" !7."x^(2)+(12)/(35)x+(1)/(35)

Factorize: x^(2)+(12)/(35)x+(1)/(35)

Factorize: x^2+(12)/(35)x+1/(35)

The sum and product of two number are 12 and 35 respectively,the sum of their reciprocals will be (12)/(35) b.(1)/(35) c.(35)/(8) d.(7)/(32)

7x+(3)/(x)=35(3)/(5)

x ^ (2) + 2x-35

int_(256)^( pi)sqrt(2)(1+cos x)^((7)/(2))dx=(A)(32)/(35)(B)(64)/(35)(C)(256)/(35)(D)(512)/(35)

Verify the following: (i) (-12)/(5)+(2)/(7)=(2)/(7)+(-12)/(5) (ii) (-5)/(8)+(-9)/(13)= (-9)/(13)+(-5)/(8) (iii) 3+(-7)/(12)=(-7)/(12)+3 (iv) (2)/(-7)+(12)/(-35)=(12)/(-35)+(2)/(-7)

Solve: (3x)/(7)-(4)/(35)=(2x)/(5)

If ((1-3x)^(1//2)+(1-x)^(5//3))/(sqrt(4-x)) is approximately equal to a+bx for small values of x , then (a,b) is (a) (1,35/24) (b) (1/-35/24) (c) (2,35/12) (d) (2,-35/12)

Lt_(x to oo)((1)/(3.5)+(1)/(5.7)+.....+(1)/((2n+1)(2n+3)))=