1. Average velocity.
It is at $t=3\sqrt3$. The total distance travelled is

$$d=\int_0^93t^2dt=t^3\vert _0^9=9^3,$$ so the average velocity is 9²=81. Its velocity is 81 when 3t²=81; i.e., when $t=\sqrt{27}=3\sqrt3$.
 

2. Smoking and heart disease.
Let the total number of people be 1000x. Then 20x  have heart disease, and of these, 12x  are smokers. There are 980x  with no heart disease, of which 196x  are smokers. Thus 12x  of the 208x  smokers have heart disease, and the fraction called for is  12/208=3/52.
 
 

3. Too large a rectangle.
If the rectangle does not extend above the curve, then it is contained in some rectangle inscribed in the region between the curve and the x-axis with base on the x-axis, so it suffices to show that the maximum area of such a rectangle is less than 257. It is, in fact, 256, as we show. Let (x, 80-x⁴) be the upper right corner of a rectangle inscribed in the region described. Then the area of the rectangle is at most

$$A(x)=2x(80-x^4)=160x-2x^5,\quad 0\le x\le\root4 \of{80}.$$ Since $A(0)=A(\root4\of{80})=0$ and A(x) is positive for $0<x<\root4\of{80}$, the maximum value of A(x) on this closed interval occurs at an interior point, where $A^{\prime}(x)=0$. Now $A^{\prime}(x)=160-10x^4=10(16-x^4)=0$ only at x=2 in this interval. Therefore the maximum area is A(2)=(4)(64)=256.

Second solution, courtesy of Charles McCarthy (U.MN-TC)
It suffices to show that 257/x > 80-x⁴  for all x>0; equivalently, that

$$f(x)\colon = 257-80x+x^5>0\quad\hbox{ for } x>0.$$

If $x\ge3$, then $-80x+x^5\ge -80x+3^4x\ge3$ so $f(x)\ge260$. If 0<x<3, then $f(x)\ge 257-80\cdot3=17$.
 
 

4. Limit of a fraction.
The limit is $\sin9$. If $f(x)=\int_1^xg(t)dt$ and g is continuous, then by the Fundamental Theorem of Calculus, $f^{\prime}(x)=g(x)$. In particular here, if $f(x)=\int_1^x\sin(t^2)dt$, then we have $f^{\prime}(x)=\sin(x^2)$. Then

$$\lim_{h\to0}{\int_1^{3+h}\sin(t^2)dt-\int_1^3\sin(t^2)dt\over h}=\lim_{h\to0}{f(3+h)-f(3)\over h}=f^{\prime}(3),$$

and $f^{\prime}(3)=\sin(3^2)=\sin9$.

Second solution
If we call the numerator F(h), then $F^{\prime}(h)=\sin(3+h)^2$. By L'Hôpital's Rule, the limit of the fraction is

$${\lim_{h\to0} F^{\prime}(h)\over1}=\lim_{h\to0}{\sin(3+h)^2\over1}=\sin9.$$

5. Evenly spaced roots.
The only such k is ${7\over4}$. Let f(x)=(x²-1)(x²-4)-k. Since  f(-x)=f(x), the roots are of the form  -b, -a, a, b, with b>a>0. For these to be equally spaced we need  b-a=a-(-a), so b=3a. Thus,

f(x)=(x-a)(x+a)(x-3a)(x+3a);

i.e., x⁴-5x²+(4-k)=(x²-a²)(x²-9a²)=x⁴-10a²x²+9a⁴.

By comparing coefficients of x² we see that a²=1/2, and by comparing constant terms, that  4-k=9a⁴=9/4, so that  k=4-9/4=7/4.
 
 

6. Final ones.
Yes. We prove by induction that for every positive integer m there is an integer n ending in 1 for which n³ ends in m ones. With m=1 we may use 1³=1. Suppose that n is an integer ending in 1 for which n³ ends in m ones; say

$$n^3=r\cdot10^m+11\cdots1,$$

where there are m final ones. Then if k is any positive integer we have

$$\null\,\vcenter{\openup1\jot \ialign{\strut\hfil$\displayst......^2+n^3\cr&=A\cdot10^{2m}+(3kn^2+r)10^m+11\cdots1,\cr\crcr}}\,$$

where A is a positive integer and there are m ones in the final group. This number will end in m+1 ones provided that 3kn²+r ends in 1. Since n ends in 1, 3kn² has the same final digit as 3k, so it suffices that 3k+r end in 1. For every positive integer r such a k exists, since 3 has an inverse modulo 10, ($3\cdot7\equiv1\pmod{10})$, and the proof is complete.
 
 

7. Simplify this sum.
The sum is ${n(n+1)\over2}$.

8. Trigonometric inequality. (Solution by Ron Rietz, Gustavus Adolphus College)

By the AM,GM inequality,

$$\null\,\vcenter{\openup1\jot \ialign{\strut\hfil$\displayst......]\cr&={1\over2}[4-(1-\cos x)^2-(1-\cos y)^2]\le2.\cr\crcr}}\,$$

Second solution, courtesy of Charles McCarthy
$\cos x+\cos y+\sin x\sin y=(\cos x,1,\sin x)\cdot(1,\cos y,\sin y)\le\vert(\cos x,1,\sin x)\vert\vert(1,\cos y,\sin y)\vert=\sqrt2\sqrt2=2$.
 
 

9. Perimeter 6 and integral area.
There is just one such triangle. Its legs are $a=(5+\sqrt{7})/3$and, and its area is ab/2=1. Here is a proof. For a right triangle with legs a and b to have perimeter 6 and integral area it is necessary and sufficient that

$$a+b+\sqrt{a^2+b^2}=6\eqno(1)$$ and $$ab=2n,\eqno(2)$$

where n is a positive integer. From $\sqrt{a^2+b^2}=6-(a+b)$ we have

a²+b²=36-12(a+b)+a²+2ab+b²,

which simplifies to

$$3(a+b)=9+n.\eqno(3)$$

There are various ways to obtain an upper bound on n. E.g., 3<a+b<4because the two legs must account for more than half and less than 2/3 of the perimeter. Then (3) yields 9<9+n<12, so n=1 or 2. But if n=2 then $a+b={11\over3}$, and substituting $b={11\over3}-a$ into (1) yields only nonreal values for a. Thus n=1 is the only possibility. Then from (3) and (2) we have a+b=10/3 and ab=2, which leads to $a+{2\over a}=10/3.$  The roots are . When  then ,  so both yield the same triangle.
 
 

10. Bigger than 2000 for large n. Solution by Macalester Team 4.
Any integer  n>e²⁰⁰⁰ suffices. For,

$$\sum_{k=n}^{n^2}{1\over k}=\int_n^{n^2+1}{1\over\lfloor x\rfloor}dx>\int_n^{n^2}{1\over x}dx=\ln n,$$

and $\ln n>2000$ when n>e²⁰⁰⁰.

Second solution, by Carleton A team
We claim that  n=2⁴⁰⁰⁰ suffices. With this value of n, (omitting the term ${1\over n}$),

$$\null\,\vcenter{\openup1\jot \ialign{\strut\hfil$\displayst......ver2}+{1\over2}+\cdots+{1\over2}={4000\over2}=2000.\cr\crcr}}\,$$