both oxidizing and reducing agent, so it can oxidize Mn²? to Mn? and reduce I? to I?

New answer posted

7 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

F = (qQ)/ (4πε? r? ²) + (qQ)/ (4πε? r? ²) + (qQ)/ (4πε? r? ²) + (qQ)/ (4πε? r? ²) + .

⇒ F = (qQ)/ (4πε? ) [1/1² + 1/2² + 1/4² + 1/8² + .]

⇒ F = (qQ)/ (4πε? ) [1/ (1 - 1/4)]

⇒ F = (qQ)/ (4πε? ) [4/3] = 10? * 1 * 9 * 10? * (4/3) = 12 * 10³ N

New answer posted

7 months ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

Δλ = (v/c)λ = (286/3000000) * 630 * 10? = 6.006 * 10? ¹? ≈ 6 * 10? ¹? m

New answer posted

7 months ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

Both C? H? OH and AgCN can generate nucleophile.
KCN generate nitrile as nucleophile while AgCN generate isonitrile as nucleophile in nucleophilic substitution reaction.

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