Let y=safe load in pounds and x= depth in inches for a certain type of rectangular horizontal beam. A constant of proportionality k exists such that y=kx^2.

What kind of variation is this relationship?

Question

Let y=safe load in pounds and x= depth in inches for a certain type of rectangular horizontal beam. A constant of proportionality k exists such that y=kx^2.

What kind of variation is this relationship?

anonymous · Answer

exponentially proportional?

anonymous · Answer

Huh?

anonymous · Answer

A: Direct 
B: Inverse
C: Joint
D: Combined
E: None

anonymous · Answer

I will have to go for direct, I think.

anonymous · Answer

Ok. & the next question ask to determine the constant K for a beam with y=800lbs, and x=4 inches.

anonymous · Answer

Well, that seems straightforward, just plug the numbers into the formula

800 = k * 4^2  -> k = 800/16 or 50

anonymous · Answer

The next one says: what is a safe laod if x=6 inches. So it would be 800=k*6^2 right?

anonymous · Answer

Um, no...y is the safe load.

anonymous · Answer

Then how would i do it?

anonymous · Answer

Well, u know what x and k are, so you can calculate y.

anonymous · Answer

x= 6 k=50 so   50(6^2) so 50*36 = 1800lbs?

anonymous · Answer

Yup, the whole idea of a constant of proportionality is that it is constant for whatever x and y.

anonymous · Answer

Oh okay. i get it. So the last question ask, which of the conic sections correctly describes this relationship for x>0?
circle, hyperbola, ellipse, or half parabola?