(2)/(x)-(3)/(4)=15;(8)/(x)+(5)/(4)=17

x+7-(8x)/(3)=(17)/(6)-(5x)/(2)

X+7-(8X)/(3)=(17)/(6)-(5X)/(2)

((x)/(3)-(2)/(5))/((3)/(4)-2x)=(16)/(15)

The area bounded by the region between curves y=x^(2)+2,x=0,x=3,y=x+1 is s(A)(15)/(2)(B)(15)/(4)(C)(15)/(3)(D)(17)/(2)

Area bounded by the curve y=x^3 , the x-axis and the ordinates x" "=" "" "2 and x" "=" "1 is (A) -9 (B) (-15)/4 (C) (15)/4 (D) (17)/4

Solve : {:((34)/(3x+4y)+(15)/(3x-2y)=5),((25)/(3x-2y)-(8.50)/(3x+4y)=4.5):}

The lines (x-4)/(15)=(y-17)/(0)=(z-11)/(8) and (x-15)/(4)=(y-9)/(17)=(z-8)/(11) intersect at the point P then square of the distance of P from the origin is

If x= (3)/(4.8) + (3.5)/(4.8.12)+ (3.5.7)/(4.812.16) + …., then 2x^2+5x=

The angle formed by the positive Y-axis and the tangent to y=x^(2)+4x-17 at ((5)/(2),-(3)/(4))