azsun

Wait till CPhill comes. He's great at this.

How did you get 54/3=13.5?

Coach Grunt has to choose 3 players from the 10 players that are remaining after Ace and Zeppo have been placed in the lineup. The order in which the players are chosen doesn't matter, so the answer is \(\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \boxed{120}.\)

Completing the square gives us \((x - 5)^2 + (y + 3)^2 = 34 - c\) . Since we want the radius to be 1, we must have \(34 - c = 1^2\) . It follows that \(c = \boxed{33}\).

The sides of the rectangle are parallel to the axes, so the fourth point must make a vertical line with (5,11) and a horizontal one with (16,-2); this means that the fourth point is (5,-2). The graph of the region inside the equation is a circle with radius 3 and center (5,-2):

Since each angle of a rectangle is \(90^{\circ}\) and the corner coincides with the center of the circle, the rectangle covers exactly a quarter of the circle. The area of the intersection is thus \(\frac{1}{4}r^2\pi=\frac14\cdot3^2\pi=\boxed{\frac94\pi}\) .

Yay! I'm back! 

Let \(a\)  be the first term, and let \(d\)  be the common difference. Then \(a_n = a + (n - 1)d\) for all \(n\) . In particular, \(a_4 = a + 3d\)  and \(a_2 = a + d\) ,so \(\frac{a + 3d}{a + d} = 3.\) Multiplying both sides by \(a+d\) , we get \(a + 3d = 3a + 3d\),   so \(a=0\)  .

Then, \(\frac{a_5}{a_3} = \frac{a + 4d}{a + 2d} = \frac{4d}{2d} = \boxed{2}.\)

Solution: Let the first term be \(a\). Because the sum of the series is \(16\) , we have \(16= \frac{a}{1-(-1/5)} = \frac{a}{6/5} = \frac{5a}{6}\) . Therefore, \(a=\boxed{\frac{96}{5}}\) .

.If the n days have passed since Sunday, then the total number of cents in her bank account is \(1+2+\cdots+2^n\). This is a geometric series with first term 1, common ratio 2 and n+1 terms. Hence the sum is: \(1+2+\cdots+2^n = \frac{1-2^{n+1}}{1-2} = 2^{n+1}-1.\) If this is greater than 500  (i.e. if the total amount of money in the account is more than $5 ) then \(2^{n+1}-1\ge 500\) , so \(2^{n+1}\ge 501\). The smallest power of 2 that is greater than 501 is  \(2^9\). Thus the first time there is more than $5  in the bank account occurs after n=8 days. This is 8 days away from Sunday, so the day of the week is \(\boxed{\text{Monday}}\) .

For this problem, we make use of the correspondence between sums/products of roots and coefficients of a polynomial.

Denote the two roots of the equation \(\alpha\)  and \(\beta\) . We know that \(\alpha\beta = 48\) , and \(\alpha/\beta = 3 \implies \alpha = 3\beta\) .

So \( b = -\alpha - \beta = -4\beta\) . To maximize \(b\), we want to make \(\beta\) negative and as large as possible. Given the relationship that \(\alpha = 3\beta\) and that \(\alpha*\beta = 48\), we see that \(\beta=4\) or \(-4\) . Clearly \(-4\)  maximizes \(b\) , and \(b = \boxed{16}\) .

Setting \(y\) to zero, we find the following:

\(\begin{align*} 0& = -4.9t^2 + 42t + 18.9\\ & = -49t^2 + 420t + 189\\ & = 7t^2 - 60t - 27\\ & = (7t + 3)(t - 9) \end{align*}\)

As \(t\) must be positive, we can see that \(t = \boxed{9}.\)