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2

Let age of A=xLet age of B=y

x=y+5x-y=5or -4x+4y=-20

3(x-7)=4(y-7)3x-4y=-7

Adding the two equations,-x=-27x=27y=22

let age of P Q and R be x y and zthen 5 years agoage of P=x-5age of Q=y-5

average=15yrs(x-5)+(y-5) /2=15x-5+y-5=30x+y=40........(1)

x+y+z /3=20x+y+z=60from eqⁿ1 put the value of x+y40+z=60z=20yrs

age of R=20yrsafter 10 years age of R=z+10=30yrs

It lies between1/2 and 1/1

270 - 270×15/100 = 270 - 40.5 = 229.5

Answer :

Given : First four digit of ten digit mobile number = 9979

We make the smallest number by using only one digit twice from 8 , 3 , 5 , 6 and 0 , As :

=9979003568 ( Ans )

1+cot²a+2cota+ 1+cot²a-2cota2+2cot²2(1+cot²a)2cosec²a

hope it helps

tq but I have 1 doubt .here 5 not come with root 45 and root 15

sorry answer is right. iam only confused

thanks for your answer

P+Q=R+25 & R-Q=5SO,P=R+25-QP=5+25 P=30 years

thanks

Thanks

thanks

multiplying(2a+a)=3ahence3a^2*3a^2=9a^4

not understand

-16/24= -8/12= -2/3thanks

2√3+3√12-√30/12

Please like the solution 👍 ✔️

Thnks

Let denominator of fraction be x then numerator = x - 1

As per given condition(x - 1 + 3)/(x + 3) =(x - 1)/x + 3/28

(x + 2)/(x + 3) - (x-1)/x = 3/28

28[(x + 2)x - (x + 3)(x - 1)]= 3x(x + 3)

28(x^2 + 2x - x^2 + x - 3x +3)= 3x^2 + 9x

3x^2 + 9x - 84 = 0

3x^2 + 21x - 12x - 84 = 0

3x(x + 7) - 12(x + 7) = 0

(3x - 12)(x + 7) = 0

x = 4, - 7

Therefore, For x = 4, fraction = 3/4For x = - 7 fraction = - 8/-7

3/4, -8/-7

5A + 34----------- B2

For given additionValue of A = 8Value of B = 9

between 348 and 349348×10=3480349×10=34903480<3486<3490

3486 lies between 3480 and 3490

between 348 and 349348×10 =3480349×10=34903480<3486<3490

As we knowsum of all angles will be 180°so here90+30+ A= 180A= 180-120= 60°

90+30+A=180A=180-120A=60

A triangle have 180 degree. So angle A = 180-90-30 ie 60 degree

180=90+30+?so,90+30=120then,180~120=60.....and some of all triangles is always 180.......

Hence, the HCF (GCF or GCD) of 4052 and 12576 is 4.

1 Can't understand

no. of boxes is 14Each box have 28 cookiestherefore 14 boxes have 14×28= 392 cookies.

(3a² + 5ab + 7b²) × (7a + 2b)

21a³ + 6a²b + 35a²b + 10ab² + 49ab²+14b³

1st question answer is 4 and 2nd question answer is 1

oops so sorry my answer is not write because I have seen the number 8 is number 6

thank

this is correct answer and sorry for 1st my answer

[-23,5) and (6,12] here option 2 is less than option 1

sum of two negative numbers is all ways negative

thnkewww

2,4,6,8,10,.... a = 2 d = 4 - 2 = 2 Sn = 210 Sn = n/2(2a+(n-1)d) 210 = n/2(2×2+(n-1)2) 210 = n(2+n-1) 210 = n*n + n n*n + n - 210 = 0 (n+15)(n-14) = 0 n = 14 14 terms required.

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24.

prime factorization of 120= 2 × 2 × 2 × 3 × 5prime factorization of 144=2×2×2×2×3×3HCF=2×2×2×3=24