Maths-

General

Easy

### Question

#### The equation represents

- a parabola
- an ellipse
- a circle
- a straight line

#### The correct answer is: a circle

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### Related Questions to study

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#### The polar equation of the circle whose end points of the diameter are and is

#### The polar equation of the circle whose end points of the diameter are and is

maths-General

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#### The radius of the circle is

#### The radius of the circle is

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#### The adjoining figure shows the graph of Then –

#### The adjoining figure shows the graph of Then –

Maths-General

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#### Graph of y = ax^{2} + bx + c = 0 is given adjacently. What conclusions can be drawn from this graph –

As we can see from the graph we have a parabola curve and since it is opening in an upward direction. So we can say that a > 0 and

Hence, the option (a) is correct.

Here, we can see that the vertex of the parabola is located in the fourth quadrant , therefore it will be =

On further solving this, we get

Therefore, the option (b) is also correct.

Since, at x=0 , the y intercept will be positive and from this, we can conclude that c < 0 and

Hence, the option (c) will also be correct

On checking all the options, and we can see all options are correct and

Therefore, we conclude that all the options available are correct.

Hence, the option (a) is correct.

Here, we can see that the vertex of the parabola is located in the fourth quadrant , therefore it will be =

On further solving this, we get

Therefore, the option (b) is also correct.

Since, at x=0 , the y intercept will be positive and from this, we can conclude that c < 0 and

Hence, the option (c) will also be correct

On checking all the options, and we can see all options are correct and

Therefore, we conclude that all the options available are correct.

#### Graph of y = ax^{2} + bx + c = 0 is given adjacently. What conclusions can be drawn from this graph –

Maths-General

As we can see from the graph we have a parabola curve and since it is opening in an upward direction. So we can say that a > 0 and

Hence, the option (a) is correct.

Here, we can see that the vertex of the parabola is located in the fourth quadrant , therefore it will be =

On further solving this, we get

Therefore, the option (b) is also correct.

Since, at x=0 , the y intercept will be positive and from this, we can conclude that c < 0 and

Hence, the option (c) will also be correct

On checking all the options, and we can see all options are correct and

Therefore, we conclude that all the options available are correct.

Hence, the option (a) is correct.

Here, we can see that the vertex of the parabola is located in the fourth quadrant , therefore it will be =

On further solving this, we get

Therefore, the option (b) is also correct.

Since, at x=0 , the y intercept will be positive and from this, we can conclude that c < 0 and

Hence, the option (c) will also be correct

On checking all the options, and we can see all options are correct and

Therefore, we conclude that all the options available are correct.

maths-

#### For the quadratic polynomial f (x) = 4x^{2} – 8kx + k, the statements which hold good are

#### For the quadratic polynomial f (x) = 4x^{2} – 8kx + k, the statements which hold good are

maths-General

Maths-

#### The graph of the quadratic polynomial y = ax^{2} + bx + c is as shown in the figure. Then :

Clearly, y = represent a parabola opening downwards. Therefore, a < 0

y = cuts negative y- axis , Putting x = 0 in the given equation

-y = c

y = -c

c < 0

Thus, from the above graph c < 0.

#### The graph of the quadratic polynomial y = ax^{2} + bx + c is as shown in the figure. Then :

Maths-General

Clearly, y = represent a parabola opening downwards. Therefore, a < 0

y = cuts negative y- axis , Putting x = 0 in the given equation

-y = c

y = -c

c < 0

Thus, from the above graph c < 0.

Maths-

#### The greatest possible number of points of intersections of 8 straight line and 4 circles is :

Complete step-by-step answer:

The number point of intersection between two circles can be counted by finding the number of ways in which one circle and one line can be selected out of the lot multiplied by 2 as one circle and one line can intersect at most two points.

For selecting r objects from n objects can be done by using the formula as follows

As mentioned in the question, we have to find the total number of intersection points.

For calculating the points of intersection between two lines, we can use the formula which is mentioned in the hint as follows = = 28

For calculating the points of intersection between two circles, we can use the formula which is mentioned in the hint as follows = = 12

For calculating the points of intersection between one line and one circle, we can use the formula which is mentioned in the hint as follows = 64

Hence, the total number of points of intersection is = 28 + 12 + 64 = 104

The number point of intersection between two circles can be counted by finding the number of ways in which one circle and one line can be selected out of the lot multiplied by 2 as one circle and one line can intersect at most two points.

For selecting r objects from n objects can be done by using the formula as follows

As mentioned in the question, we have to find the total number of intersection points.

For calculating the points of intersection between two lines, we can use the formula which is mentioned in the hint as follows = = 28

For calculating the points of intersection between two circles, we can use the formula which is mentioned in the hint as follows = = 12

For calculating the points of intersection between one line and one circle, we can use the formula which is mentioned in the hint as follows = 64

Hence, the total number of points of intersection is = 28 + 12 + 64 = 104

#### The greatest possible number of points of intersections of 8 straight line and 4 circles is :

Maths-General

Complete step-by-step answer:

The number point of intersection between two circles can be counted by finding the number of ways in which one circle and one line can be selected out of the lot multiplied by 2 as one circle and one line can intersect at most two points.

For selecting r objects from n objects can be done by using the formula as follows

As mentioned in the question, we have to find the total number of intersection points.

For calculating the points of intersection between two lines, we can use the formula which is mentioned in the hint as follows = = 28

For calculating the points of intersection between two circles, we can use the formula which is mentioned in the hint as follows = = 12

For calculating the points of intersection between one line and one circle, we can use the formula which is mentioned in the hint as follows = 64

Hence, the total number of points of intersection is = 28 + 12 + 64 = 104

The number point of intersection between two circles can be counted by finding the number of ways in which one circle and one line can be selected out of the lot multiplied by 2 as one circle and one line can intersect at most two points.

For selecting r objects from n objects can be done by using the formula as follows

As mentioned in the question, we have to find the total number of intersection points.

For calculating the points of intersection between two lines, we can use the formula which is mentioned in the hint as follows = = 28

For calculating the points of intersection between two circles, we can use the formula which is mentioned in the hint as follows = = 12

For calculating the points of intersection between one line and one circle, we can use the formula which is mentioned in the hint as follows = 64

Hence, the total number of points of intersection is = 28 + 12 + 64 = 104

Maths-

#### How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even position ?

Complete step-by-step answer:

Here we need to find the total number of nine digit numbers that can be formed using the given digits i.e. 2, 2, 3, 3, 5, 5, 8, 8, 8.

Number of ways to arrange the odd digits in 4 even places =

On finding the value of the factorials, we get

Number of ways to arrange the odd digits in 4 even places = 6

Now, we have to arrange the even digits in odd places.

Number of ways to arrange the even digits in 5 odd places =

On finding the value of the factorials, we get

Number of ways to arrange the even digits in 5 odd places =

On further simplification, we get

Number of ways to arrange the even digits in 5 odd places =10

Total number of 9 digits number = 6×10 = 60

Hence, the required number of 9 digit numbers = 60

Here we need to find the total number of nine digit numbers that can be formed using the given digits i.e. 2, 2, 3, 3, 5, 5, 8, 8, 8.

X−X−X−X−X

Here, symbol ( − ) is for the even places and (X) is for the odd places of the digit number.

The digits which are even are 2, 2, 8, 8 and 8.

Number of even digits = 5

The digits which are odd are 3, 3, 5 and 5.

Number of odd digits = 4

Number of odd digits = 4

We have to arrange the odd digits in even places.

On finding the value of the factorials, we get

Number of ways to arrange the odd digits in 4 even places = 6

Now, we have to arrange the even digits in odd places.

Number of ways to arrange the even digits in 5 odd places =

On finding the value of the factorials, we get

Number of ways to arrange the even digits in 5 odd places =

On further simplification, we get

Number of ways to arrange the even digits in 5 odd places =10

Total number of 9 digits number = 6×10 = 60

Hence, the required number of 9 digit numbers = 60

#### How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even position ?

Maths-General

Complete step-by-step answer:

Here we need to find the total number of nine digit numbers that can be formed using the given digits i.e. 2, 2, 3, 3, 5, 5, 8, 8, 8.

Number of ways to arrange the odd digits in 4 even places =

On finding the value of the factorials, we get

Number of ways to arrange the odd digits in 4 even places = 6

Now, we have to arrange the even digits in odd places.

Number of ways to arrange the even digits in 5 odd places =

On finding the value of the factorials, we get

Number of ways to arrange the even digits in 5 odd places =

On further simplification, we get

Number of ways to arrange the even digits in 5 odd places =10

Total number of 9 digits number = 6×10 = 60

Hence, the required number of 9 digit numbers = 60

Here we need to find the total number of nine digit numbers that can be formed using the given digits i.e. 2, 2, 3, 3, 5, 5, 8, 8, 8.

X−X−X−X−X

Here, symbol ( − ) is for the even places and (X) is for the odd places of the digit number.

The digits which are even are 2, 2, 8, 8 and 8.

Number of even digits = 5

The digits which are odd are 3, 3, 5 and 5.

Number of odd digits = 4

Number of odd digits = 4

We have to arrange the odd digits in even places.

On finding the value of the factorials, we get

Number of ways to arrange the odd digits in 4 even places = 6

Now, we have to arrange the even digits in odd places.

Number of ways to arrange the even digits in 5 odd places =

On finding the value of the factorials, we get

Number of ways to arrange the even digits in 5 odd places =

On further simplification, we get

Number of ways to arrange the even digits in 5 odd places =10

Total number of 9 digits number = 6×10 = 60

Hence, the required number of 9 digit numbers = 60

Maths-

#### A person predicts the outcome of 20 cricket matches of his home team. Each match can result either in a win, loss or tie for the home team. Total number of ways in which he can make the predictions so that exactly 10 predictions are correct, is equal to :

Matches whose prediction are correct can be selected in ways.

Since each match can result either in a win, loss or tie for the home team.

Now, each wrong prediction can be made in two ways (i.e. the correct result is win and the person predicts either lose or tie)

and there are 10 matches for which he predicted wrong.

Total number of ways in which he can make the predictions so that exactly 10 predictions are correct =

Thus, total number of ways in which he can make the predictions so that exactly 10 predictions are correct, is equal to

Since each match can result either in a win, loss or tie for the home team.

Now, each wrong prediction can be made in two ways (i.e. the correct result is win and the person predicts either lose or tie)

and there are 10 matches for which he predicted wrong.

Total number of ways in which he can make the predictions so that exactly 10 predictions are correct =

Thus, total number of ways in which he can make the predictions so that exactly 10 predictions are correct, is equal to

#### A person predicts the outcome of 20 cricket matches of his home team. Each match can result either in a win, loss or tie for the home team. Total number of ways in which he can make the predictions so that exactly 10 predictions are correct, is equal to :

Maths-General

Matches whose prediction are correct can be selected in ways.

Since each match can result either in a win, loss or tie for the home team.

Now, each wrong prediction can be made in two ways (i.e. the correct result is win and the person predicts either lose or tie)

and there are 10 matches for which he predicted wrong.

Total number of ways in which he can make the predictions so that exactly 10 predictions are correct =

Thus, total number of ways in which he can make the predictions so that exactly 10 predictions are correct, is equal to

Since each match can result either in a win, loss or tie for the home team.

Now, each wrong prediction can be made in two ways (i.e. the correct result is win and the person predicts either lose or tie)

and there are 10 matches for which he predicted wrong.

Total number of ways in which he can make the predictions so that exactly 10 predictions are correct =

Thus, total number of ways in which he can make the predictions so that exactly 10 predictions are correct, is equal to

maths-

#### The foot of the perpendicular from the point on the line is

#### The foot of the perpendicular from the point on the line is

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#### The point of intersection of the lines is

#### The point of intersection of the lines is

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#### The line passing through and perpendicular to is

#### The line passing through and perpendicular to is

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

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

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#### The length of the perpendicular from (-1, π/6) to the line is

#### The length of the perpendicular from (-1, π/6) to the line is

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#### The sum of all the numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time is (repetition is not allowed) :

Complete step-by-step answer:

According to the problem, we need to find the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time.

Let us first fix a number in a unit place and find the total number of words possible due on fixing this number.

We need to arrange the remaining three places with three digits.

We know that the number of ways of arranging n objects in n places is n! ways.

So, we get 3!=6 numbers on fixing the unit place with a particular digit.

Now, let us find the sum of all digits. We get sum as 2+3+4+5=14.

Now, we get a sum of digits in units place for all the numbers as 14×6=84.

We use the same digits in ten, hundred and thousand places also. So, the sum of those digits will also be 84 but with the multiplication of its place value.

i.e., We multiply the sum of the digits in tenth place with 10, hundredth place with 100 and so on. We then add these sums.

So, we get the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time is

(84×1000)+(84×100)+(84×10)+(84×1)

Sum = 84000+8400+840+84

Sum = 93324

We have found the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time as 93324.

According to the problem, we need to find the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time.

Let us first fix a number in a unit place and find the total number of words possible due on fixing this number.

We need to arrange the remaining three places with three digits.

We know that the number of ways of arranging n objects in n places is n! ways.

So, we get 3!=6 numbers on fixing the unit place with a particular digit.

Now, let us find the sum of all digits. We get sum as 2+3+4+5=14.

Now, we get a sum of digits in units place for all the numbers as 14×6=84.

We use the same digits in ten, hundred and thousand places also. So, the sum of those digits will also be 84 but with the multiplication of its place value.

i.e., We multiply the sum of the digits in tenth place with 10, hundredth place with 100 and so on. We then add these sums.

So, we get the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time is

(84×1000)+(84×100)+(84×10)+(84×1)

Sum = 84000+8400+840+84

Sum = 93324

We have found the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time as 93324.

#### The sum of all the numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time is (repetition is not allowed) :

Maths-General

Complete step-by-step answer:

According to the problem, we need to find the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time.

Let us first fix a number in a unit place and find the total number of words possible due on fixing this number.

We need to arrange the remaining three places with three digits.

We know that the number of ways of arranging n objects in n places is n! ways.

So, we get 3!=6 numbers on fixing the unit place with a particular digit.

Now, let us find the sum of all digits. We get sum as 2+3+4+5=14.

Now, we get a sum of digits in units place for all the numbers as 14×6=84.

We use the same digits in ten, hundred and thousand places also. So, the sum of those digits will also be 84 but with the multiplication of its place value.

i.e., We multiply the sum of the digits in tenth place with 10, hundredth place with 100 and so on. We then add these sums.

So, we get the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time is

(84×1000)+(84×100)+(84×10)+(84×1)

Sum = 84000+8400+840+84

Sum = 93324

We have found the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time as 93324.

According to the problem, we need to find the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time.

Let us first fix a number in a unit place and find the total number of words possible due on fixing this number.

We need to arrange the remaining three places with three digits.

We know that the number of ways of arranging n objects in n places is n! ways.

So, we get 3!=6 numbers on fixing the unit place with a particular digit.

Now, let us find the sum of all digits. We get sum as 2+3+4+5=14.

Now, we get a sum of digits in units place for all the numbers as 14×6=84.

We use the same digits in ten, hundred and thousand places also. So, the sum of those digits will also be 84 but with the multiplication of its place value.

i.e., We multiply the sum of the digits in tenth place with 10, hundredth place with 100 and so on. We then add these sums.

So, we get the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time is

(84×1000)+(84×100)+(84×10)+(84×1)

Sum = 84000+8400+840+84

Sum = 93324

We have found the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time as 93324.