Melody

As you know, they are both right angled triangles.

For 45 degree ratios use a right angles isosceles triangle where the 2 equal sides are both 1 unit.

  Find the hypotenuse using pythagoras's theorem and then read off the ratio values from it.

 For both 60degrees and 30degrees.

Start with a equilateral triangle with side length 2 units.

Then cut it in half to get two congruent  triangles, you will only use one of them.

  it has a right angle, a 30degree angle and a 60 degree angle,

  the hypotenuse is 2 units and one side is 1 unit.

   Use pyth to find the 3rd side and that is it.

I get the same as Chris

One machine added  11 scoops      the other machine took 38 scoops

Do you know what the correct answer is?  

If you do you should share it with us.  

Thanks Alan,

I did think there might be a higher answer than mine.

I guess I should have said so   :)

I also should have thought to use calculus. That never even crossed my mind  

\(\frac {1}{x} + \frac {1}{y} = \frac {1}{12}\\~\\ \frac{1}{x}<\frac{1}{12}\qquad \qquad \frac{1}{y}<\frac{1}{12} \\ x>12\qquad \qquad y>12 \\ so\\ x+y>24 \)

That is not correct Nickolas.

Why did you post the same question twice!     How annoying!

1.    2 0[1/((sqrt1+x^3)) dx

\(\displaystyle\int_0^2\;\frac{1}{\sqrt1+x^3} dx\)

This is what you asked for, I suspect it is not what you really want.  Be careful with your brackets.

Is this what you want?

Integral from 0 to 2 of    [1/(sqrt[1+x^3]) dx

\(\displaystyle\int_0^2\;\frac{1}{\sqrt{1+x^3}} \;dx\)

and  You vote me down for answering your questions.

HOW  RUDE!