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  • 2 answers

Arshpreet Anand 1 year, 7 months ago

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Arshpreet Anand 1 year, 7 months ago

You are late Should've started when you were one
  • 4 answers

D.Manoj Kumar 1 year, 7 months ago

It is not a Quadratic equation but then the value of 'x'will be 3/7

Khushpreet Kaur 1 year, 7 months ago

2 + 3 x + 1 = ( x − 2 ) 2 x 2 + 3 x + 1 = x 2 − 4 x + 4 ⇒ x 2 + 3 x + 1 − x 2 + 4 x − 4 = 0 ⇒ 7 x − 3 = 0 has its highest power of variable as 1 Hence it is not a quadratic equation

Khushpreet Kaur 1 year, 7 months ago

by using identity (a-b)²=a²-2ab+b² =X²–2(x)(2)+2² =x²–4x+4=x²+3x+1 By applying hit and trial method (1)²-4(1)+4=0 0=0 x-1 is zero of polynomial

Khushpreet Kaur 1 year, 7 months ago

by using identity (a-b)²=a²-2ab+b² =X²–2(x)(2)+2² =x²–4x+4 By splitting the mid term =x²-2x-2x+4 =x(x-2)-2(x-2) (x-2)(x-2)=
  • 1 answers

Arush Saraswat 1 year, 7 months ago

Assume that .............................. That is answer find it
  • 2 answers

Not Fine Bro 🎭 1 year, 7 months ago

Let a and b irrational number where a and b are odd numbers....

Shreyansh Singh 1 year, 7 months ago

Hii sir
  • 4 answers

Not Fine Bro 🎭 1 year, 7 months ago

It will be very tough!! Because I know But I think you will pass by scoring around 30-40 marks

D.Manoj Kumar 1 year, 7 months ago

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Hari Shankar 1 year, 7 months ago

IT MAY VE EASY OR TOUGH DOESNT MATTER THE MATTER IS HOW U PREPARE FOR EXAMS

Srikanth Marre 1 year, 7 months ago

Find the sum of exponents of prime factor in the prime factorisation
  • 1 answers

Not Fine Bro 🎭 1 year, 7 months ago

This type of questions will not be asked in board exam
  • 3 answers

Vandit Sharma 1 year, 7 months ago

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Bhargav Kulkarni 1 year, 7 months ago

We know that , 1+tan2Φ=sec2Φ 1+sin2Φ/cos2Φ=1/cos2Φ Multiply cos2Φ on both sides We get cos2Φ+sin2Φ=1 sin2Φ+cos2Φ=1

Indrajeet Bera 1 year, 7 months ago

Let a, b, c be lengths of right angled triangle By definition sinθ=b/c(opposite sidehypotenuse) cosθ=a/c(adjacent sidehypotenuse) sin2θ+cos2θ=b2c2+a2c2=a2+b2c2 From Pythagoras theorem c2=a2+b2 ∴a2+b2c2=1 sin2θ+cos2θ=1 Hence, proved.
  • 1 answers

Dikha Sarkar 1 year, 7 months ago

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  • 1 answers

Sachin Prajapati 1 year, 7 months ago

It depends on its highest power like If the power is- 2-then there are two zeroes 3-then there are three zero And so on.
  • 1 answers

Arun Choudhary 1 year, 7 months ago

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  • 2 answers

Naincy Agarwal 1 year, 7 months ago

sol a2=a+d=14 ➡️a=14-d. i a+2d=18 ➡️14-d+2d=18. From i ➡️d=4 Now put d=4 in I a=14-4 a=10 Now to find sum of first 51 term Sn=n/2(2a+{n-1}d) S51=51/2(2×10+{51-1}×4) S51=51/2(20+50×4) S51=51/2(20+220) S51=51/2×240 S51=51×120 S51=6120 So sum of first 51 term of an AP will be 6120 where a=10 and d=4

Rudra King 1 year, 7 months ago

5610
  • 1 answers

Satyadeep Yadav 1 year, 7 months ago

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  • 1 answers

Zeel Patel 1 year, 7 months ago

√3/2 x √3/2 + 1/2 x 1/2 3/4 + 1/4 4/4 1
  • 1 answers

Manya Yadav 1 year, 7 months ago

Firt let 2root3 + root 5 is rational no. Then written in the form of p/q where p and q are co prime no. Then equalise The both side
  • 1 answers

Tanya Patel 1 year, 7 months ago

The perimeter of each semicircle will be 44 cm.
  • 1 answers

Khushi Tiwari 1 year, 8 months ago

Let the radius of smaller circle be (r) 2cm And, the radius of bigger circle be(R), 5cm Now, Area of circle is πr^2 So, πR^2 -πr^2 = π(R^2 - r^2) =22/7[(5)^2 - (2)^2] =22/7[25 - 9] = 22/7 ×16 = 3.14 × 16 = 50.24cm^2
  • 0 answers
  • 1 answers

Lionel Messi 1 year, 7 months ago

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  • 2 answers

Manya Yadav 1 year, 7 months ago

X = 1 and y= -2

Tarnika S 1 year, 7 months ago

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  • 0 answers
  • 1 answers

Nitin Kumar 1 year, 8 months ago

3x-⁵/3 then multiply by 3, 9x-5 9x=5 x=5/9
  • 2 answers

Arpan Das 1 year, 8 months ago

17×19×23

Nitin Kumar 1 year, 8 months ago

17*19*23
  • 2 answers

Fateh Gupta 1 year, 8 months ago

We have to prove that √5 is an irrational number It can be proved using the contradiction method Assuming √5 as a rational number, i.e., can be written in the form a/b where a and b are integers with no common factors other than 1 and b is not equal to zero. √5/1 = a/b √5b = a By squaring on both sides (5b)2 = (a)2 b2 = a2/5 …. (1) It means that 5 divides a2. It means that it also divides a a/5 = c a = 5c By squaring on both sides a2 = 25c2 Substituting the value of a2 in equation (1) 5b2 = 25c2 b2 = 5c2 b2/5 = c2 As b2 is divisible by 5, b is also divisible by 5 a and b have a common factor as 5 It contradicts the fact that a and b are coprime This has arisen due to the incorrect assumption as √5 is a rational number. Therefore, √5 is irrational.

Akeet Deep 1 year, 8 months ago

We have to prove that √5 is an irrational number It can be proved using the contradiction method Assuming √5 as a rational number, i.e., can be written in the form a/b where a and b are integers with no common factors other than 1 and b is not equal to zero. √5/1 = a/b √5b = a By squaring on both sides (5b)2 = (a)2 b2 = a2/5 …. (1) It means that 5 divides a2. It means that it also divides a a/5 = c a = 5c By squaring on both sides a2 = 25c2 Substituting the value of a2 in equation (1) 5b2 = 25c2 b2 = 5c2 b2/5 = c2 As b2 is divisible by 5, b is also divisible by 5 a and b have a common factor as 5 It contradicts the fact that a and b are coprime This has arisen due to the incorrect assumption as √5 is a rational number. Therefore, √5 is irrational.
  • 4 answers

Nitin Kumar 1 year, 8 months ago

a³+b³+c³+3(a+b)(b+c)(c+a). Right answer

Dashrath Singh Sisodiya 1 year, 8 months ago

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Harinie S 1 year, 8 months ago

a+b+c+3a²b+3ab²

Brajmohan Brajmohan 1 year, 8 months ago

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  • 4 answers

Himanshi Thakur 1 year, 8 months ago

6 is right answer on SQP but I don't know how to Solve it

Himanshi Thakur 1 year, 8 months ago

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Pramith Fernandes A 1 year, 8 months ago

2 CM IS THE ANSWER

Pramith Fernandes A 1 year, 8 months ago

2 CM IS THE ANSWER
  • 1 answers

Himanshi Thakur 1 year, 8 months ago

Let √5 be a rational number. then it must be in form of pq where, q≠0 ( p and q are co-prime) √5=pq √5×q=p Suaring on both sides, 5q2=p2 --------------(1) p2 is divisible by 5. So, p is divisible by 5. p=5c Suaring on both sides, p2=25c2 --------------(2) Put p2 in eqn.(1) 5q2=25(c)2 q2=5c2 So, q is divisible by 5. . Thus p and q have a common factor of 5. So, there is a contradiction as per our assumption. We have assumed p and q are co-prime but here they a common factor of 5. The above statement contradicts our assumption. Therefore, √5 is an irrational number.
  • 3 answers

Alfiya💫💫💫 . 1 year, 8 months ago

Yes, it is true as you can check it by taking 1&2 or 3&4 or any other pair 1x2=2. ( divisible by 2) 3x4=12(it is also divisible by 2).. Hence the statement is true. Clear💫💫💫..

Bhakti Mehta 1 year, 8 months ago

It is true. Because in any two consecutive positive integers, one will always by divisible by two.For example, 17 and 18 are consecutive numbers;here 18 is divisible by 2 and 17 × 18 equals 306 which is divisible by 2.

Sai Pansare 1 year, 8 months ago

No, As in the two consecutive no. One of them is going to be even and the other would odd so the sum a odd no. and a even no. is always odd which is not divisible by 2

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