Class 12th

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New answer posted

6 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

The circuit can be redrawn as an equivalent circuit given below

i = 5 10 = 0.5 A

New answer posted

6 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

  3 μ F and 3 μ F  in parallel

 

New answer posted

6 months ago

0 Follower 6 Views

R
Raj Pandey

Contributor-Level 9

(bonus)

Using the equation

e V = h v - ?

or 

e V 2 = h v 2 - h v T h

e V s = h u - h u T h

Data Incorrect

New answer posted

6 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

n = ? r u r

n = c v v = c n

V = c ? r μ r

New answer posted

6 months ago

0 Follower 6 Views

R
Raj Pandey

Contributor-Level 9

B = 0.5 T

Angle between B ?  & A A ?  is zero

? = B.A. c o s ? 0

= 0.5 * ( 1 ) * 1

= 0.5 W b

New answer posted

6 months ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

(P? ¹AP - I)²
= (P? ¹AP - I) (P? ¹AP - I)
= P? ¹A (PP? ¹)AP - P? ¹AP - P? ¹AP + I
= P? ¹A²P - 2P? ¹AP + I
= P? ¹ (A² - 2A + I)P = P? ¹ (A - I)²P
| (P? ¹AP - I)²| = |P? ¹ (A - I)²P| = |P? ¹| | (A - I)²| |P| = | (A - I)²| = |A - I|²
A - I = [1, 7, w²], [-1, w², 1], [0, -w, -w]
|A - I| = 1 (-w³ + w) - 7 (w) + w² (w) = -w³ + w - 7w + w³ = -6w.
|A - I|² = (-6w)² = 36w².

New answer posted

6 months ago

0 Follower 6 Views

R
Raj Pandey

Contributor-Level 9

Let A = [a? ]? Sum of diagonal elements of A.A? is Tr (A.A? ) = ∑? ∑? a? ² = 9.
where each a? ∈ {0, 1, 2, 3}.
Case I: One of a? = 3 and rest are 0. (3²=9). There are? C? = 9 ways.
Case II: Two of a? are 2, one is 1, and rest are 0. (2² + 2² + 1² = 9). There are? C? *? C? = 36 * 7 = 252 ways.
Case III: One of a? = 2, five are 1, and rest are 0. (2² + 1²+1²+1²+1²+1² = 9). There are? C? *? C? = 9 * 56 = 504 ways.
Case IV: All nine a? = 1. (1² * 9 = 9). There is 1 way.
Total = 9 + 252 + 504 + 1 = 766.

New answer posted

6 months ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

A = lim (n→∞) (2/n) ∑ (r=1 to n) f (r/n + n/ (n²)
(The term n/n² seems intended to be part of the function argument, not simply added. The solution proceeds as if it's f (r/n)
A = lim (n→∞) (2/n) ∑ (r=1 to n) [ f (r/n) + f (1/n) + . + f (n-1)/n) ]
The expression in the image seems to be: A = lim (n→∞) (2/n) [ f (1/n) + f (2/n) + . + f (n-1)/n) ]
A = 2 ∫? ¹ f (x) dx = 2 ∫? ¹ log? (1 + tan (πx/4) dx
put πx/4 = t ⇒ dx = 4/π dt
A = 2 ∫? ^ (π/4) log? (1 + tan (t) * (4/π) dt = (8/π) ∫? ^ (π/4) log? (1 + tan (t) dt
Using the property ∫? f (x)dx = ∫? f (a-x)dx, the integral ∫? ^ (π/4) log (1 + tan (t)dt ev

...more

New answer posted

6 months ago

0 Follower 16 Views

A
alok kumar singh

Contributor-Level 10

P will be the centroid of triangle ABC.
The centroid P is (x? +x? +x? )/3, (y? +y? +y? )/3).
The coordinates of P are given as (17/6, 8/3).
The coordinates of Q are not given, but a calculation is shown.
PQ = √ (24/6)² + (9/3)²) = √ (4² + 3²) = √ (16+9) = √25 = 5.
This implies the coordinates of Q are such that the difference in coordinates with P leads to this result. For example if P= (x? , y? ) and Q= (x? , y? ), then x? -x? =4 and y? -y? =3.

New answer posted

6 months ago

0 Follower 8 Views

R
Raj Pandey

Contributor-Level 9

dy/dx = 2 (x + 1)
dy = 2 (x + 1)dx.
y = (x + 1)² - c (1)
For y = 0, (x+1)² = c ⇒ x = -1 ± √c.
Area A = ∫? √c? ¹? √c (0 - [ (x+1)² - c])dx = ∫? √c? ¹? √c (c - (x+1)²)dx
A = [cx - (x+1)³/3] from -1-√c to -1+√c
= (c (-1+√c) - (√c)³/3) - (c (-1-√c) - (-√c)³/3)
= -c+c√c - c√c/3 + c+c√c - c√c/3 = 2c√c - 2c√c/3 = (4/3)c√c.
Given A = 4√8 / 3 = 8√2 / 3.
(4/3)c^ (3/2) = 8√2 / 3 ⇒ c^ (3/2) = 2√2 = 2^ (3/2) ⇒ c = 2.
∴ y = (x + 1)² - 2.
∴ y (1) = (1 + 1)² - 2 = 2.

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