Solution:

Factoring out $2x+3$ gives $(2x+3)\big[(x-4)+(x-6)\big] = (2x+3)(2x-10) = 0.$

The roots are $-\dfrac{3}{2}$ and $5,$ whose sum is $-\dfrac{3}{2} + 5 = \dfrac{7}{2}.$

Thus, the correct answer is A.

Solution:

Let $x$ be the number. Cindy computed $\dfrac{x-9}{3} = 43,$ so $x - 9 = 129$ and $x = 138.$

The correct computation gives $\dfrac{138-3}{9} = \dfrac{135}{9} = 15.$

Thus, the correct answer is A.

Solution:

The five parenthesizations of $2^{2^{2^2}}$ give $$\left(\left(2^2\right)^2\right)^2 = 2^8,\quad \left(2^{2^2}\right)^2 = 2^8,\quad \left(2^2\right)^{2^2} = 2^8,$$ $$2^{\left(2^2\right)^2} = 2^{16},\quad 2^{2^{2^2}} = 2^{16}.$$

So the only values are $2^{16} = 65{,}536$ and $2^8 = 256.$ Besides the standard value there is exactly $1$ other.

Thus, the correct answer is B.

Solution:

Let the angle be $x.$ Then $90 - x = \dfrac14(180 - x),$ so $360 - 4x = 180 - x.$

This gives $3x = 180,$ so $x = 60.$

Thus, the correct answer is B.

Solution:

Each of the six outer unit circles is tangent to the central unit circle, so its center is $2$ units from the center. Adding one more radius, the large circle has radius $3$ and area $9\pi.$

The seven unit circles have total area $7\pi,$ so the shaded region has area $9\pi - 7\pi = 2\pi.$

Thus, the correct answer is C.

Solution:

Taking $n = 1,$ the inequality becomes $m \le m + 1,$ which holds for every positive integer $m.$

So every positive integer $m$ works, and there are infinitely many.

Thus, the correct answer is E.

Solution:

Equal arc lengths give $\dfrac{45}{360}\cdot 2\pi R_A = \dfrac{30}{360}\cdot 2\pi R_B,$ so $\dfrac{R_A}{R_B} = \dfrac{30}{45} = \dfrac{2}{3}.$

The ratio of areas is $\left(\dfrac{R_A}{R_B}\right)^2 = \dfrac{4}{9}.$

Thus, the correct answer is A.

Solution:

Drawing the diagonals shown, the entire flag is tiled by congruent triangles. Counting, there are $24$ triangles in the blue region, $24$ in the white region, and $16$ in the red square.

Since the blue and white regions contain the same number of triangles, $B = W.$

Thus, the correct answer is A.

Solution:

The files need $3(0.8)+12(0.7)+15(0.4) = 16.8$ mb, so at least $\dfrac{16.8}{1.44} = 11\tfrac{2}{3}$ disks by volume alone.

A disk containing a $0.8$-mb file has room for only one more $0.4$-mb file, leaving at least $0.24$ mb unused. Across the three $0.8$-mb files this wastes at least $3(0.24) = 0.72$ mb, over half a disk, forcing at least $13$ disks.

Thirteen suffice: six disks each hold two $0.7$-mb files, three disks each hold one $0.8$-mb file plus one $0.4$-mb file, and four disks each hold three $0.4$-mb files.

Thus, the correct answer is B.

Solution:

After transferring half the coffee, cup $1$ has $2$ oz coffee and cup $2$ has $2$ oz coffee and $4$ oz cream, a total of $6$ oz.

Transferring half of cup $2$ back moves $1$ oz coffee and $2$ oz cream. Cup $1$ then holds $2 + 1 = 3$ oz coffee and $2$ oz cream.

The fraction that is cream is $\dfrac{2}{2+3} = \dfrac{2}{5}.$

Thus, the correct answer is D.

Solution:

Let $t$ be the time in hours to arrive on time. Since three minutes is $0.05$ hours, $40(t+0.05) = 60(t-0.05).$

This gives $40t + 2 = 60t - 3,$ so $t = 0.25.$ The distance is $40(0.25+0.05) = 12$ miles, and the required speed is $\dfrac{12}{0.25} = 48$ miles per hour.

Thus, the correct answer is B.

Solution:

If the roots are primes $p$ and $q,$ then by Vieta's formulas $p + q = 63$ and $pq = k.$

Since $63$ is odd, one prime must be even, namely $2,$ and the other is $61.$ Both are prime, so $k = 2\cdot 61 = 122$ is the only possible value.

Thus, the correct answer is B.

Solution:

A positive number $x$ differs from its reciprocal by $1$ when $x - \dfrac1x = 1$ or $x - \dfrac1x = -1,$ i.e. $x^2 - x - 1 = 0$ or $x^2 + x - 1 = 0.$

The positive roots are $\dfrac{1+\sqrt5}{2}$ and $\dfrac{-1+\sqrt5}{2},$ which are reciprocals of each other. Their sum is $a + b = \sqrt5.$

Thus, the correct answer is C.

Solution:

Using $\log a^2 = 2\log a$ and adding logs, $$N = \log_{2002} 11^2 + \log_{2002} 13^2 + \log_{2002} 14^2 = \log_{2002}(11\cdot 13\cdot 14)^2.$$

Since $11\cdot 13\cdot 14 = 2002,$ this is $\log_{2002} 2002^2 = 2.$

Thus, the correct answer is D.

Solution:

The collection $6, 6, 6, 8, 8, 8, 8, 14$ has mean, median, unique mode, and range all equal to $8,$ so $14$ is attainable.

Suppose the largest were $15.$ The range $8$ forces the smallest to be $7,$ and the median $8$ fixes the two middle values as $8, 8.$ Then $7 + 8 + 8 + 15 = 38,$ so the remaining four values sum to $64 - 38 = 26,$ averaging $6.5.$ At least one would be below $7,$ contradicting the minimum. So $15$ is impossible.

Thus, the correct answer is D.

Solution:

Tina's ten pairs have sums $3, 4, 5, 5, 6, 6, 7, 7, 8, 9.$ For a sum $s,$ Sergio's number exceeds it with probability $\dfrac{10 - s}{10}.$

The corresponding values of $10 - s$ are $7, 6, 5, 5, 4, 4, 3, 3, 2, 1,$ totaling $40.$ The overall probability is $\dfrac{40}{10\cdot 10} = \dfrac{2}{5}.$

Thus, the correct answer is A.

Solution:

The even digits $4, 6, 8$ cannot be the units digit of a multi-digit prime, so each must appear in a tens place or higher, contributing at least $40 + 60 + 80 = 180.$ The other six digits contribute at least $1 + 2 + 3 + 5 + 7 + 9 = 27,$ so the sum is at least $207.$

This bound is achieved, for example by $\{2, 3, 5, 41, 67, 89\},$ whose sum is $207.$

Thus, the correct answer is B.

Solution:

The centers are $A = (10, 0)$ and $B = (-15, 0),$ with radii $6$ and $9,$ so $AB = 25.$ The shortest tangent is the internal one, meeting $\overline{AB}$ at a point $D$ that splits it in the ratio $6 : 9,$ giving $D = (0, 0).$

The right triangles $APD$ and $BQD$ are similar with ratio $2 : 3.$ Then $PD = \sqrt{10^2 - 6^2} = 8$ and $QD = \sqrt{15^2 - 9^2} = 12,$ so $PQ = 8 + 12 = 20.$

Thus, the correct answer is C.

Solution:

The graph reaches $6$ at $x = -2$ and $x = 1,$ so $f(f(x)) = 6$ requires $f(x) = -2$ or $f(x) = 1.$

The horizontal line $y = -2$ meets the graph twice, and $y = 1$ meets it four times, giving $2 + 4 = 6$ solutions.

Thus, the correct answer is D.

Solution:

Since $0.\overline{ab} = \dfrac{\overline{ab}}{99},$ the reduced denominator divides $99 = 3^2\cdot 11.$ The divisors are $1, 3, 9, 11, 33, 99.$

The denominator $1$ would require $\overline{ab} = 99,$ i.e. $a = b = 9,$ which is excluded. Each of $3, 9, 11, 33, 99$ is achievable, giving $5$ possible denominators.

Thus, the correct answer is C.

Solution:

Continuing the sequence gives $4, 7, 1, 8, 9, 7, 6, 3, 9, 2, 1, 3, 4, 7, 1, \ldots,$ which repeats with period $12.$ Each block of $12$ terms sums to $60.$

The largest $k$ with $60k \le 10{,}000$ is $k = 166,$ giving $S_{12\cdot 166} = 9960.$ Adding the next terms $4, 7, 1, 8, 9, 7, 6$ contributes $42,$ pushing the total past $10{,}000.$ So $n = 12\cdot 166 + 7 = 1999.$

Thus, the correct answer is B.

Solution:

Since $AB = 10$ and $\angle ABC = 60^\circ,$ the $30$-$60$-$90$ triangle has $BC = 5$ and $AC = 5\sqrt3.$

Place $E$ on $\overline{AC}$ with $CE = 5;$ then $BE = \sqrt{5^2 + 5^2} = 5\sqrt2.$ As $D$ moves along $\overline{AC},$ $BD = \sqrt{25 + CD^2}$ exceeds $5\sqrt2$ exactly when $CD \gt 5,$ i.e. when $D$ lies beyond $E,$ which happens iff $P$ is inside $\triangle ABE.$

The probability is $\dfrac{[ABE]}{[ABC]} = \dfrac{EA}{CA} = \dfrac{5\sqrt3 - 5}{5\sqrt3} = \dfrac{3 - \sqrt3}{3}.$

Thus, the correct answer is C.

Solution:

Since $D$ lies on the perpendicular bisector of $\overline{BC},$ $DB = DC = 7.$ The angle bisector $\overline{BD}$ gives $\dfrac{AB}{BC} = \dfrac{AD}{DC} = \dfrac{9}{7},$ so write $AB = 9x$ and $BC = 7x.$

Let $\theta = \angle ABD = \angle DBC.$ In isosceles $\triangle BDC,$ the foot of the perpendicular is the midpoint $M$ of $\overline{BC},$ so $\cos\theta = \dfrac{BM}{BD} = \dfrac{7x/2}{7} = \dfrac{x}{2}.$

Applying the Law of Cosines in $\triangle ABD:$ $$9^2 = (9x)^2 + 7^2 - 2(9x)(7)\cdot\dfrac{x}{2},$$ which simplifies to $81 = 18x^2 + 49,$ so $x = \dfrac43$ and $AB = 12.$

Now $\triangle ABD$ has sides $9, 7, 12.$ By Heron's formula with $s = 14,$ the area is $\sqrt{14\cdot 5\cdot 7\cdot 2} = \sqrt{980} = 14\sqrt5.$

Thus, the correct answer is D.

Solution:

Let $z = a + bi.$ The equation is $z^{2002} = \overline{z}.$ Taking magnitudes, $|z|^{2002} = |z|,$ so $|z|\big(|z|^{2001} - 1\big) = 0,$ giving $|z| = 0$ or $|z| = 1.$

If $|z| = 0,$ then $(a, b) = (0, 0),$ one solution. If $|z| = 1,$ then $\overline{z} = \dfrac1z,$ so $z^{2002} = \dfrac1z,$ i.e. $z^{2003} = 1,$ which has $2003$ distinct roots.

Altogether there are $1 + 2003 = 2004$ ordered pairs.

Thus, the correct answer is E.

Solution:

Replacing the nonzero coefficients by their mean keeps the total of the coefficients unchanged, so $P$ and $Q$ have the same coefficient sum. Since $P(1)$ and $Q(1)$ each equal that sum, $P(1) = Q(1).$

Therefore the graphs of $y = P(x)$ and $y = Q(x)$ must cross at $x = 1.$ The only choice showing an intersection at $x = 1$ is graph B. (There, $P(x) = 2x^4 - 3x^2 - 3x - 4$ and $Q(x) = -2x^4 - 2x^2 - 2x - 2.$)

Thus, the correct answer is B.