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General
Easy

Question

The point of intersection of the lines 2 c o s space theta plus s i n space theta equals 1 over r comma c o s space theta plus s i n space theta equals 1 over r is

  1. open parentheses 1 comma fraction numerator pi over denominator 4 end fraction close parentheses    
  2. open parentheses 1 comma fraction numerator pi over denominator 2 end fraction close parentheses    
  3. open parentheses 2 comma fraction numerator pi over denominator 4 end fraction close parentheses    
  4. open parentheses 2 comma fraction numerator pi over denominator 2 end fraction close parentheses    

The correct answer is: open parentheses 1 comma fraction numerator pi over denominator 2 end fraction close parentheses

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General
maths-

The line passing through open parentheses negative 1 comma fraction numerator pi over denominator 2 end fraction close parentheses and perpendicular to square root of 3 s i n space theta plus 2 c o s space theta equals 4 over r is

The line passing through open parentheses negative 1 comma fraction numerator pi over denominator 2 end fraction close parentheses and perpendicular to square root of 3 s i n space theta plus 2 c o s space theta equals 4 over r is

maths-General
General
maths-

The equation of the line passing through left parenthesis negative 1 comma pi divided by 6 right parenthesis comma left parenthesis 1 comma pi divided by 2 right parenthesis is

The equation of the line passing through left parenthesis negative 1 comma pi divided by 6 right parenthesis comma left parenthesis 1 comma pi divided by 2 right parenthesis is

maths-General
General
maths-

The length of the perpendicular from (-1, π/6) to the line r left parenthesis 3 s i n space theta plus square root of 3 c o s space theta right parenthesis equals 3 is

The length of the perpendicular from (-1, π/6) to the line r left parenthesis 3 s i n space theta plus square root of 3 c o s space theta right parenthesis equals 3 is

maths-General
General
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 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.<br> </br>
Now, we get a sum of digits in units place for all the numbers as 14×6=84.<br> </br>
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)Error converting from MathML to accessible text.
Sum =  84000+8400+840+84<br> </br>
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.
Error converting from MathML to accessible text.
Error converting from MathML to accessible text.

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.<br> </br>
Now, we get a sum of digits in units place for all the numbers as 14×6=84.<br> </br>
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)Error converting from MathML to accessible text.
Sum =  84000+8400+840+84<br> </br>
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.
Error converting from MathML to accessible text.
Error converting from MathML to accessible text.
General
Maths-

Total number of divisors of 480, that are of the form 4n + 2, n greater or equal than 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 greater or equal than 0 
To solve this question, we should know that the total number of divisors of any number x of the form a to the power of m b to the power of n c to the power of p...... space w h e r e space a comma b comma c are prime numbers and is given by (m + 1) (n + 1) (p + 1)….. we know that 480 can be expressed as
480 space equals space 2 to the power of 5cross times 3 cross times 5

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, 480 space equals space 2 squared space left parenthesis space 2 to the power of 8 cross times 3 cross times 5 right parenthesis
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, n0
.
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.

Total number of divisors of 480, that are of the form 4n + 2, n greater or equal than 0, is equal to :

Maths-General
In this question, we have been asked to find the total number of divisors of 480 which are of the form 4n + 2,n greater or equal than 0 
To solve this question, we should know that the total number of divisors of any number x of the form a to the power of m b to the power of n c to the power of p...... space w h e r e space a comma b comma c are prime numbers and is given by (m + 1) (n + 1) (p + 1)….. we know that 480 can be expressed as
480 space equals space 2 to the power of 5cross times 3 cross times 5

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, 480 space equals space 2 squared space left parenthesis space 2 to the power of 8 cross times 3 cross times 5 right parenthesis
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, n0
.
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.
General
Maths-

If 9P5 + 5 9P4 = 10Pr , then r =

Given : P presuperscript 9 subscript 5 space plus space 5 space cross times P presuperscript 9 subscript 4 space equals space P presuperscript 10 subscript r
Using Formula :
P presuperscript n subscript r space equals space fraction numerator n factorial over denominator left parenthesis n minus r right parenthesis factorial end fraction
fraction numerator 9 factorial over denominator left parenthesis 9 minus 5 right parenthesis factorial end fraction space plus space 5 cross times fraction numerator 9 factorial over denominator left parenthesis 9 minus 4 right parenthesis factorial end fraction space equals space fraction numerator 10 factorial over denominator left parenthesis 10 minus r right parenthesis factorial end fraction
fraction numerator 9 factorial over denominator 4 factorial end fraction space plus space 5 space cross times space fraction numerator 9 factorial over denominator 5 factorial end fraction space equals space fraction numerator 10 cross times 9 factorial over denominator left parenthesis 10 minus r right parenthesis factorial end fraction
Dividing both sides by 9!
fraction numerator 1 over denominator 4 factorial end fraction plus 5 cross times fraction numerator 1 over denominator 5 factorial end fraction space equals space fraction numerator 10 over denominator left parenthesis 10 minus r right parenthesis factorial end fraction
fraction numerator 5 over denominator 5 factorial end fraction plus space fraction numerator 5 over denominator 5 factorial end fraction equals space fraction numerator 10 over denominator left parenthesis 10 minus r right parenthesis factorial end fraction
fraction numerator 10 over denominator 5 factorial end fraction space equals space fraction numerator 10 over denominator left parenthesis 10 minus r right parenthesis factorial end fraction
E q u a t i n g space t h e space d e n o m i n a t o r s
5 factorial space equals space left parenthesis 10 minus r right parenthesis factorial
5 space equals space 10 space minus space r
r space equals space 5

If 9P5 + 5 9P4 = 10Pr , then r =

Maths-General
Given : P presuperscript 9 subscript 5 space plus space 5 space cross times P presuperscript 9 subscript 4 space equals space P presuperscript 10 subscript r
Using Formula :
P presuperscript n subscript r space equals space fraction numerator n factorial over denominator left parenthesis n minus r right parenthesis factorial end fraction
fraction numerator 9 factorial over denominator left parenthesis 9 minus 5 right parenthesis factorial end fraction space plus space 5 cross times fraction numerator 9 factorial over denominator left parenthesis 9 minus 4 right parenthesis factorial end fraction space equals space fraction numerator 10 factorial over denominator left parenthesis 10 minus r right parenthesis factorial end fraction
fraction numerator 9 factorial over denominator 4 factorial end fraction space plus space 5 space cross times space fraction numerator 9 factorial over denominator 5 factorial end fraction space equals space fraction numerator 10 cross times 9 factorial over denominator left parenthesis 10 minus r right parenthesis factorial end fraction
Dividing both sides by 9!
fraction numerator 1 over denominator 4 factorial end fraction plus 5 cross times fraction numerator 1 over denominator 5 factorial end fraction space equals space fraction numerator 10 over denominator left parenthesis 10 minus r right parenthesis factorial end fraction
fraction numerator 5 over denominator 5 factorial end fraction plus space fraction numerator 5 over denominator 5 factorial end fraction equals space fraction numerator 10 over denominator left parenthesis 10 minus r right parenthesis factorial end fraction
fraction numerator 10 over denominator 5 factorial end fraction space equals space fraction numerator 10 over denominator left parenthesis 10 minus r right parenthesis factorial end fraction
E q u a t i n g space t h e space d e n o m i n a t o r s
5 factorial space equals space left parenthesis 10 minus r right parenthesis factorial
5 space equals space 10 space minus space r
r space equals space 5
General
Maths-

The number of proper divisors of 2 to the power of p. 6 to the power of q. 15r is-

  2 to the power of p6 to the power of q. 15 to the power of r  - We need to find proper divisors.
Suppose a to the power of n is a number then factors of a to the power of n  = (a to the power of 0 comma space a to the power of 1 comma space a squared comma a cubed comma........ space a to the power of n right parenthesis
 and a is proper
i.e. a to the power of n has total division = (n + 1)
Now,   2 to the power of pcross times 6 to the power of qcross times 15 to the power of r = 2 to the power of p cross times 2 to the power of q cross times 3 to the power of q cross times 3 to the power of r cross times 5 to the power of r
We know that x to the power of m cross times space x to the power of n space equals x to the power of m space plus space n end exponent
Thus,   2 to the power of pcross times 6 to the power of qcross times 15 to the power of r = 2 to the power of p cross times 2 to the power of q cross times 3 to the power of q cross times 3 to the power of r cross times 5 to the power of r = 2 to the power of left parenthesis p plus q right parenthesis end exponent cross times 3 to the power of left parenthesis q plus r right parenthesis end exponent cross times 5 to the power of r
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 2 to the power of p. 6 to the power of q. 15r is-

Maths-General
  2 to the power of p6 to the power of q. 15 to the power of r  - We need to find proper divisors.
Suppose a to the power of n is a number then factors of a to the power of n  = (a to the power of 0 comma space a to the power of 1 comma space a squared comma a cubed comma........ space a to the power of n right parenthesis
 and a is proper
i.e. a to the power of n has total division = (n + 1)
Now,   2 to the power of pcross times 6 to the power of qcross times 15 to the power of r = 2 to the power of p cross times 2 to the power of q cross times 3 to the power of q cross times 3 to the power of r cross times 5 to the power of r
We know that x to the power of m cross times space x to the power of n space equals x to the power of m space plus space n end exponent
Thus,   2 to the power of pcross times 6 to the power of qcross times 15 to the power of r = 2 to the power of p cross times 2 to the power of q cross times 3 to the power of q cross times 3 to the power of r cross times 5 to the power of r = 2 to the power of left parenthesis p plus q right parenthesis end exponent cross times 3 to the power of left parenthesis q plus r right parenthesis end exponent cross times 5 to the power of r
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.
General
Maths-

Ifx squared minus 11 x plus a text  and  end text x squared minus 14 x plus 2 a have a common factor then 'a' is equal to

Ifx squared minus 11 x plus a text  and  end text x squared minus 14 x plus 2 a have a common factor then 'a' is equal to

Maths-General
General
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

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 ?

physics-General
Using conservation of linear momentum, we
have

Or
Using conservation of energy, we have

Where  compression in the spring

Or
General
Maths-

If fraction numerator 1 over denominator left parenthesis 1 plus 2 x right parenthesis open parentheses 1 minus x squared close parentheses end fraction equals fraction numerator A over denominator 1 plus 2 x end fraction plus fraction numerator B over denominator 1 plus x end fraction plus fraction numerator C over denominator 1 minus x end fraction then assending order of A,B,C

If fraction numerator 1 over denominator left parenthesis 1 plus 2 x right parenthesis open parentheses 1 minus x squared close parentheses end fraction equals fraction numerator A over denominator 1 plus 2 x end fraction plus fraction numerator B over denominator 1 plus x end fraction plus fraction numerator C over denominator 1 minus x end fraction then assending order of A,B,C

Maths-General
General
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 seven-digit 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 =C presuperscript 7 subscript 2

Now, the remaining five digits can be written using two digits 1 and 3 in 2 to the power of 5ways.
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 = C presuperscript 7 subscript 2 space cross times space 2 to the power of 5
Now by using the formula 
C presuperscript n subscript r space equals space begin inline style fraction numerator n factorial over denominator r factorial space left parenthesis n minus r right parenthesis factorial end fraction end style end subscript , we get
 Total number of seven digit number = fraction numerator 7 factorial over denominator left parenthesis 7 minus 2 right parenthesis factorial space 2 factorial end fraction cross times space 2 to the power of 5
We know that the factorial can be written by the formula  n! = ncross times(n-1)! , so we get
Total number of seven digit number = fraction numerator 7 space cross times 6 cross times 5 factorial over denominator 5 factorial space 2 factorial end fraction cross times 2 to the power of 5
Total number of seven digit number  = fraction numerator 7 space cross times 6 over denominator 2 factorial end fraction cross times 2 to the power of 5
Simplifying the expression, we get
Total number of seven digit number  = 7 cross times 6 cross times 2 to the power of 4

Multiplying the terms, we get
Total number of seven digit number  672
Therefore, the number of different seven-digit 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-

Maths-General
Complete step by step solution:
We are given the number of different seven-digit 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 =C presuperscript 7 subscript 2

Now, the remaining five digits can be written using two digits 1 and 3 in 2 to the power of 5ways.
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 = C presuperscript 7 subscript 2 space cross times space 2 to the power of 5
Now by using the formula 
C presuperscript n subscript r space equals space begin inline style fraction numerator n factorial over denominator r factorial space left parenthesis n minus r right parenthesis factorial end fraction end style end subscript , we get
 Total number of seven digit number = fraction numerator 7 factorial over denominator left parenthesis 7 minus 2 right parenthesis factorial space 2 factorial end fraction cross times space 2 to the power of 5
We know that the factorial can be written by the formula  n! = ncross times(n-1)! , so we get
Total number of seven digit number = fraction numerator 7 space cross times 6 cross times 5 factorial over denominator 5 factorial space 2 factorial end fraction cross times 2 to the power of 5
Total number of seven digit number  = fraction numerator 7 space cross times 6 over denominator 2 factorial end fraction cross times 2 to the power of 5
Simplifying the expression, we get
Total number of seven digit number  = 7 cross times 6 cross times 2 to the power of 4

Multiplying the terms, we get
Total number of seven digit number  672
Therefore, the number of different seven-digit numbers that can be written using only three digits 1,2 and 3 is 672.

 
General
maths-

The centre and radius of the circle r equals c o s space theta minus s i n space theta are respectively

The centre and radius of the circle r equals c o s space theta minus s i n space theta are respectively

maths-General
General
maths-

The centre of the circle r squared minus 2 r left parenthesis 3 c o s space theta plus 4 s i n space theta right parenthesis minus 24 is

The centre of the circle r squared minus 2 r left parenthesis 3 c o s space theta plus 4 s i n space theta right parenthesis minus 24 is

maths-General
General
maths-

The equation of the circle with centre at open parentheses 1 comma 0 to the power of 0 end exponent close parentheses, which passes through the point open parentheses 0 comma fraction numerator pi over denominator 2 end fraction close parentheses is

The equation of the circle with centre at open parentheses 1 comma 0 to the power of 0 end exponent close parentheses, which passes through the point open parentheses 0 comma fraction numerator pi over denominator 2 end fraction close parentheses is

maths-General
General
maths-

The foot of the perpendicular from left parenthesis negative 1 comma pi divided by 6 right parenthesis on the line r left parenthesis 3 s i n space theta plus square root of 3 c o s space theta right parenthesis equals 3 is

The foot of the perpendicular from left parenthesis negative 1 comma pi divided by 6 right parenthesis on the line r left parenthesis 3 s i n space theta plus square root of 3 c o s space theta right parenthesis equals 3 is

maths-General