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    • Posted by Fareha Manzar 1 year, 11 months ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Amishi Agarwal 1 year, 11 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Muskaan Syed 1 year, 11 months ago

    Let's assume that √2 + √5 is a rational number. √2+√5=a/b Sq. On both sides =2+5+2√10= a^2/b^2 2√10=a^2/b^2 -7 2√10=a^2-7/b^2. √10=a^2-7/2b^2 . It's a fact that √10 is irrational. Hence our assumption that √2+√5 is rational is incorrect..... Hence, √2+√5 is irrational

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    • Posted by Amishi Agarwal 1 year, 11 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    B 01 Rojalin Behera 1 year, 11 months ago

    Let assume that √2 is rational So, √2=a/b (a and b are positive co-prime number and b is not equal to 0) i.e. HCF (a,b)=1 b√2=a Squaring both the sides, we get (b√2)^2=a^2 =2b^2=a^2....(i) =b^2=a^2/2 =a^2/2 (if P is a prime number, if P divides a^2 then P divides a) =a/2 So, a=3c for some integer c Putting a=2c in equation (i) 2b^2=(2c)^2 =2b^2=4c^2 =b^2=2c^2 =b^2/2=c^2 =b^2/2 (If P is a prime number, if P divides a^2 then P divides a) =b/2 So, a and b have another prime factor 2 But this contradicts the fact that a and b are co-prime number. This contradiction has arisen due to our incorrect assumption. So, √2 is an irrational number. Hence, proved.

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    • Posted by Cccc Bbb 1 year, 11 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Sourav Singh 1 year, 11 months ago

    -0.133974596

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    • Posted by Paridhi Saxena 1 year, 11 months ago

    CBSE > Class 10 > Mathematics
    • 3 answers

    Sourav Singh 1 year, 11 months ago

    α2+β2=a2b2−2ac

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    Ashish Ashish 1 year, 11 months ago

    Paper to easy hota hai ya hard

    2Thank You

    Ashish Ashish 1 year, 11 months ago

    Sir mai 10th compartment kar rha hu

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    • Posted by Kaushik Krishna 1 year, 11 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Aditya D 1 year, 11 months ago

    LCM of 48 and 60 = 240 bookd

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    • Posted by Mohammed Ali Rahib 1 year, 11 months ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Prashasti Jk 2 years ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    B 01 Rojalin Behera 1 year, 11 months ago

    6=2×3 72=2×2×2×3×3 120=2×2×2×3×5 HCF(6,72,120)=2×3 =6 LCM(6,72,120)=2^3×3^2×5 =8×9×5 =360

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    • Posted by Ananya Tiwari 2 years ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Cute Boy 2 years ago

    CBSE > Class 10 > Mathematics
    • 2 answers

    Sourav Singh 1 year, 11 months ago

    Let x^2 =t then the given equation reduces to t^2+4t+6=0 D=b^2 −4ac D=(4)^2−4(6)(1) D=16−24=−8 D<0 Discriminant is less than zero,so the given equation has imaginary roots.  As all the 3 terms in this polynomial are all positive, this polynomial can never become zero for real values of x. Hence, the given polynomial has no real

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    Parv Jain 2 years ago

    Let x^2 =t then the given equation reduces to t^2+4t+6=0 D=b^2 −4ac D=(4)^2−4(6)(1) D=16−24=−8 D<0 Discriminant is less than zero,so the given equation has imaginary roots.  As all the 3 terms in this polynomial are all positive, this polynomial can never become zero for real values of x. Hence, the given polynomial has no real zeros.

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    • Posted by Kavya Singla 2 years ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Prateek Singh 2 years ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Shreya Agnihotri 2 years ago

    CBSE > Class 10 > Mathematics
    • 2 answers

    Yukti - 1 year, 11 months ago

    Prime factors=3²*5²*17

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    Shreya Agnihotri 2 years ago

    3×3×5×5×17

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    • Posted by Pawan Verma 2 years ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Pallavi Maravi 2 years ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Parv Jain 2 years ago

    16ⁿ = (2×2×2×2)ⁿ Prime factorisation of 16ⁿ does not contain the prime factor 5. Therefore, 16ⁿ cannot end with digit 0 for any natural number n

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    • Posted by Mayank 1234 2 years ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Mayank 1234 1 year, 11 months ago

    Hi

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    • Posted by Piyush Rajpurohit 2 years ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Shivam Kumar 2 years ago

    Let, assume that ✓17 is rational number. We will prove it by contradiction ✓=p/q, ( where p and q are coprime integer & q does not = 0) Squaring on both sides (✓17)2 = (p/q)2 17= p2/q2 P2 = 17 q2 < first equation Here 17 divide p and 17 divide p2 also P2/17= c , ( where c is any positive integer) P2 = 17c Squaring on both sides P2 = 289c2 < second equ From 1 and 2 17c = 289 c2 c = 17 c2 17 divide c and 17 divide c2 also We can observe that 17 is common factor of p and q This contradict the fact that p and q are coprime So, our assumption that✓17 is rational is wrong Hence, proved that ✓17 is irrational

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    • Posted by Abuzar Turkei 2 years ago

    CBSE > Class 10 > Mathematics
    • 0 answers
    ANSWER

    • Posted by Devershika Nayak 2 years ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Thari S 2 years ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Bhuvaneshwari Shivapur 2 years ago

    CBSE > Class 10 > Mathematics
    • 0 answers
    ANSWER

    • Posted by Mahi Singh ❤️‍🔥❤️‍🔥 2 years ago

    CBSE > Class 10 > Mathematics
    • 0 answers
    ANSWER

    • Posted by Dasani Ia I Sungoh 2 years ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Bhuvaneshwari Shivapur 2 years 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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    • Posted by Shz Khatri 2 years ago

    CBSE > Class 10 > Mathematics
    • 0 answers
    ANSWER

    • Posted by Bhuvaneshwari Shivapur 2 years ago

    CBSE > Class 10 > Mathematics
    • 3 answers

    Kaushik Krishna 1 year, 11 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

    2Thank You

    Sureksha Sp 2 years 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

    2Thank You

    Joshi Priya Dharmeshkumar 2 years 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.

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    • Posted by Damor Dhruv 2 years ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Rohit Kumar 2 years ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Vanshika Choudhary 2 years ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Divyanshu Oraon 2 years ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Pratha Sharma 2 years ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Khushi Kushwaha 2 years ago

    204 = 2 x 2 x 3 x 17

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