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The polar equation of the straight line parallel to the initial line and at a distance of 4 units above the initial line is



 r
The correct answer is:
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The polar equation of axy is
The polar equation of axy is
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If x, y, z are integers and x 0, y 1, z 2, x + y + z = 15, then the number of values of the ordered triplet (x, y, z) is 
Let y = p + 1 and z = q + 2.
Then x 0, p 0, q 0 and x + y + z = 15
x + p + q = 12
The reqd. number of values of (x, y, z) and hence of (x, p, q)
= No. of nonnegative integral solutions of x + p + q= 12
= Coeff. of x^{12} in (x^{0} + x^{1} + x^{2} + ……)^{3}
= Coeff. of x^{12} in (1 – x)^{–3}
= Coeff. of x^{12} in [^{2}C_{0} + ^{3}C_{1} x + ^{4}C_{2} x^{2} + ….]
= ^{14}C_{12} = = = 91.
Then x 0, p 0, q 0 and x + y + z = 15
x + p + q = 12
The reqd. number of values of (x, y, z) and hence of (x, p, q)
= No. of nonnegative integral solutions of x + p + q= 12
= Coeff. of x^{12} in (x^{0} + x^{1} + x^{2} + ……)^{3}
= Coeff. of x^{12} in (1 – x)^{–3}
= Coeff. of x^{12} in [^{2}C_{0} + ^{3}C_{1} x + ^{4}C_{2} x^{2} + ….]
= ^{14}C_{12} = = = 91.
If x, y, z are integers and x 0, y 1, z 2, x + y + z = 15, then the number of values of the ordered triplet (x, y, z) is 
mathsGeneral
Let y = p + 1 and z = q + 2.
Then x 0, p 0, q 0 and x + y + z = 15
x + p + q = 12
The reqd. number of values of (x, y, z) and hence of (x, p, q)
= No. of nonnegative integral solutions of x + p + q= 12
= Coeff. of x^{12} in (x^{0} + x^{1} + x^{2} + ……)^{3}
= Coeff. of x^{12} in (1 – x)^{–3}
= Coeff. of x^{12} in [^{2}C_{0} + ^{3}C_{1} x + ^{4}C_{2} x^{2} + ….]
= ^{14}C_{12} = = = 91.
Then x 0, p 0, q 0 and x + y + z = 15
x + p + q = 12
The reqd. number of values of (x, y, z) and hence of (x, p, q)
= No. of nonnegative integral solutions of x + p + q= 12
= Coeff. of x^{12} in (x^{0} + x^{1} + x^{2} + ……)^{3}
= Coeff. of x^{12} in (1 – x)^{–3}
= Coeff. of x^{12} in [^{2}C_{0} + ^{3}C_{1} x + ^{4}C_{2} x^{2} + ….]
= ^{14}C_{12} = = = 91.
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If then the equation whose roots are
If then the equation whose roots are
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Let p, q {1, 2, 3, 4}. Then number of equation of the form px^{2} + qx + 1 = 0, having real roots, is
Let p, q {1, 2, 3, 4}. Then number of equation of the form px^{2} + qx + 1 = 0, having real roots, is
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ax^{2} + bx + c = 0 has real and distinct roots null. Further a > 0, b < 0 and c < 0, then –
ax^{2} + bx + c = 0 has real and distinct roots null. Further a > 0, b < 0 and c < 0, then –
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The cartesian equation of is
A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as .
Now we know that:
.
So applying this, we get:
Now lets substitute x=$rcosθ andy=rsinθ, we get:$
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as .
Now we know that:
.
So applying this, we get:
Now lets substitute x=$rcosθ andy=rsinθ, we get:$
The cartesian equation of is
MathsGeneral
A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as .
Now we know that:
.
So applying this, we get:
Now lets substitute x=$rcosθ andy=rsinθ, we get:$
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as .
Now we know that:
.
So applying this, we get:
Now lets substitute x=$rcosθ andy=rsinθ, we get:$
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The castesian equation of is
A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as .
Now we know that:
.
So applying this, we get:
Now lets substitute x=$rcosθ andy=rsinθ, we get:$
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as .
Now we know that:
.
So applying this, we get:
Now lets substitute x=$rcosθ andy=rsinθ, we get:$
The castesian equation of is
MathsGeneral
A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as .
Now we know that:
.
So applying this, we get:
Now lets substitute x=$rcosθ andy=rsinθ, we get:$
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as .
Now we know that:
.
So applying this, we get:
Now lets substitute x=$rcosθ andy=rsinθ, we get:$
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The equation of the directrix of the conic whose length of the latusrectum is 5 and eccenticity is 1/2 is
The equation of the directrix of the conic whose length of the latusrectum is 5 and eccenticity is 1/2 is
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The equation of the directrix of the conic is
The equation of the directrix of the conic is
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The equation of the circle touching the initial line at pole and radius 2 is
The equation of the circle touching the initial line at pole and radius 2 is
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The equation of the circle passing through pole and centre at (4,0) is
The equation of the circle passing through pole and centre at (4,0) is
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The polar equation of the circle with pole as centre and radius 3 is
The polar equation of the circle with pole as centre and radius 3 is
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(Area of GPL) to (Area of ALD) is equal to
(Area of GPL) to (Area of ALD) is equal to
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A small source of sound moves on a circle as shown in the figure and an observer is standing on Let and be the frequencies heard when the source is at and respectively. Then
At point source is moving away from observer so apparent frequency (actual frequency) At point source is coming towards observer so apparent frequency and point source is moving perpendicular to observer so
Hence
Hence
A small source of sound moves on a circle as shown in the figure and an observer is standing on Let and be the frequencies heard when the source is at and respectively. Then
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At point source is moving away from observer so apparent frequency (actual frequency) At point source is coming towards observer so apparent frequency and point source is moving perpendicular to observer so
Hence
Hence
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In a triangle ABC, if a : b : c = 7 : 8 : 9, then cos A : cos B equals to
In a triangle ABC, if a : b : c = 7 : 8 : 9, then cos A : cos B equals to
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