The polar equation of the circle with pole as centre and radius 3 is

physics-

A small source of sound moves on a circle as shown in the figure and an observer is standing on $O.$ Let $n subscript 1 end subscript comma n subscript 2 end subscript$ and $n subscript 3 end subscript$ be the frequencies heard when the source is at $A comma blank B blank$ and $C$ respectively. Then

physics-General

In a triangle ABC, if a : b : c = 7 : 8 : 9, then cos A : cos B equals to

physics-

Which of the following curves represents correctly the oscillation given by $y equals y subscript 0 end subscript sin invisible function application open parentheses omega t minus ϕ close parentheses comma blank w h e r e blank 0 less than ϕ less than 90$

physics-General

A is a set containing n elements. A subset P₁ is chosen, and A is reconstructed by replacing the elements of P₁. The same process is repeated for subsets P₁, P₂, … , P_m, with m > 1. The Number of ways of choosing P₁, P₂, …, P_m so that P₁ $union$ P₂ $union$ … $union$ P_m= A is -

The number of points in the Cartesian plane with integral co-ordinates satisfying the inequalities |x| $less or equal than$ k, |y| $less or equal than$ k, |x – y| $less or equal than$ k ; is-

The angle between the lines $r left square bracket 2 C o s space theta plus 5 S i n space theta right square bracket equals 3$ and $r left square bracket 2 s i n space theta minus 5 C o s space theta right square bracket plus 4 equals 0$ is

Here we used the concept of cartesian lines and some trigonometric terms to solve. With the help of slope we identified the angle between them. Hence, these lines are perpendicular so the angle between them is $9 0 degrees.$

The angle between the lines $r left square bracket 2 C o s space theta plus 5 S i n space theta right square bracket equals 3$ and $r left square bracket 2 s i n space theta minus 5 C o s space theta right square bracket plus 4 equals 0$ is

The polar equation of the straight line passing through $open parentheses 6 comma fraction numerator pi over denominator 3 end fraction close parentheses$ and perpendicular to the initial line is

The polar equation of the straight line passing through $open parentheses 2 comma fraction numerator pi over denominator 6 end fraction close parentheses$ and parallel to the initial line is

The equation of the line passing through pole and $open parentheses 2 comma fraction numerator pi over denominator 3 end fraction close parentheses$ is

The polar equation of $y to the power of 2 end exponent equals 4 x$ is

Here we used the concept of polar coordinate system and also the trigonometric ratios to find the solution. So the equation is $r equals 4 space c o t space theta space cos e c space theta$ .

The polar equation of $y to the power of 2 end exponent equals 4 x$ is

Here we used the concept of polar coordinate system and also the trigonometric ratios to find the solution. So the equation is $r equals 4 space c o t space theta space cos e c space theta$ .

The cartesian equation of $r equals a s i n space 2 theta$ is

Here we used the concept of the polar coordinate system and also the trigonometric ratios to find the solution. So the equation is $4 a squared x squared y squared equals left parenthesis x squared plus y squared right parenthesis cubed$ .

The cartesian equation of $r equals a s i n space 2 theta$ is

physics-

Two tuning forks $P$ and $Q$ are vibrated together. The number of beats produced are represented by the straight line $O A$ in the following graph. After loading $Q$ with wax again these are vibrated together and the beats produced are represented by the line $O B.$ If the frequency of $P$ is $341 H z comma$ the frequency of $Q$ will be

physics-General

If a hyperbola passing through the origin has $3 x minus 4 y minus 1 equals 0$ and $4 x minus 3 y minus 6 equals 0$ as its asymptotes, then the equation of its tranvsverse and conjugate axes are

Here we used the concept of polar coordinate system and also the trigonometric ratios to find the solution. $T h e space e q u a t i o n space o f space t r a n s v e r s e space i s space x plus y minus 5 equals 0. space T h e space e q u a t i o n space o f space c o n j u g a t e space a x i s space i s space x minus y minus 1 equals 0.$

If a hyperbola passing through the origin has $3 x minus 4 y minus 1 equals 0$ and $4 x minus 3 y minus 6 equals 0$ as its asymptotes, then the equation of its tranvsverse and conjugate axes are

Five distinct letters are to be transmitted through a communication channel. A total number of 15 blanks is to be inserted between the two letters with at least three between every two. The number of ways in which this can be done is -