Maths-

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Easy

### Question

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

#### The correct answer is:

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### Related Questions to study

physics-

#### Which of the following curves represents correctly the oscillation given by

Given equation

At

This is case with curve marked

At

This is case with curve marked

#### Which of the following curves represents correctly the oscillation given by

physics-General

Given equation

At

This is case with curve marked

At

This is case with curve marked

maths-

#### A is a set containing n elements. A subset P_{1} is chosen, and A is reconstructed by replacing the elements of P_{1}. The same process is repeated for subsets P_{1}, P_{2}, … , P_{m}, with m > 1. The Number of ways of choosing P_{1}, P_{2}, …, P_{m} so that P_{1} P_{2} … P_{m}= A is -

Let A = {a

For each a

Also there is exactly one choice, viz., a

Therefore, a

P

_{1}, a_{2},…..a_{n}}.For each a

_{i}(1 i n), either a_{i}P_{j}or a_{i}P_{j}(1 j m) . Thus, there are 2^{m}choices in which a_{i}(1 j n) may belong to the P_{j}s.Also there is exactly one choice, viz., a

_{i}P_{j}for j = 1, 2, …, m, for which a_{i}P_{1}P_{2}... P_{m}.Therefore, a

_{i}P_{1}P_{2}…. P_{m}in (2^{m}– 1) ways . Since there are n elements in the set A, the number of ways of constructing subsetsP

_{1}, P_{2}, ….. , P_{m}is (2^{m}– 1)^{n}#### A is a set containing n elements. A subset P_{1} is chosen, and A is reconstructed by replacing the elements of P_{1}. The same process is repeated for subsets P_{1}, P_{2}, … , P_{m}, with m > 1. The Number of ways of choosing P_{1}, P_{2}, …, P_{m} so that P_{1} P_{2} … P_{m}= A is -

maths-General

Let A = {a

For each a

Also there is exactly one choice, viz., a

Therefore, a

P

_{1}, a_{2},…..a_{n}}.For each a

_{i}(1 i n), either a_{i}P_{j}or a_{i}P_{j}(1 j m) . Thus, there are 2^{m}choices in which a_{i}(1 j n) may belong to the P_{j}s.Also there is exactly one choice, viz., a

_{i}P_{j}for j = 1, 2, …, m, for which a_{i}P_{1}P_{2}... P_{m}.Therefore, a

_{i}P_{1}P_{2}…. P_{m}in (2^{m}– 1) ways . Since there are n elements in the set A, the number of ways of constructing subsetsP

_{1}, P_{2}, ….. , P_{m}is (2^{m}– 1)^{n}maths-

#### The number of points in the Cartesian plane with integral co-ordinates satisfying the inequalities |x| k, |y| k, |x – y| k ; is-

|x| k –k x k ….(1)

& |y| k –k y k ….(2)

& |x – y| k |y – x| k –k y – x k x – k y x + k ….(3)

Number of points having integral coordinates

= (2k + 1)

= (3k

& |y| k –k y k ….(2)

& |x – y| k |y – x| k –k y – x k x – k y x + k ….(3)

Number of points having integral coordinates

= (2k + 1)

^{2}– 2= (3k

^{2}+ 3k + 1).#### The number of points in the Cartesian plane with integral co-ordinates satisfying the inequalities |x| k, |y| k, |x – y| k ; is-

maths-General

|x| k –k x k ….(1)

& |y| k –k y k ….(2)

& |x – y| k |y – x| k –k y – x k x – k y x + k ….(3)

Number of points having integral coordinates

= (2k + 1)

= (3k

& |y| k –k y k ….(2)

& |x – y| k |y – x| k –k y – x k x – k y x + k ….(3)

Number of points having integral coordinates

= (2k + 1)

^{2}– 2= (3k

^{2}+ 3k + 1).Maths-

#### The angle between the lines and is

The intersection of two perpendicular lines results in the formation of the cartesian plane, a two-dimensional coordinate plane. The X-axis is the horizontal line, and the Y-axis is the vertical line. The Cartesian coordinate point (x, y) indicates that the distance from the origin is x in the horizontal direction and y in the vertical direction.

Now the given lines are:

and

The cartesian form will be:

Now the given lines are:

and

The cartesian form will be:

$2x+5y=3$

$2y−5x=−4$

Slopes of these lines are $−5/2 $ and $2/5 $

Here, we can say that the product of slopes is $−1$

Hence, these lines are perpendicular so the angles between them is $90 degrees.$

#### The angle between the lines and is

Maths-General

The intersection of two perpendicular lines results in the formation of the cartesian plane, a two-dimensional coordinate plane. The X-axis is the horizontal line, and the Y-axis is the vertical line. The Cartesian coordinate point (x, y) indicates that the distance from the origin is x in the horizontal direction and y in the vertical direction.

Now the given lines are:

and

The cartesian form will be:

Now the given lines are:

and

The cartesian form will be:

$2x+5y=3$

$2y−5x=−4$

Slopes of these lines are $−5/2 $ and $2/5 $

Here, we can say that the product of slopes is $−1$

Hence, these lines are perpendicular so the angles between them is $90 degrees.$

maths-

#### The polar equation of the straight line passing through and perpendicular to the initial line is

#### The polar equation of the straight line passing through and perpendicular to the initial line is

maths-General

maths-

#### The polar equation of the straight line passing through and parallel to the initial line is

#### The polar equation of the straight line passing through and parallel to the initial line is

maths-General

maths-

#### The equation of the line passing through pole and is

#### The equation of the line passing through pole and is

maths-General

Maths-

#### The polar equation of is

Using the formula, we may generate an endless number of polar coordinates for a single coordinate point.

(r, +2n) or (-r, +(2n+1)) are possible formulas, where n is an integer.

If measured in the opposite direction, the value of is positive.

When calculated counterclockwise, the value of is negative.

If laid off at the terminal side of, the value of r is positive.

If r is terminated at the prolongation through the origin from the terminal side of, it has a negative value.

The equation is given as:

Now lets xonsider: x=rcosθ and y=rsinθ

x

Now putting the values, we get:

(r, +2n) or (-r, +(2n+1)) are possible formulas, where n is an integer.

If measured in the opposite direction, the value of is positive.

When calculated counterclockwise, the value of is negative.

If laid off at the terminal side of, the value of r is positive.

If r is terminated at the prolongation through the origin from the terminal side of, it has a negative value.

The equation is given as:

Now lets xonsider: x=rcosθ and y=rsinθ

x

^{2}+y^{2}=r^{2}Now putting the values, we get:

#### The polar equation of is

Maths-General

Using the formula, we may generate an endless number of polar coordinates for a single coordinate point.

(r, +2n) or (-r, +(2n+1)) are possible formulas, where n is an integer.

If measured in the opposite direction, the value of is positive.

When calculated counterclockwise, the value of is negative.

If laid off at the terminal side of, the value of r is positive.

If r is terminated at the prolongation through the origin from the terminal side of, it has a negative value.

The equation is given as:

Now lets xonsider: x=rcosθ and y=rsinθ

x

Now putting the values, we get:

(r, +2n) or (-r, +(2n+1)) are possible formulas, where n is an integer.

If measured in the opposite direction, the value of is positive.

When calculated counterclockwise, the value of is negative.

If laid off at the terminal side of, the value of r is positive.

If r is terminated at the prolongation through the origin from the terminal side of, it has a negative value.

The equation is given as:

Now lets xonsider: x=rcosθ and y=rsinθ

x

^{2}+y^{2}=r^{2}Now putting the values, we get:

Maths-

#### The cartesian equation of is

Using the formula, we may generate an endless number of polar coordinates for a single coordinate point.

(r, +2n) or (-r, +(2n+1)) are possible formulas, where n is an integer.

If measured in the opposite direction, the value of is positive.

When calculated counterclockwise, the value of is negative.

If laid off at the terminal side of, the value of r is positive.

If r is terminated at the prolongation through the origin from the terminal side of, it has a negative value.

The equation is given as

#### The cartesian equation of is

Maths-General

Using the formula, we may generate an endless number of polar coordinates for a single coordinate point.

(r, +2n) or (-r, +(2n+1)) are possible formulas, where n is an integer.

If measured in the opposite direction, the value of is positive.

When calculated counterclockwise, the value of is negative.

If laid off at the terminal side of, the value of r is positive.

If r is terminated at the prolongation through the origin from the terminal side of, it has a negative value.

The equation is given as

physics-

#### Two tuning forks and are vibrated together. The number of beats produced are represented by the straight line in the following graph. After loading with wax again these are vibrated together and the beats produced are represented by the line If the frequency of is the frequency of will be

or

On waxing the number of beats decreases hence

On waxing the number of beats decreases hence

#### Two tuning forks and are vibrated together. The number of beats produced are represented by the straight line in the following graph. After loading with wax again these are vibrated together and the beats produced are represented by the line If the frequency of is the frequency of will be

physics-General

or

On waxing the number of beats decreases hence

On waxing the number of beats decreases hence

Maths-

#### If a hyperbola passing through the origin has and as its asymptotes, then the equation of its tranvsverse and conjugate axes are

a line or curve that serves as the boundary of another line or curve in mathematics. An example of an asymptotic curve is a descending curve that approaches but does not reach the horizontal axis, which is the asymptote of the curve.

The transverse axis is the bisector containing origin, and the hyperbola's axes are the bisectors of the pair of asmptodes.

So we have:

The transverse axis is the bisector containing origin, and the hyperbola's axes are the bisectors of the pair of asmptodes.

So we have:

#### If a hyperbola passing through the origin has and as its asymptotes, then the equation of its tranvsverse and conjugate axes are

Maths-General

a line or curve that serves as the boundary of another line or curve in mathematics. An example of an asymptotic curve is a descending curve that approaches but does not reach the horizontal axis, which is the asymptote of the curve.

The transverse axis is the bisector containing origin, and the hyperbola's axes are the bisectors of the pair of asmptodes.

So we have:

The transverse axis is the bisector containing origin, and the hyperbola's axes are the bisectors of the pair of asmptodes.

So we have:

maths-

#### 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 -

For 1 i 4, let x

x

The number of solutions of (1)

= coefficient of t

= coefficient of t

= coefficient of t

=

But 5 letters can be permuted in 5! = 120 ways.

Thus, the required number arrangements

= (120) (20) = 2400.

Hence

_{i}( 3) be the number of blanks between i^{th}and (i + 1)^{th}letters. Then,x

_{1}+ x_{2}+ x_{3}+ x_{4}= 15 …..(1)The number of solutions of (1)

= coefficient of t

^{15}in (t^{3}+ t^{4}+….)^{4}= coefficient of t

^{3}in (1 – t)^{–4}= coefficient of t

^{3}in [1 +^{4}C_{1}+^{5}C_{2}t^{2}+^{6}C_{3 }t^{3}+ …..]=

^{6}C_{3}= 20.But 5 letters can be permuted in 5! = 120 ways.

Thus, the required number arrangements

= (120) (20) = 2400.

Hence

#### 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 -

maths-General

For 1 i 4, let x

x

The number of solutions of (1)

= coefficient of t

= coefficient of t

= coefficient of t

=

But 5 letters can be permuted in 5! = 120 ways.

Thus, the required number arrangements

= (120) (20) = 2400.

Hence

_{i}( 3) be the number of blanks between i^{th}and (i + 1)^{th}letters. Then,x

_{1}+ x_{2}+ x_{3}+ x_{4}= 15 …..(1)The number of solutions of (1)

= coefficient of t

^{15}in (t^{3}+ t^{4}+….)^{4}= coefficient of t

^{3}in (1 – t)^{–4}= coefficient of t

^{3}in [1 +^{4}C_{1}+^{5}C_{2}t^{2}+^{6}C_{3 }t^{3}+ …..]=

^{6}C_{3}= 20.But 5 letters can be permuted in 5! = 120 ways.

Thus, the required number arrangements

= (120) (20) = 2400.

Hence

maths-

#### The number of ordered pairs (m, n), m, n {1, 2, … 100} such that 7^{m} + 7^{n} is divisible by 5 is -

Note that 7

Thus, 7

For 7

m n

1 4r 4s + 2

2 4r + 1 4s + 3

3 4r + 2 4s

4 4r + 3 4s + 1

Thus, for a given value of m there are just 25 values of n for which 7

There are 100 × 25 = 2500 ordered pairs (m, n) for which 7

Hence

^{r}(r N) ends in 7, 9, 3 or 1 (corresponding to r = 1, 2, 3 and 4 respectively).Thus, 7

^{m}+ 7^{n}cannot end in 5 for any values of m, n N. In other words, for 7^{m}+ 7^{n}to be divisible by 5, it should end in 0.For 7

^{m}+ 7^{n}to end in 0, the forms of m and n should be as follows :m n

1 4r 4s + 2

2 4r + 1 4s + 3

3 4r + 2 4s

4 4r + 3 4s + 1

Thus, for a given value of m there are just 25 values of n for which 7

^{m}+ 7^{n}ends in 0. [For instance, if m = 4r, then = 2, 6, 10,….. , 98] There are 100 × 25 = 2500 ordered pairs (m, n) for which 7

^{m}+ 7^{n}is divisible by 5.Hence

#### The number of ordered pairs (m, n), m, n {1, 2, … 100} such that 7^{m} + 7^{n} is divisible by 5 is -

maths-General

Note that 7

Thus, 7

For 7

m n

1 4r 4s + 2

2 4r + 1 4s + 3

3 4r + 2 4s

4 4r + 3 4s + 1

Thus, for a given value of m there are just 25 values of n for which 7

There are 100 × 25 = 2500 ordered pairs (m, n) for which 7

Hence

^{r}(r N) ends in 7, 9, 3 or 1 (corresponding to r = 1, 2, 3 and 4 respectively).Thus, 7

^{m}+ 7^{n}cannot end in 5 for any values of m, n N. In other words, for 7^{m}+ 7^{n}to be divisible by 5, it should end in 0.For 7

^{m}+ 7^{n}to end in 0, the forms of m and n should be as follows :m n

1 4r 4s + 2

2 4r + 1 4s + 3

3 4r + 2 4s

4 4r + 3 4s + 1

Thus, for a given value of m there are just 25 values of n for which 7

^{m}+ 7^{n}ends in 0. [For instance, if m = 4r, then = 2, 6, 10,….. , 98] There are 100 × 25 = 2500 ordered pairs (m, n) for which 7

^{m}+ 7^{n}is divisible by 5.Hence

maths-

#### Consider the following statements:

1. The number of ways of arranging m different things taken all at a time in which p m particular things are never together is m! – (m – p + 1)! p!.

2. A pack of 52 cards can be divided equally among four players in order in ways.

Which of these is/are correct?

(1) Total number of ways of arranging m things = m!.

To find the number of ways in which p particular things are together, we consider p particular things as a group.

Number of ways in which p particular things are together = (m – p + 1)! p!

So, number of ways in which p particular things are not together

= m! – (m – p + 1)! p!

Total number of ways =

Hence, both of statements are correct.

To find the number of ways in which p particular things are together, we consider p particular things as a group.

Number of ways in which p particular things are together = (m – p + 1)! p!

So, number of ways in which p particular things are not together

= m! – (m – p + 1)! p!

Total number of ways =

Hence, both of statements are correct.

#### Consider the following statements:

1. The number of ways of arranging m different things taken all at a time in which p m particular things are never together is m! – (m – p + 1)! p!.

2. A pack of 52 cards can be divided equally among four players in order in ways.

Which of these is/are correct?

maths-General

(1) Total number of ways of arranging m things = m!.

To find the number of ways in which p particular things are together, we consider p particular things as a group.

Number of ways in which p particular things are together = (m – p + 1)! p!

So, number of ways in which p particular things are not together

= m! – (m – p + 1)! p!

Total number of ways =

Hence, both of statements are correct.

To find the number of ways in which p particular things are together, we consider p particular things as a group.

Number of ways in which p particular things are together = (m – p + 1)! p!

So, number of ways in which p particular things are not together

= m! – (m – p + 1)! p!

Total number of ways =

Hence, both of statements are correct.

maths-

#### The total number of function ‘ƒ’ from the set {1, 2, 3} into the set {1, 2, 3, 4, 5} such that ƒ(i) ƒ(j), i < j, is equal to-

Let ‘l’ is associated with ‘r’ ,

r {1, 2, 3, 4, 5} then ‘2’ can be associated with r, r + 1, ….., 5.

Let ‘2’ is associated with ‘j’ then 3 can be associated with j, j + 1, …., 5. Thus required number of functions

= =

=

= = 35

Hence (a) is correct answer.

r {1, 2, 3, 4, 5} then ‘2’ can be associated with r, r + 1, ….., 5.

Let ‘2’ is associated with ‘j’ then 3 can be associated with j, j + 1, …., 5. Thus required number of functions

= =

=

= = 35

Hence (a) is correct answer.

#### The total number of function ‘ƒ’ from the set {1, 2, 3} into the set {1, 2, 3, 4, 5} such that ƒ(i) ƒ(j), i < j, is equal to-

maths-General

Let ‘l’ is associated with ‘r’ ,

r {1, 2, 3, 4, 5} then ‘2’ can be associated with r, r + 1, ….., 5.

Let ‘2’ is associated with ‘j’ then 3 can be associated with j, j + 1, …., 5. Thus required number of functions

= =

=

= = 35

Hence (a) is correct answer.

r {1, 2, 3, 4, 5} then ‘2’ can be associated with r, r + 1, ….., 5.

Let ‘2’ is associated with ‘j’ then 3 can be associated with j, j + 1, …., 5. Thus required number of functions

= =

=

= = 35

Hence (a) is correct answer.