Need help
Do you know the formula for the volume of such a cone? Find the volume, multiply by the density and you have the target scale reading...
\[V=\frac{1}{3}\pi r^{2} h\]
You'll have to work out the radius from the angle and the height
But I don't have the radius just the apex
Oh okay
Hint: that angle is 2x what you're going to use in your trig formula...
@whpalmer4 I got 6.4 as the radius
I don't think that's correct. What was your formula?
Did you use 40 degrees or 20 degrees as your angle?
radius = h sin (angle)
I used 40 degrees
No, that's the whole angle, and your triangle only uses half of the angle — look closely at the diagram...
I'm afraid that might not have been your only error, however.
Mkay
isn't it SOHCAHTOA Sin = Opposite over Hypotenuse Cos = Adjacent over Hypotenuse Tan = Opposite over Adjacent We don't know the hypotenuse here, so sin won't work. We do know adjacent, however, and we know the angle, so \[\tan (20 ^\circ) = \frac{r}{10}\] right?
yeah so the 20 degrees is because I'm only using half of the angle?
Exactly.
and the sneaky SOB who wrote the question stuck in the answer you would get if you made the mistake of using 40 degrees, so you wouldn't even realize your error until you got the graded problem back :-)
Oh okay lolz anyway this helps a lot thanks!
Do you have an answer?
I think it's C
How did you get that?
What was your new, improved value for the radius?
3
Show me your work to find the radius...
tan(20) is equal to 0.36 right?
Yep, approximately.
So wouldn't r = 3.6
Yes, r =3.6 What do you get for the volume of the cone with r = 3.6 and h = 10?
135.7
Yep. Now you know the volume of the cone in cm^3. What is the mass of the gold?
Interesting factoid: all the gold ever mined would make up a cube about 20 or 21 meters on a side...
Sorry my computer powered off for some reason >.<
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