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If possible,let √p be a rational number.also a and b is rational.then,√p = a/bon squaring both sides,we get,(√p)²= a²/b²→p = a²/b²→b² = a²/p [p divides a² so,p divides a]Let a= pr for some integer r→b² = (pr)²/p→b² = p²r²/p→b² = pr²→r² = b²/p [p divides b² so, p divides b]Thus p is a common factor of a and b.But this is a contradiction, since a and b have no common factor.This contradiction arises by assuming √p a rational number.Hence,√p is irrational

tyssm

3/4 is the answer of the following

hence the other number is 3/4

the answer is almost 3/4

mam polar form to last second step hoga na jisme COS +iSIN THETA... HOTA HAI

yes..I have mentioned the values for further reference.

let i-1/ cos(180/3)+isin(180/3)= i-1/ cos60+ isin60=i-1/1/2+V3/2i= i-1/1+V3i/2=2(i-1)/1+V3i; rationalizing =2(i-1)/1+V3i x 1- V3i/1-V3i =2[i(i-V3i)-1(1-V3i)/(1-V3i)(1-V3i) using (a-b)(a+b)=a^2-b^2 =2(-1+u+V3i-V3i(-1)/1-3(-1)= 2(-1+i + V3i+V3/1+3= 2((-1+V3)+(i+V3i))/4 =(-1+V3)+i(1+V3)/2= V3-1/2 + I v3+1/2; z=V3-1/2 + I V3+1/2

c) is the correct answer

c is correct answer of this question

Product of 2 number =x*-8=10X=10/-8=-1.25

-1.25 is the other number

-5/4 is the correct answer of the given question

It's always great to have someone that you're close to in order to go through life's woes with.

rationalising the denominator-16(1-i√3)/(1+i√3)(1-i√3)hence-16+16√3i/1+3hence-16/4+i4√3-4+4√3ihence in polar formfind thetahence theta is tan^-1(b/a)=tan^-1(-√3)hence theta is (-π/3)and r=√a^2+b^2r=√-4^2+(4√3)^2r=8hence in.polar form8(cos{-π/3)+isin(-π/3)

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x^2+x+1 > (x^2-1) and also > 2x+1

Cosθ= ((x^2-1)^2 + (2x+1)^2- (x2 + x + 1)^2)/ (2 (x2 – 1)(2x + 1)= ((x^2-1)^2 - (x2 + x + 1)^2)+ (2x+1)^2)/ (2 (x2 – 1)(2x + 1)= (2x^2+x)(-x-2)+ (2x+1)^2/(2 (x2 – 1)(2x + 1)=-x (2x+1)(x+2)+ (2x+1)^2/(2 (x2 – 1)(2x + 1)=(-x (x+2)+ (2x+1))/(2 (x2 – 1)=(-x^2 +1)/(2 (x2 – 1)= -1/2θ= π- π/3= 2π/3= 120 degdl

i havent got uh m sry can u explain this again...

k(kx+3y=k-23(12x+ky=k

k^2x+3ky=k^2-2k36x+3ky=3k

For no solution, lines must be parallelSo, k^2=36; k=6,-6and k^2-2k not equal to 3k

k^2-2k=3kk^2-5k=0k(k-5)=0k=0,5(no contradiction) So k should be 6,-6

nahi

a = 2/5 = 0.4 m/s²

Let , 7-√5 is a rational no. is equals to r, where r is a rational no.So,7-√ 5 = r 7-r = √5.Now,LHS is a rational but RHS is an irrational.

So, our assumption was wrong; 7-√5 is not a rational no. Therefore, 7-√5 is a irrational no. Hence, proved.

NP13 :(n+1)P 12 = 3 : 4

4( nP3) = 3 { (n+1) P 12)

4 ( n !/ (n-13) ! ) = 3 { (n+1)! / (n+1-12)!}4 n !/ (n-13) ! = 3 ( (n+1)!/(n-11)!)4.(n-11)! / (n-13)! = 3 (n+1)! / n!4. (n-11)(n-12) = 3(n+1)4 (n² - 23n + 132) = 3n + 34n² - 92n + 528 - 3n -3 = 04x² - 95n + 525 = 0(4n - 35)(n- 15)= 04n -35 = 0n = 35/4 n -15 =0n= 15

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Rs.520

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Let the number be x.x-12=50

x=50+12x=62

Therefore, the number is 62

let the number be x

x - 12% of x = 66

x - 12x/100 = 66

x - 3x/25 = 66

22x/25 = 66

x = (66 × 25)/22

x = 3 × 25

x = 75

The number is 75

x=-41+root(1681-230)/2 or -41-root(1681-230)/2=-41+root(1451)/2 or -41-root(1451)/2

You didn't send the full question! Please post it again neatly!

Actually its full Q only! From [Find to....(3,4)]. Next line contains another Q!

12=2×2×315=3×521=3×7HCF=1LCM=2×2×3×5×7=420 right ans

LCM i) 12, 15, and 21; 12/2=6/2=3;;15/3=5; 21/3=7; 12)15-12=3)12-12=0, 3x5=15 tuition , ; 17/1=17; 23/1/23; 29/1=29; (iii)8, 9,25, 8/2=4/2=2; 25/5=5; 9/3=3; (iv)72/2=36/2=18/2=9/3=3; 108/2=54/2=27/3=9/3=3; (v)306/2=153/3=51/3=17/1=17; 657/3=219/3=73;

b + (b+1) + ( b+2) --------------------------- = 21 4

3b + 3 = 4 × 21

b + 1 = 28

b = 28 - 1 = 27

1-chloro-4- methyl benzene

No. of stickers = 7635No. of schools = 16No. of stickers each school get = No. of stickers ÷ No. of school = 7635 ÷ 16 = 477 sticker and 3 are left

No of stickers distributed in school =7635No of schools=16 so we simply divide total no of stickers by total no of schools. so stickers would each school get is=7635/16=477(approx)stickers left =3

f=7s(f-2)=13(s-2)7s-2=13s-26so 6s=24 so s=4f=28