14*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)

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

3-(1)/(2)of(4)/(7)+(5)/(7)xx(14)/(35)=

x ^ (2) + 2x-35

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)

((-16)/35)-:((-15)/14)

log_((1)/(4))((35-x^(2))/(x))>=-(1)/(2)

If the sum of the coefficient in the expansion of (alpha x^(2) - 2x + 1)^(35) is equal to the sum of the coefficient of the expansion of (x - alpha y)^(35) , then alpha =