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Tanvi Chauhan?? 6 years, 7 months ago
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Sia ? 6 years, 7 months ago
Total Number of students in class = 102
Students Failed in Maths = 50
Students Failed in Physics= 45
Students Failed in Chemistry = 40
Students Failed in two or three subjects = 32
Total number of students failed = 167 students
Students Failed in two or three Subjects = 167 - 102
Students failed in all three Subjects = 65
Posted by Tanvi Chauhan?? 6 years, 7 months ago
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Sia ? 6 years, 7 months ago
sin11asina+sin7asin3a/cos11asina+cos7asin3a
=2sin11asina+2sin7asin3a/2cos11asina+2cos7asin3a
=[{cos(11a-a)-cos(11a+a)}+{cos(7a-3a)-cos(7a+3a)}]/
[{sin(11a+a)-sin(11a-a)}+{sin(7a+3a)-sin(7a-3a)}]
=(cos10a-cos12a+cos4a-cos10a)/(sin12a-sin10a+sin10a-sin4a)
=(cos4a-cos12a)/(sin12a-sin4a)
=[2sin(4a+12a)/2sin(12a-4a)/2]/[2cos(12a+4a)/2sin(12a-4a)/2]
=sin8asin4a/cos8asin4a
=sin8a/cos8a
=tan 8a
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Tanvi Chauhan?? 6 years, 7 months ago
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Sia ? 6 years, 7 months ago
A relation from a set A to another set B by definition is a subset of the Cartesian product of the two sets A and B i.e.( A x B ) . If A has 2 elements and B has 3 elements then (A x B ) has (2 x 3 ) = 6 elements in it.
Therefore no. of subsets of (A x B) = 2^(6) = 64 ; that are the required numbers of relations from A to B.
Posted by Abhay Dwivedi 6 years, 7 months ago
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Sia ? 6 years, 7 months ago
tan(13π/12)
= tan(4π/3 - π/4),
= [tan(4π/3) - tan(π/4)]/[1 + tan(4π/3)tan(π/4)]
= (-1 + √3)/(1 + √3)
= -(1 - √3)/(1 + √3)
= -(1 - √3)^2/[(1 + √3)(1 - √3)],
= -(1 - √3)^2/(1 - 3)
= (1/2)(1 - √3)^2
= (1/2)(1 - 2√3 + 3)
= (1/2)(4 - 2√3)
= 2 - √3.
Posted by Prateek Thakur 6 years, 7 months ago
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Sia ? 6 years, 7 months ago
Here f (x) = (x - a)(x - b)
{tex}\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}(x - a)(x - b){/tex}
{tex} = (x - a)\frac{d}{{dx}}(x - b) + (x - b)\frac{d}{{dx}}(x - a){/tex}
= (x - a) × 1 + (x - b) × 1
= x - a + x - b = 2x - a - b
Posted by Palak Agrawal 6 years, 7 months ago
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Sia ? 6 years, 7 months ago
A-B=A∩Bc (A intersect B complement)
pick an element x
let x∈(A-B)
therefore x∈A but x∉B
x∉B means x∈Bc
x∈A and x∈Bc
x∈(A∩Bc)
x∈(A-B)
therefore A-B=A∩Bc
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