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  • 1 answers

Bhuvaneshwari Shivapur 2 years, 5 months ago

Solution:- Euclid's division lemma : Let a and b be any two positive Integers . Then there exist two unique whole numbers q and r such that a=bq+r , 0≤r<b Now , start with a larger integer , that is 337554, Apply the division lemma to 337554 and 26676, 337554=26676×12+17442 26676=17442×1+9234 17442=9234×1+8208 9234=8208×1+1026 8208=1026×8+0 The remainder has now become zero , so our procedure stops. Since the divisor at this stage is 1026 . ∴HCF(337554,26676)=1026
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Kaushik Krishna 2 years, 5 months ago

Let √5 a rational number Where √5=a/b and a and b are co primes and b not equal to 0 Square them 5=a square / b square Cross multiply 5b square is equal to a square B is a factor of a B is also a factor of a square Now take A is equal to 5 c Square them A square is equal to 25 c square We know that a square is equal to 5 b square So now 5 b square is equal to 25 c square Simplify them We get b square is equal to 5 c square Now 5 is a factor of b square and b It contradict to our assumption that a and b are co primes Hence our assumption is wrong Therefore √5 is an irrational number Hence proved

Sureksha Sp 2 years, 5 months ago

Let √5 a rational number Where √5=a/b and a and b are co primes and b not equal to 0 Square them 5=a square / b square Cross multiply 5b square is equal to a square B is a factor of a B is also a factor of a square Now take A is equal to 5 c Square them A square is equal to 25 c square We know that a square is equal to 5 b square So now 5 b square is equal to 25 c square Simplify them We get b square is equal to 5 c square Now 5 is a factor of b square and b It contradict to our assumption that a and b are co primes Hence our assumption is wrong Therefore √5 is an irrational number Hence proved

Joshi Priya Dharmeshkumar 2 years, 5 months ago

Let √5 =a/b (where a/b is an rational number where a/b are co-primes) Squaring both the sides 5 = a2/b2 5b2 = a2 ---1 Thus 5 divides a2 , 5 divides a Now take any other integer as c such that a=5c squaring both the sides a2=25c2 --- 2 put a value from eq-1 into eq-2 5 b2 =25 c 2 b2=5c2 5 divides b2 and 5 divide b We let a&b are co-primes but here we are getting 5 as a factor of both a &b. So our prediction that a&b are co-primes is wrong Thus ✓5 is an Irrational number.
  • 1 answers

Khushi Kushwaha 2 years, 5 months ago

204 = 2 x 2 x 3 x 17
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Purba Chowdhary 2 years, 5 months ago

(17 × 11 × 2 + 17 × 11 × 5 ) = 17 × 11 ( 2 + 5 ) {explanation : taking 17 and 11 as common } = 187 ( 2 + 5 ) here 187 is a factor of (17 × 11 × 2 + 17 × 11 × 5 ) so we can say that (17 × 11 × 2 + 17 × 11 × 5 ) is composite number because it has factor other than 1
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Aditya Singh 2 years, 5 months ago

Prove that root 2 is an irrational number. Rational numbers are integers that are expressed in the form of p / q where p and q are both co-prime numbers and q is non-zero. Answer: Hence proved that √2 is an irrational number. Let's find if √2 is irrational. Explanation: To prove that √2 is an irrational number, we will use the contradiction method. Let us assume that √2 is a rational number with p and q as co-prime integers and q ≠ 0 ⇒ √2 = p/q On squaring both sides we get, ⇒ 2q2 = p2 ⇒ Here, 2q2 is a multiple of 2 and hence it is even. Thus, p2 is an even number. Therefore, p is also even. So we can assume that p = 2x where x is an integer. By substituting this value of p in 2q2 = p2, we get ⇒ 2q2 = (2x)2 ⇒ 2q2 = 4x2 ⇒ q2 = 2x2 ⇒ q2 is an even number. Therefore, q is also even. Since p and q both are even numbers, they have 2 as a common multiple which means that p and q are not co-prime numbers as their HCF is 2. This leads to the contradiction that root 2 is a rational number in the form of p/q with "p and q both co-prime numbers" and q ≠ 0. Thus, √2 is an irrational number by the contradiction method.
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  • 2 answers

Vedshree Patil 2 years, 5 months ago

X=9y=4

Sagar Kuntal 2 years, 5 months ago

4x+3y=24 -[(-2x)+3y]=-6 4x+3y=24 2x-3y=-6 4x+2x+3y-3y=24-6 6x=18 x=18/6 x= 3 2(3)-3y=-6 6-3y=-6 -3y=-12 y=-12/-3 y=4
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Ankit Class 2 years, 6 months ago

Give space and use bracket because not understand
876
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  • 2 answers

Akshita Akshita 2 years, 5 months ago

5×7×11×13

G.Charan Charan Teja Raju 2 years, 5 months ago

Prime factor 5005
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Akshita Akshita 2 years, 5 months ago

638
  • 1 answers

Sadhvi Tiwari 2 years, 6 months ago

A very easy method first u have to make coefficient same of any one veriable by multiplying the coefficient of 2nd in 1st or 1st in 2nd (in whole equation) then add or subtract to cancel one variable then find the value of another variable now put the value of another variable in eq then u get the value of other that is an elimination method ..........hope understood
  • 3 answers

Yash Thakur 2 years, 6 months ago

What is elimination method?

Shreya Pandey 2 years, 6 months ago

You can use substitution and elimination method also

Kanika Mehra 2 years, 6 months ago

Pls help me I am confused 😕😕
  • 1 answers

Aryan Bhatt 2 years, 6 months ago

डीजीवीजी
  • 1 answers

Sureksha Sp 2 years, 5 months ago

1 square 1 2 square 4 3 square 9 4 square 16 5 square 25 6 square 36 7 square 49 8 square 64 9 square 81 10 square 100
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Pari Agrahari 2 years, 6 months ago

Yessss
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Akshita Akshita 2 years, 5 months ago

4x^2 - 4x - 3 = 4x^2- 6x+ 2x- 3 = 2x(2x-3)+1(2x-3) = (2x-3)(2x+1) So, zeroes are -1/2 and 3/2

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