Class 12th
Get insights from 12k questions on Class 12th, answered by students, alumni, and experts. You may also ask and answer any question you like about Class 12th
Follow Ask QuestionQuestions
Discussions
Active Users
Followers
New answer posted
a year agoContributor-Level 10
Required area (above x-axis)
A? = 2∫? (8/2 - x - √x)dx
= 2 [16 - 16/4 - 8/3*2] = 40/3
and A? = 4 (1/2 k²) = 2k²
∴ 27 * (40/3) = 5 * (2k²)
=> k = 6
for above x-axis.
New answer posted
a year agoContributor-Level 10
|adj (adj (A)| = |A|? ¹² = |A|?
|A| = | (14, 28, -14), (-14, 28), (25, -14, 14) |
= (14)² | (1, 2, -1), (-1, 2), (25/14, -1, 1) |
= (14)² (3–2 (–5)–1 (–1)
|A|? = (14)? => |A| = 14²
New answer posted
a year agoContributor-Level 10
Y = A·B + B·C
(i) o to t? A = 0, B = 0, C = 1
(ii) Y = 0.0 + 0.1 = 0 + 1 = 1
(ii) t? to t? A = 1, B = 0, C = 1 Y = 1.0 + 0.1 = 0 + 1 = 1
(iii) t? to t? A = 0, B = 1, C = 0 Y = 0.1 + 0.1 = 0 + 1 = 1
New answer posted
a year agoContributor-Level 10
f (x) = 2e²? / (e²? +e? ) and f (1-x) = 2e²? / (e²? +e¹? )
∴ f (x) + f (1-x) = 1/2
i.e. f (x) + f (1-x) = 2
∴ f (1/100) + f (2/100) + . + f (99/100)
Σ? f (x/100) + f (1-x/100) + f (1/2)
= 49 x 2 + 1 = 99
New answer posted
a year agoContributor-Level 10
sin? ¹ (√3/2) + cos? ¹ (-√3/2) + tan? ¹ (-1)
= π/3 + 5π/6 – π/4
= (4π+10π–3π)/12 = 11π/12
New answer posted
a year agoContributor-Level 10
(cos (2π/7) + cos (4π/7) + cos (6π/7)/ (sin (3π/7)cos (π/7) = (2sin (π/7)cos (2π/7) + 2sin (π/7)cos (4π/7) + 2sin (π/7)cos (6π/7)/ (2sin (π/7)
= (sin (3π/7)-sin (π/7) + sin (5π/7)-sin (3π/7) + sin (π)-sin (5π/7)/ (2sin (π/7) = (-sin (π/7)/ (2sin (π/7) = -1/2
New answer posted
a year agoContributor-Level 10
Given P (X=3) = 5P (X=4) and n=7
=>? C? p³q? = 5? C? p? q³
=> q = 5p and also p + q = 1
=> p = 1/6 and q = 5/6
Mean = 7/6 and variance = 35/36
Mean + Variance
= 7/6 + 35/36 = 77/36
New answer posted
a year agoContributor-Level 10
V = ir
i = V/r
i = (1/r) [v is same for r? & r? ]
i? /i? = r? /r?
i? = (r? / (r? +r? )i?
i? /i? = r? / (r? +r? )
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 67k Colleges
- 1.2k Exams
- 716k Reviews
- 1850k Answers





