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# Ask a Mathematics(Class XI-XII Tuition (PUC)) Question

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S

Shashikant 10/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

What is the value of (sin 45)(sin 60)?

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S

Sample 10/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

What is the value of cos 45 - sin 60?

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U

Uma 10/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

What do you mean by trigonometry?

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N

Neeraj 10/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

What is the value of cos 50 + sec 60?

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K

Karunanidhi 10/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

What is the value of sin 56 + tan 90 + sin 45 + cos 45?

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P

Pradeep 10/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

What is the antiderivative of sin x with respect to x?

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M

Momin 10/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

What is the integral value of 2 tan x + 2 cot x?

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G

Gowrikanth 10/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

Give an example 2nd order integration?

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S

Srinitha 10/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

Solve integral function of: f(x) = 2 cos x + 3 sin x + 2 cot x?

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B

Benjamin 10/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

Solve by 2nd order derivative of the function: f(x) = 10 - 3x?

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A

Ajith 10/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

Find the integral of the function: f(x) = 2 cos x + sec 4x?

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D

Dhruv 10/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

Simplify: f(x) = 2 - t + (square of t) - (cube of t) + (cube of t) with respect to t?

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M

Mrs 10/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

Solve: integration of (1 + tan 2x)/(1 - 2 tan x) with respect to x?

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J

Jyoti 10/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

What is the antiderivative of (1 + cot 2x) with respect to x?

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M

Mohan 08/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

Solve: 5x + y < 100
x + y < 60
x > 0
y > 0?

R

Rahul replied | 20/11/2016

Solving them simultaneously,
4x <40
x<10
y<50

Hussam Uddin replied | 22/11/2016

It is nothing but area of quad with points (0,0), (0,50), (10,0), (10,50).

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A

Aditi 09/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

What do you mean by adjoint of matrix?

Sagar Khandelwal replied | 19/11/2016

In linear algebra, the adjugate, classical adjoint, or adjunct of a square matrix is the transpose of its co factor matrix. The adjugate has sometimes been called the "adjoint", but today the "adjoint" of a matrix normally refers to its corresponding adjoint operator, which is its conjugate transpose.

S

Shiv replied | 19/11/2016

It is the transpose of co factor matrix.

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S

Swati 08/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

Find the equation of the circle having (1, -2) as its centre and passing through 3x + y = 14, 2x + 5y = 18?

Hussam Uddin replied | 22/11/2016

Equation of circle is (x-1)2 + (y +2)2 = r2 , we need to find r , point of intersection of 2 lines is 13x = 52, x= 4 , y = 2, therefore r2 = 25.

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S

Sriram 08/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

Find the distance between the directrices of the ellipse: square of (x) / 36 + square of (y) / 20 = 1?

Sarvajeet replied | 03/12/2016

18.

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S

Selvi 08/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

Find the equation of the hyperbola with eccentricity 3 / 2 and foci at (2, 0)?

Sarvajeet replied | 03/12/2016

square of (x) / (16/9) - square of (y) / (20/9) = 1.

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N

Nasreen 08/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

Define a circle as the locus of a point?

Hussam Uddin replied | 22/11/2016

Circle is locus of a point , whose distance from a fixed point i.e centre is constant i.e radius

Sarvajeet replied | 03/12/2016

Locus of a point whose distance from a fixed point is remain constant.

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B

Bivash 08/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

Find the centre and the radius of the circle from the given equation: (square of x) + (square of y) = 16?

Tapas replied | 28/11/2016

Circle: x^2 + y^2 = 16 = 4^2.
Center of the circle: (0, 0). Radius = 4 units.

Sarvajeet replied | 03/12/2016

The center = (0, 0), Radius = 4.

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N

Neetu 08/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

Find the equation of the circle whose centre is (3, 6) and which passes through the point (-1, 1)?

Sarvajeet replied | 03/12/2016

Square of (x-3) + Square of (x-6) = 41.

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N

Neha 08/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

Find the whether (1, 3) lies inside, or outside or on the circle (square of x) + (square of y) = 16?

Gunturi replied | 03/12/2016

Substitute those values in the L.H.S of the equation. if
LHS < RHS --- Lies Inside
LHS = RHS ---- Lies on the circle
LHS > RHS ---- Lies outside of the circle.

Sarvajeet replied | 03/12/2016

Inside.

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R

Rakhi 08/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

Show that the point (5, 6) outside the circle: (square of x) + (square of y) = 4?

Sarvajeet replied | 03/12/2016

S'=(square of 5) + (square of 6) - 4 = -57.

Tapas replied | 04/12/2016

The equation of the circle is:
x^2 + y^2 = 4 = 2^2.
Therefore the radius of the circle is 2 units with its center at (0, 0). So, any point at a distance bigger than 2 units must be outside the circle.
Distance between the center (0, 0) and point (5, 6) is:
Sq. root of [(5-0)^2 + (6-0)^2] = sq root of (25 + 36) = sq root of 61 = 7.8. This is greater than the radius...  more»
The equation of the circle is:
x^2 + y^2 = 4 = 2^2.
Therefore the radius of the circle is 2 units with its center at (0, 0). So, any point at a distance bigger than 2 units must be outside the circle.
Distance between the center (0, 0) and point (5, 6) is:
Sq. root of [(5-0)^2 + (6-0)^2] = sq root of (25 + 36) = sq root of 61 = 7.8. This is greater than the radius of the circle. So the point (5, 6) is outside the circle. «less

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R

Raj 08/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

Find the equation of the circle which passes through the points (1, 2), (2, 2) and (4, -1)?

Sarvajeet replied | 03/12/2016

square of (x-3/2) + square of (y-3/2) = 1/2.

Tapas replied | 04/12/2016

Take the general equation for a circle:
x^2 + y^2 + 2gx + 2fy + c = 0, where the center is at (-g, -f) and the radius is sq root of (g^2 +f^2 -c).
Now use the given points to get three equations to solve for three unknowns, g, f, and c. The equations with point (1, 2) becomes:
1^2 + 2^2 + 2xgx1 + 2xfx2 + c = 0
=> 2g + 4f + c = - 5 .............. (1)
Similarly,...  more»
Take the general equation for a circle:
x^2 + y^2 + 2gx + 2fy + c = 0, where the center is at (-g, -f) and the radius is sq root of (g^2 +f^2 -c).
Now use the given points to get three equations to solve for three unknowns, g, f, and c. The equations with point (1, 2) becomes:
1^2 + 2^2 + 2xgx1 + 2xfx2 + c = 0
=> 2g + 4f + c = - 5 .............. (1)
Similarly, the equation with the point (2, 2) is:
=> 4g + 4f + c = - 8 .............. (2)
And, the equation with the point (4, -1) is:
=> 8g - 2f + c = - 17 .............. (3)

From (1) - (2) we get: -2g = 3 --> g = -3/2.
Using this value of g in (1) we get:
4f + c = - 2 ............ (4)
Using this value of g in (3) we get:
-2f + c = - 5 ............ (5)
From (4) - (5) we can get:
6f = 3 --> f = 1/2.
Using this value of f in (5) we get: c = - 4.
So, the equation of the circle is:
x^2 + y^2 + 2(-3/2)x + 2(1/2)y + (- 4) = 0.
After simplification:
--> x^2 + y^2 -3x +y - 4 = 0 (Answer) «less

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