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Ask QuestionPosted by Harshita Bansal❤️ 4 years, 10 months ago
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Posted by Harshita Bansal❤️ 4 years, 10 months ago
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Posted by Ishika Dixit 4 years, 10 months ago
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Gaurav Seth 4 years, 10 months ago
T
he square root of a negative real number is called an imaginary quantity or imaginary number. e.g., √-3, √-7/2
The quantity √-1 is an imaginary number, denoted by ‘i’, called iota.
Integral Powers of Iota (i)
i=√-1, i2 = -1, i3 = -i, i4=1
So, i4n+1= i, i4n+2 = -1, i4n+3 = -i, i4n+4 = i4n = 1
In other words,
in = (-1)n/2, if n is an even integer
in = (-1)(n-1)/2.i, if is an odd integer
Posted by Nik . 4 years, 10 months ago
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Posted by Aleena Philip Shaji 4 years, 10 months ago
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Gaurav Seth 4 years, 10 months ago
Total no. of student=25
Let n(M)=student who studying mathematics
n(C)=student who studying chemistry
n(P)=student who studying Physics
∴n(M)=15,n(P)=12,n(C)=11,n(M∩P)=9,n(M∩C)=5,n(P∩C)=4,n(M∩P∩C)=3
Number of student who studying mathematics and physic but not chemistry
⇒n(M∩P)−n(M∩P∩C)
⇒9−3=6
Number of student who studying chemistry and physic but not mathematics
⇒n(P∩C)−n(M∩P∩C)
⇒4−3=1
Number of student who studying mathematics and chemistry but not physics
⇒n(M∩C)−n(M∩P∩C)
⇒5−3=2
∴No.of student who had taken two of the three subjects=6+1+2=9
Posted by Tilak Gupta 4 years, 10 months ago
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Posted by Tilak Gupta 4 years, 10 months ago
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Posted by Siddiqua Rahman 4 years, 10 months ago
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Posted by Gaur Saab?? 4 years, 10 months ago
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Posted by Gaur Saab?? 4 years, 10 months ago
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Posted by Soham Biswas 4 years, 10 months ago
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Posted by Dikshant Shokeen 4 years, 10 months ago
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Gaurav Seth 4 years, 10 months ago
Given : a circle of diameter 40cm,the length of a chord is 20 cm,
To Find : the length of the minor arc of the chord
Solution:
Chord Length = 20 cm
Diameter = 40 cm
Perpendicular from center on chord bisect the chord
Hence Half of chord = 10 cm
Radius = 40/2 = 20 cm
Sin ( 1/2 of chord angle ) = 10 /20
=> Sin ( 1/2 of chord angle ) = 1/2
=> Sin ( 1/2 of chord angle ) = Sin 30°
=> 1/2 of chord angle = 30°
=> chord angle = 60°
Minor arc angle = 60°
or another way to get angle
as Radius = 20 cm
Chord length = 20 cm
Hence it forms an Equilateral triangle
Hence angle formed by chord at center = 60°
Minor arc angle = 60°
length of the major arc of the chord = (60/360) * 2π * Radius
= ( 1/6 ) * 2π * 10
= 10π/3
= 50 * 3.14/3
= 10.47 cm
Yogita Ingle 4 years, 10 months ago
Diameter of the circle =40 cm
∴ Radius (r) of the circle = 40/2 =20 cm
Let AB be a chord (length = 20 cm) of the circle.
In △OAB, OA = OB = Radius of circle = 20 cm
Also, AB = 20 cm
Thus, △OAB is an equilateral triangle.
∴θ=60∘=π/3 radian
We know that in a circle of radius r unit, if an arc of length I unit subtends an angle θ radian at the centre, then θ=l/r
∴π/3 = arc AB/20
⟹arc AB = 20/ 3π cm
Posted by सत्य सनातन? 4 years, 10 months ago
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Posted by Priyanka Gurjar 4 years, 10 months ago
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Gaurav Seth 4 years, 10 months ago
In order to show that points (3, 0), (–2, –2), and (8, 2) are collinear, it suffices to show that the line passing through points (3, 0) and (–2, –2) also passes through point (8, 2).
The equation of the line passing through points (3, 0) and (–2, –2) is
It is observed that at x = 8 and y = 2,
L.H.S. = 2 × 8 – 5 × 2 = 16 – 10 = 6 = R.H.S.
Therefore, the line passing through points (3, 0) and (–2, –2) also passes through point (8, 2). Hence, points (3, 0), (–2, –2), and (8, 2) are collinear.
Posted by Yashansh😎😇😇 Jii 4 years, 10 months ago
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Posted by Shubham Bisht 4 years, 10 months ago
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Yogita Ingle 4 years, 10 months ago
We have this equation:
x + y = 1
x + y = 1
y = - x + 1
The slope of the tangent here is -1
And the slope of the line perpendicular to the tangent is 1
The line will pass through the focus y = x + c which will pass through (1,1)
The equation for this would be: y = x
The point of intersection of x + y = 1 and x = y can be obtained by:
y + y = 1
2y = 1
y = 1/2
The vertex for this is : (1/2, 1/2)
The distance between the focus and vertex is √[2x(1/2)²] = √2/2
The equation of parabola would be (y -1/2)² = (2x√2) x (x-1/2) y² + 1/4 - y = 2√2*x - √2
= 4y² - 4y - 8 x √2 *x + 4*√2 + 1 = 0
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