Maths
General
Easy
Question
The length of the perpendicular from (1, π/6) to the line is

 3


The correct answer is:
Related Questions to study
Maths
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 stepbystep 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) :
MathsGeneral
Complete stepbystep 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.
Maths
Total number of divisors of 480, that are of the form 4n + 2, n 0, is equal to :
In this question, we have been asked to find the total number of divisors of 480 which are of the form 4n + 2,n 0
To solve this question, we should know that the total number of divisors of any number x of the form are prime numbers and is given by (m + 1) (n + 1) (p + 1)….. we know that 480 can be expressed as
So, according to the formula, the total number of divisors of 480 are (5 + 1) (1 + 1) (1 + 1) = 6×2×2=24
Now, we have been asked to find the number of divisors which are of the form 4n + 2 = 2 (2n + 1), which means odd divisors cannot be a part of the solution. So, the total number of odd divisors that are possible are (1 + 1) (1 + 1) = 2×2=4, according to the property.
Now, we can say the total number of even divisors are = all divisors – odd divisor
= 24 – 4
= 20
Now, we have been given that the divisor should be of 4n + 2, which means they should not be a multiple of 4 but multiple of 2. For that, we will subtract the multiple of 4 which are divisor of 480 from the even divisors
And, we know that,
So, the number of divisors that are multiples of 4 are (3 + 1) (1 + 1) (1 + 1) = 4×2×2 = 16.
Hence, we can say that there are 16 divisors of 480 which are multiple of 4.
So, the total number of divisors which are even but not divisible by 2 can be given by 20 – 16 = 4.
Hence, we can say that there are 4 divisors of 480 that are of 4n + 2 form, n≥0
.
To solve this question, we should know that the total number of divisors of any number x of the form are prime numbers and is given by (m + 1) (n + 1) (p + 1)….. we know that 480 can be expressed as
So, according to the formula, the total number of divisors of 480 are (5 + 1) (1 + 1) (1 + 1) = 6×2×2=24
Now, we have been asked to find the number of divisors which are of the form 4n + 2 = 2 (2n + 1), which means odd divisors cannot be a part of the solution. So, the total number of odd divisors that are possible are (1 + 1) (1 + 1) = 2×2=4, according to the property.
Now, we can say the total number of even divisors are = all divisors – odd divisor
= 24 – 4
= 20
Now, we have been given that the divisor should be of 4n + 2, which means they should not be a multiple of 4 but multiple of 2. For that, we will subtract the multiple of 4 which are divisor of 480 from the even divisors
And, we know that,
So, the number of divisors that are multiples of 4 are (3 + 1) (1 + 1) (1 + 1) = 4×2×2 = 16.
Hence, we can say that there are 16 divisors of 480 which are multiple of 4.
So, the total number of divisors which are even but not divisible by 2 can be given by 20 – 16 = 4.
Hence, we can say that there are 4 divisors of 480 that are of 4n + 2 form, n≥0
.
.
Total number of divisors of 480, that are of the form 4n + 2, n 0, is equal to :
MathsGeneral
In this question, we have been asked to find the total number of divisors of 480 which are of the form 4n + 2,n 0
To solve this question, we should know that the total number of divisors of any number x of the form are prime numbers and is given by (m + 1) (n + 1) (p + 1)….. we know that 480 can be expressed as
So, according to the formula, the total number of divisors of 480 are (5 + 1) (1 + 1) (1 + 1) = 6×2×2=24
Now, we have been asked to find the number of divisors which are of the form 4n + 2 = 2 (2n + 1), which means odd divisors cannot be a part of the solution. So, the total number of odd divisors that are possible are (1 + 1) (1 + 1) = 2×2=4, according to the property.
Now, we can say the total number of even divisors are = all divisors – odd divisor
= 24 – 4
= 20
Now, we have been given that the divisor should be of 4n + 2, which means they should not be a multiple of 4 but multiple of 2. For that, we will subtract the multiple of 4 which are divisor of 480 from the even divisors
And, we know that,
So, the number of divisors that are multiples of 4 are (3 + 1) (1 + 1) (1 + 1) = 4×2×2 = 16.
Hence, we can say that there are 16 divisors of 480 which are multiple of 4.
So, the total number of divisors which are even but not divisible by 2 can be given by 20 – 16 = 4.
Hence, we can say that there are 4 divisors of 480 that are of 4n + 2 form, n≥0
.
To solve this question, we should know that the total number of divisors of any number x of the form are prime numbers and is given by (m + 1) (n + 1) (p + 1)….. we know that 480 can be expressed as
So, according to the formula, the total number of divisors of 480 are (5 + 1) (1 + 1) (1 + 1) = 6×2×2=24
Now, we have been asked to find the number of divisors which are of the form 4n + 2 = 2 (2n + 1), which means odd divisors cannot be a part of the solution. So, the total number of odd divisors that are possible are (1 + 1) (1 + 1) = 2×2=4, according to the property.
Now, we can say the total number of even divisors are = all divisors – odd divisor
= 24 – 4
= 20
Now, we have been given that the divisor should be of 4n + 2, which means they should not be a multiple of 4 but multiple of 2. For that, we will subtract the multiple of 4 which are divisor of 480 from the even divisors
And, we know that,
So, the number of divisors that are multiples of 4 are (3 + 1) (1 + 1) (1 + 1) = 4×2×2 = 16.
Hence, we can say that there are 16 divisors of 480 which are multiple of 4.
So, the total number of divisors which are even but not divisible by 2 can be given by 20 – 16 = 4.
Hence, we can say that there are 4 divisors of 480 that are of 4n + 2 form, n≥0
.
.
Maths
If ^{9}P_{5} + 5 ^{9}P_{4} = ^{10}P_{r }, then r =
Given :
Using Formula :
Dividing both sides by 9!
Using Formula :
Dividing both sides by 9!
If ^{9}P_{5} + 5 ^{9}P_{4} = ^{10}P_{r }, then r =
MathsGeneral
Given :
Using Formula :
Dividing both sides by 9!
Using Formula :
Dividing both sides by 9!
Maths
The number of proper divisors of . . 15^{r} is
. .  We need to find proper divisors.
Suppose is a number then factors of = ( and a is proper
i.e. has total division = (n + 1)
Now, =
We know that
Thus, = =
Total factors = (p+q+1)(q+r+1)(r+1)
However, proper divisors exclude $1$ and the number itself.
Hence, the answer is $(p+q+1)(q+r+1)(r+1)−2.$
Suppose is a number then factors of = ( and a is proper
i.e. has total division = (n + 1)
Now, =
We know that
Thus, = =
Total factors = (p+q+1)(q+r+1)(r+1)
However, proper divisors exclude $1$ and the number itself.
Hence, the answer is $(p+q+1)(q+r+1)(r+1)−2.$
The number of proper divisors of . . 15^{r} is
MathsGeneral
. .  We need to find proper divisors.
Suppose is a number then factors of = ( and a is proper
i.e. has total division = (n + 1)
Now, =
We know that
Thus, = =
Total factors = (p+q+1)(q+r+1)(r+1)
However, proper divisors exclude $1$ and the number itself.
Hence, the answer is $(p+q+1)(q+r+1)(r+1)−2.$
Suppose is a number then factors of = ( and a is proper
i.e. has total division = (n + 1)
Now, =
We know that
Thus, = =
Total factors = (p+q+1)(q+r+1)(r+1)
However, proper divisors exclude $1$ and the number itself.
Hence, the answer is $(p+q+1)(q+r+1)(r+1)−2.$
Maths
If have a common factor then 'a' is equal to
If have a common factor then 'a' is equal to
MathsGeneral
physics
A block C of mass is moving with velocity and collides elastically with block of mass and connected to another block of mass through spring constant .What is if is compression of spring when velocity of is same ?
Using conservation of linear momentum, we
have
Or
Using conservation of energy, we have
Where compression in the spring
Or
have
Or
Using conservation of energy, we have
Where compression in the spring
Or
A block C of mass is moving with velocity and collides elastically with block of mass and connected to another block of mass through spring constant .What is if is compression of spring when velocity of is same ?
physicsGeneral
Using conservation of linear momentum, we
have
Or
Using conservation of energy, we have
Where compression in the spring
Or
have
Or
Using conservation of energy, we have
Where compression in the spring
Or
Maths
If then ascending order of A, B, C.
If then ascending order of A, B, C.
MathsGeneral
Maths
The number of different seven digit numbers that can be written using only the three digits 1, 2 and 3 with the condition that the digit 2 occurs twice in each number is
Complete step by step solution:
We are given the number of different sevendigit numbers that can be written using only three digits 1,2 and 3. Therefore,
Total number of Digits = 7
We are given that the digit two occurs exactly twice in each number.
Thus, the digit two occurs twice in the seven digit number.
Now, we will find the number of ways of arrangement of the digit two in the seven digit number by using combination.
Total number of ways that the digit two occurs exactly twice in each number =
Now, the remaining five digits can be written using two digits 1 and 3 in ways.
We will now find the total number of seven digit number by multiplying the number of ways of arrangement in both the cases. Therefore
Total number of seven digit number =
Now by using the formula
, we get
Total number of seven digit number =
We know that the factorial can be written by the formula n! = n(n1)! , so we get
Total number of seven digit number =
Total number of seven digit number =
Simplifying the expression, we get
Total number of seven digit number =
Multiplying the terms, we get
Total number of seven digit number = 672
Therefore, the number of different sevendigit numbers that can be written using only three digits 1,2 and 3 is 672.
We are given the number of different sevendigit numbers that can be written using only three digits 1,2 and 3. Therefore,
Total number of Digits = 7
We are given that the digit two occurs exactly twice in each number.
Thus, the digit two occurs twice in the seven digit number.
Now, we will find the number of ways of arrangement of the digit two in the seven digit number by using combination.
Total number of ways that the digit two occurs exactly twice in each number =
Now, the remaining five digits can be written using two digits 1 and 3 in ways.
We will now find the total number of seven digit number by multiplying the number of ways of arrangement in both the cases. Therefore
Total number of seven digit number =
Now by using the formula
, we get
Total number of seven digit number =
We know that the factorial can be written by the formula n! = n(n1)! , so we get
Total number of seven digit number =
Total number of seven digit number =
Simplifying the expression, we get
Total number of seven digit number =
Multiplying the terms, we get
Total number of seven digit number = 672
Therefore, the number of different sevendigit numbers that can be written using only three digits 1,2 and 3 is 672.
The number of different seven digit numbers that can be written using only the three digits 1, 2 and 3 with the condition that the digit 2 occurs twice in each number is
MathsGeneral
Complete step by step solution:
We are given the number of different sevendigit numbers that can be written using only three digits 1,2 and 3. Therefore,
Total number of Digits = 7
We are given that the digit two occurs exactly twice in each number.
Thus, the digit two occurs twice in the seven digit number.
Now, we will find the number of ways of arrangement of the digit two in the seven digit number by using combination.
Total number of ways that the digit two occurs exactly twice in each number =
Now, the remaining five digits can be written using two digits 1 and 3 in ways.
We will now find the total number of seven digit number by multiplying the number of ways of arrangement in both the cases. Therefore
Total number of seven digit number =
Now by using the formula
, we get
Total number of seven digit number =
We know that the factorial can be written by the formula n! = n(n1)! , so we get
Total number of seven digit number =
Total number of seven digit number =
Simplifying the expression, we get
Total number of seven digit number =
Multiplying the terms, we get
Total number of seven digit number = 672
Therefore, the number of different sevendigit numbers that can be written using only three digits 1,2 and 3 is 672.
We are given the number of different sevendigit numbers that can be written using only three digits 1,2 and 3. Therefore,
Total number of Digits = 7
We are given that the digit two occurs exactly twice in each number.
Thus, the digit two occurs twice in the seven digit number.
Now, we will find the number of ways of arrangement of the digit two in the seven digit number by using combination.
Total number of ways that the digit two occurs exactly twice in each number =
Now, the remaining five digits can be written using two digits 1 and 3 in ways.
We will now find the total number of seven digit number by multiplying the number of ways of arrangement in both the cases. Therefore
Total number of seven digit number =
Now by using the formula
, we get
Total number of seven digit number =
We know that the factorial can be written by the formula n! = n(n1)! , so we get
Total number of seven digit number =
Total number of seven digit number =
Simplifying the expression, we get
Total number of seven digit number =
Multiplying the terms, we get
Total number of seven digit number = 672
Therefore, the number of different sevendigit numbers that can be written using only three digits 1,2 and 3 is 672.
Maths
The centre and radius of the circle are respectively
The centre and radius of the circle are respectively
MathsGeneral
maths
The centre of the circle is
The centre of the circle is
mathsGeneral
maths
The equation of the circle with centre at , which passes through the point is
The equation of the circle with centre at , which passes through the point is
mathsGeneral
maths
The foot of the perpendicular from on the line is
The foot of the perpendicular from on the line is
mathsGeneral
maths
The foot of the perpendicular from the pole on the line is
The foot of the perpendicular from the pole on the line is
mathsGeneral
maths
The equation of the line parallel to and passing through is
The equation of the line parallel to and passing through is
mathsGeneral
Maths
The line passing through the points , (3,0) is
The line passing through the points , (3,0) is
MathsGeneral