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  • 2 answers

Fateh Gupta 1 year 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 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 ago

a³+b³+c³+3(a+b)(b+c)(c+a). Right answer
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Harinie S 1 year ago

a+b+c+3a²b+3ab²
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  • 4 answers
6 is right answer on SQP but I don't know how to Solve it

Himanshi Thakur 1 year ago

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2 CM IS THE ANSWER
2 CM IS THE ANSWER
  • 1 answers
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
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 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 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
  • 3 answers
The distance between the points (0,2/5)and (-2/5,0)is
Sn = n/2 [2a + (n-1)d] Substituting values :- 0 = n/2 [2(50) + (n-1)(-4)] 0 = n/2 [100 - 4n + 4] 0 = n/2 [104 - 4n] 0 = n/2 × 4 [ 26 - n ] 0 = 2n [ 26 - n ] 0 = 26 - n - 26 = - n 26 = n. This should be the answer according to me.

Its Vicky 1 year ago

Sn = n/2 [2a + (n-1)d] Substituting values :- 0 = n/2 [2(50) + (n-1)(-4)] 0 = n/2 [100 - 4n + 4] 0 = n/2 [104 - 4n] 0 = n/2 × 4 [ 26 - n ] 0 = 2n [ 26 - n ] 0 = 26 - n - 26 = - n 26 = n. This should be the answer according to me.
  • 3 answers
(x-1) (x-2) (x-7)
(x-2)(x-7)(x-1)

Aditya Sinha 1 year ago

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  • 3 answers
LCM will be (x-2)(x-2)(x-7)

Sarthak Malviya 1 year ago

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Ritika Chauhan 1 year ago

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  • 2 answers
Solution - Sin30° = 1/2 , tan 45° = 1 , sec 30° = 2/√3 , cot45° = 1 = 1/2 + 1 ÷ √3/2 + 1 = 1+2 / 2 ÷ √3 + 2 / 2 = 3 / 2 × 2 / √3 + 2 = 3 / √3 + 2 = √3 × √3 / √3 + 2 = √3 / 2 = Sin 60° answer I hope this is helpful to you...~
1/2+1÷ 2/√3+1 = 1+2/3 ÷ 2+√3/2√3 = 3√3(2-√3)/2 is a answer
  • 3 answers

Sai Pansare 1 year ago

Let the ratio in which the line segment joining A(- 3, 10) and B(6, - 8) be divided by point C(- 1, 6) be k : 1. By Section formula, C(x, y) = [(mx₂ + nx₁) / m + n, (my₂ + ny₁) / m + n] m = k, n = 1 Therefore, - 1 = (6k - 3) / (k + 1) - k - 1 = 6k - 3 7k = 2 k = 2 / 7 Hence, the point C divides line segment AB in the ratio 2 : 7.

Raunak Singh 1 year ago

2:7

Parimal Katpatal 1 year ago

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  • 2 answers
TO PROVE:-(1+tan²A)/(1+cot²A) = sec²A-1 =>(1+tan²A)/(1+1/tan²A){ we know that cot²A = 1/tan²A } =>(1+ tan²A)/ {(tan²A + 1)/tan²A} =>{(tan²A)(1+ tan²A)}/ (tan²A+ 1) => tan²A => sec²A-1 { from the identity sec²A-tan²A = 1}

Shagun Sehra 1 year ago

TO PROVE:-(1+tan²A)/(1+cot²A) = sec²A-1 =>(1+tan²A)/(1+1/tan²A){ we know that cot²A = 1/tan²A } =>(1+ tan²A)/ {(tan²A + 1)/tan²A} =>{(tan²A)(1+ tan²A)}/ (tan²A+ 1) => tan²A => sec²A-1 { from the identity sec²A-tan²A = 1}
  • 1 answers
Because it has non terminating ana non repeating decimal expansion
  • 1 answers

Shagun Sehra 1 year ago

x = m + n y = m - n
  • 2 answers
0.4

Fjak Nka 1 year ago

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  • 1 answers

Raunak Singh 1 year ago

x= 10 or x= 21/2
  • 2 answers

Manoj Kumar 10 months, 2 weeks ago

(Tan theta)^2 - (sin theta)^2

Yug Purohit 1 year ago

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