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3x+2y=7--------(1)2x+3y=3---------(2)

(1)-(2)x-y=4

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sin(x+y)=sinxcosy +cosxsinysin(x-y) =sinxcosy-cosxsinyNow sin(x+y)/sin(x-y) = sinxcosy +cosxsiny / sinxcosy-cosxsinyDividing numerator and denominatorby cos x cosy and solving we gettanx+ tany /tanx -tany

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mujhe mathematic bahut kam ata hai

root(13 - a*root(10)) = root(8) + root(5)

13 - a*root(10) = (root(8) + root(5))^2

13 - a*root(10) = 8 + 5 + 2*2root(2*5)

13 - a*root(10) = 13 + 4root(10)

a = - 4

Total terms =37first term=a, common difference=d

3 middle terms are 18,19,20th termsa+17d+a+18d+a+19d=2253a+54d=225a+18d=75.....(1)

last three terms are 35,36,37th termsa+34d+a+35d+a+36d=4293a+105d=429a+35d=143.....(2)

Subtract equation 1 from 2

17d=68d=4a=75-4(35)a=75-140a=-65

AP is -65,-61,-57....

a+17d+a+18d+a+19d=2253a+54d=225a+18d=75…...1

a+34d+a+35d+a+36d=4293a+105d=429a+35d=143.....2

Subtract equation 1 from 217d=68d=4a=75-72=3

AP is 3,7,11,15...

5m=500cmand 3m=300cmso (500-50)cm=450cm, (300-50)cm=250cmso the new length and breadth are 4.5m and 2.5marea =l*b =4.5m*2.5m =1125m^2

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81 = 3×3×3×3 237 = 3×79

HCF = 3

So it can be written as (6(3)+2)/3 = 20/3.

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5/3V2 - 2V3 x 3V2 + 2V3/ 3V2 + 2V3 = 5(3V2 - 2V3)/(3V2)^2 - (2V3)^2 = 15V2 - 10V3/9(2)-(4×3) = 5(3V2 -2V3)/18-12 = 5(3V2 - 2V3)/6 = 5V2/3 -5V3/3

Let us assume, to the contrary, that 7-√5 is rational

That is, we can find Coprime a and b (b≠ 0) such that

7-√5 = a/b

Therefore, 7 - a/b = √5

Rearranging this equation √5 = (7b -a)/b

since a and b are integers,so (7b -a)/b is an rational.

And so √5 is rational

But this contradicts the fact that √5 is irrational.

This contradiction has arisen because of our incorrect assumption that 7-√5 is rational.

Therefore we can conclude that

7-√5 is irrational.

Let us assume, to the contrary, that 7-√5 is rational

That is, we can find Coprime a and b (b≠ 0) such that

7-√5 = a/b

Therefore, 7 - a/b = √5

Rearranging this equation √5 = (7b -a)/b

since a and b are integers,so (7b -a)/b is an rational.

And so √5 is rational

But this contradicts the fact that √5 is irrational.

This contradiction has arisen because of our incorrect assumption that 7-√5 is rational.

Therefore we can conclude that

7-√5 is irrational.

Given series is 3, 8, 13, ...... 253 here, first term , a = 3 common difference , d = 5 Let us find total number of terms at first.use tn = a + (n-1)d=> 253 = 3 + (n - 1)5 => 250 = 5(n - 1) => n - 1 = 50 => n = 51 so, there are 51 terms in given series.now we know, mth term from last =last term - mth term + 1 so, 20th term from last = 51 - 20 + 1 = 32 hence, 20th from last = 32th term from first

use , tn = a+ (32-1)d= 3 + 31 × 5 = 3 + 155 = 158

hence , 20th term from last = 158

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2⁵⁵ +2⁵⁴ +2⁵³=2⁵³(4+1+1)=2⁵³×6=2⁵³×2×3=2⁵⁴×3

a=3d=8-3=5n=20an=a+(n-1)da20=3+19*5=3+95=98

don't cheat another's answers

your is wrong1.309 is the answer

give your reply

the answer is right. checked in calculator

By trail and error, x= 3 satisfies it.divide the polynomial by x-3You will get 9x²-100 = 3x+10 * 3x-10So factors are (x-3)(3x+10)(3x-10)

We know √5 is irrational so

Let 7 - √5 be rational

So, 7 - √5 = r, where r is rational

7 - r = √5

This is absurd as LHS is rational and RHS is irrational so a a contradiction to our assumption.Hence 7 - √5 is irrational

c is right answer

Let us assume, to the contrary, that 7-√5 is rational

That is, we can find Coprime a and b (b≠ 0) such that

7-√5 = a/b

Therefore, 7 - a/b = √5

Rearranging this equation √5 = (7b -a)/b

since a and b are integers,so (7b -a)/b is an rational.

And so √5 is rational

But this contradicts the fact that √5 is irrational.

This contradiction has arisen because of our incorrect assumption that 7-√5 is rational.

Therefore we can conclude that

7-√5 is irrational.

Let , 7-root5 is a rational no. is equals to r, where r is a rational no.So,7-root 5 = r 7-r = root 5.Now,LHS is a rational but RHS is an irrational.So, our assumption was wrong; 7-root5 is not a rational no. Therefore, 7-root5 is a irrational no. Hence, proved.

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100 ( x - 3) = 1000

x - 3 = 1000/100

x - 3 = 10

x = 13

50 is the correct answer of the given question

50 is right answer its right answer

5*10=50 is the correct answer

5×10_____ 50your answer is 50

a^2-b^2a^2=(2+√5/2-√5)^2=4+5+4√5/4-+5-4√5=9+4√5/9-4√5b^2=(2-√5/2+√5)^2=4+5-4√5/4+5+4√5=9-4√5/9+4√5hence subtracting9+4√5/9-4√5-(9-4√5/9+4√5)(9+4√5)^2-(9-4√5)^2/81-80=81+80+81+80=160+162=322

A=3630

T=2yrs

R=10%

A=p(1+r/100)^t

3630=p(1+10/100)^2

3630=p(11/10)^2

3630=121p/100

P=3630*100/121

P=3000

X=k/y, k is constantk=y×6010×6=6015 ×4 =6020 ×3 =601.5×30=450.6×100=691.5 and 30 is not the possible pair as it's constant is 45 and all others are 60

nikunj

We know

Root 2 = 1.414Root 3 = 1.732

Now, we can write n number of rational number between these.

That is just greater than 1.414 and less than 1.732 and it should be terminating or not terminating but repeating.

For example1.415659756, 1.416893, 1.715644, ...

hope it helps you

Sum of 1 to 100.

Its an AP series, sum of n terms of APSn = n/2(a1 + an)

Given, a1 = 1, an = 100, n = 100

Then S = 100/2(1 + 100) = 50(101) = 5050

Therefore sum of numbers from 1 to 100 is 5050