This is a magical post with units that make the formula work out

Our Math HL midterm is coming up, and to help each other study we each wrote a question, sent it to Mrs. Hessler and put it on a wiki that I created for the class. (By the way, in English class, we just finished a book called The Sailor Who Fell from Grace with the Sea.) Jordan submitted this as his question:

1)The sum of the forces acting on a boy as he is falling from grace to an ineluctable death at the pit of math HL despair provides that his terminal velocity is: v=K((5t^3+t^2+12t)/(2t^3)) where K is a magical constant equal to 1 with units that make the formula work out. Since failing out would cause him to lose his glory and transferring classes is not an option, find his terminal velocity as time becomes an endless motif.

2)The level of sanity for the average IA student is demonstrated by the following formula: s=f(I)=ln(I)tan(5I). Deduce when a student’s sanity does not exist. In other words, find the intervals for which the function is continuous.

Really. He did. He also submitted this as an answer:

1) (5/2) Most of the question is irrelevant. The only important parts in the limits of fractions are the highest order exponents in the numerator and the denominator. Since both have exponents of 3, the answer is simply the quotient of the coefficients (5/2). In general, if the numerator has a higher order exponent then the limit is infinity, if the denominator has a higher order than the limit is zero.

2) continuous on (0+(pi/10)k,(pi/10)+(pi/10)k) for all positive integers k Keep in mind that the domains of the natural log and the tangent function do not necessarily affect each other. The natural log function merely establishes that all x-values must be greater than zero, other than that it can simply be seen as a sort of coefficient in finding the limit. The only other time the function is undefined is at the vertical asymptotes of the tangent function. Since tan usually has undefined points on every (pi/2) interval and this function is shrunk to be a fifth is original period, the new interval is (pi/10)

What is actually the best part is both answers are completely right. The second one is stated a bit awkwardly, but after thinking about it for way too long I realized that my potential simplification, using pi/10 as the starting point and pi/5 as the interval length, fails from zero to pi/10. Graph it if you don't believe me.