Calculus
posted by Rudy .
If the line 3x4y=0 is tangent in the first quadrant to the curve y=x^3+k, then k is
a: 1/2
b: 1/4
c: 0
d: 1/8
e: 1/2

if y = x^3 + k
dy/dx = 3x^2 , which of course is the slope of the tangent
but the slope of the given line is 3/4 , so ...
3x^2 = 3/4
x^2 = 1/4
x = ± 1/2 , but since we are restricted to the first quad,
x = 1/2
if x=1/2, 3(1/2)  4y = 0
y = 3/8
The point of contact of the tangent is (1/2 , 3/8)
so in the cubic ...
3/8 = (1/8) + k
k = 3/81/8 = 1/4 
Thank youu(: