Solution:

There are $2006$ terms. Pairing consecutive terms gives $(-1+1)+(-1+1)+\cdots.$ Since $2006$ is even, every term pairs off and the sum is $0.$

Thus, the correct answer is C.

Solution:

Since $x \spadesuit y = (x+y)(x-y) = x^2 - y^2,$ we have $4 \spadesuit 5 = 16 - 25 = -9.$

Then $3 \spadesuit (-9) = 9 - 81 = -72.$

Thus, the correct answer is A.

Solution:

Let $c$ and $p$ be the Cougars' and Panthers' scores. Then $c+p=34$ and $c-p=14.$ Subtracting gives $2p=20,$ so $p=10.$

Thus, the correct answer is A.

Solution:

The red circle has area $\pi(\tfrac12)^2 = \tfrac{\pi}{4},$ and the large circle has area $\pi(\tfrac32)^2 = \tfrac{9\pi}{4}.$ The blue ring is $\tfrac{9\pi}{4}-\tfrac{\pi}{4}=2\pi.$

The ratio is $2\pi \div \tfrac{\pi}{4} = 8.$

Thus, the correct answer is D.

Solution:

Place the rectangles side by side with their $3$-length sides vertical. Their widths add to $2+3=5,$ and the heights $3$ and $4$ both fit within $5.$

The side cannot be smaller than $5,$ since the two smaller dimensions $2$ and $3$ must be accommodated. The smallest area is $5^2=25.$

Thus, the correct answer is B.

Solution:

Each side has length $\tfrac{2}{\pi},$ the diameter of a semicircular arc, so each arc has length $\tfrac12\pi\cdot\tfrac{2}{\pi}=1.$

The boundary consists of four such arcs, so the perimeter is $4\cdot1=4.$

Thus, the correct answer is D.

Solution:

The denominator simplifies: $1-\dfrac{x-1}{x}=\dfrac{x-(x-1)}{x}=\dfrac{1}{x}.$

So the expression is $\sqrt{\dfrac{x}{1/x}}=\sqrt{x^2}=|x|.$ Since $x \lt 0,$ this equals $-x.$

Thus, the correct answer is A.

Solution:

Let the square have side $s,$ so $s^2=40.$ Its base lies centered on the diameter, and a top corner at $\left(\tfrac{s}{2},s\right)$ lies on the circle.

Then $r^2=\left(\tfrac{s}{2}\right)^2+s^2=\tfrac{40}{4}+40=50.$ The semicircle area is $\tfrac12\pi r^2=\tfrac12\pi(50)=25\pi.$

Thus, the correct answer is B.

Solution:

The lemonade totals $100+100+400=600$ grams containing $25+386=411$ calories.

In $200$ grams there are $411\cdot\tfrac{200}{600}=\tfrac{411}{3}=137$ calories.

Thus, the correct answer is B.

Solution:

Let the sides be $x,$ $3x,$ and $15.$ The triangle inequality $x+15 \gt 3x$ gives $x \lt 7.5.$

The largest integer is $x=7,$ giving sides $7,$ $21,$ $15$ and perimeter $7+21+15=43.$

Thus, the correct answer is A.

Solution:

For $n\ge 10,$ $n!$ is divisible by $100,$ so it does not affect the last two digits.

The tens digit comes from $7!+8!+9!=5040+40320+362880=408240,$ whose tens digit is $4.$

Thus, the correct answer is C.

Solution:

Substituting $(1,2)$: from $1=\tfrac14(2)+a$ we get $a=\tfrac12,$ and from $2=\tfrac14(1)+b$ we get $b=\tfrac74.$

Then $a+b=\tfrac12+\tfrac74=\tfrac94.$

Thus, the correct answer is E.

Solution:

Joe adds $2$ ounces of cream and drinks nothing afterward, so he has $2$ ounces of cream.

JoAnn has $12$ ounces of coffee plus $2$ ounces of cream, making $14$ ounces of uniform mixture. After drinking $2$ ounces she keeps $\tfrac{12}{14}=\tfrac67$ of her cream, which is $\tfrac67\cdot2=\tfrac{12}{7}$ ounces.

The ratio is $2\div\tfrac{12}{7}=\tfrac{7}{6}.$

Thus, the correct answer is E.

Solution:

Since $a$ and $b$ are roots of $x^2-mx+2,$ we have $ab=2.$

The value $q$ is the product of the new roots: $$q=\left(a+\tfrac1b\right)\left(b+\tfrac1a\right)=ab+1+1+\tfrac{1}{ab}=2+2+\tfrac12=\tfrac92.$$

Thus, the correct answer is D.

Solution:

Because $AB=AD$ and $\angle BAD=60^\circ,$ triangle $ABD$ is equilateral, and so is triangle $CBD.$

Points $E$ and $F$ split the rhombus into six congruent triangles, each of area $\tfrac{24}{6}=4.$

Rhombus $BFDE$ is the union of triangles $BED$ and $BFD,$ so its area is $2\cdot4=8.$

Thus, the correct answer is C.

Solution:

From one Leap Day to the next is $3\cdot365+366=1461$ days, and $1461\equiv 5\pmod 7.$

Over the four cycles from 2004 to 2020, the weekday advances $4\cdot5=20\equiv 6\pmod 7,$ that is, $6$ days forward, which is one day back from Sunday.

So Leap Day 2020 falls on a Saturday.

Thus, the correct answer is E.

Solution:

Alice moves one ball to Bob, so Bob's bag holds $6$ balls with exactly one color appearing twice.

The two bags end up identical exactly when Bob returns one of that duplicated pair. Two of the six balls qualify, so the probability is $\tfrac26=\tfrac13.$

Thus, the correct answer is D.

Solution:

The terms are $2,\,3,\,\tfrac32,\,\tfrac12,\,\tfrac13,\,\tfrac23,$ then $2,\,3,\ldots,$ a cycle of length $6.$

Since $2006=6\cdot334+2,$ we have $a_{2006}=a_2=3.$

Thus, the correct answer is E.

Solution:

Since $OA=1$ and $OD=2$ with $D$ on the line $x=1,$ we get $\angle AOD=60^\circ,$ and likewise $\angle COE=60^\circ,$ so $\angle DOE=30^\circ.$

The sector $DOE$ has area $\tfrac{30}{360}\pi(2^2)=\tfrac{\pi}{3}.$

The region is this sector minus triangles $OBD$ and $OBE.$ With $BD=BE=\sqrt3-1,$ each triangle has area $\tfrac12(\sqrt3-1)(1),$ totaling $\sqrt3-1.$

So the shaded area is $\tfrac{\pi}{3}-(\sqrt3-1)=\tfrac{\pi}{3}+1-\sqrt3.$

Thus, the correct answer is A.

Solution:

The slope of $\overline{AB}$ is $\tfrac{178-(-22)}{2006-6}=\tfrac{200}{2000}=\tfrac1{10}.$ Since $\overline{AD}\perp\overline{AB},$ its slope is $-10,$ so $\tfrac{y+22}{8-6}=-10$ gives $y=-42.$

Then $AB=\sqrt{2000^2+200^2}=200\sqrt{101}$ and $AD=\sqrt{2^2+20^2}=2\sqrt{101}.$

The area is $200\sqrt{101}\cdot2\sqrt{101}=400\cdot101=40{,}400.$

Thus, the correct answer is E.

Solution:

Each die shows $k$ with probability $\tfrac{k}{1+2+\cdots+6}=\tfrac{k}{21}.$

For a total of $7,$ the ordered pairs $(1,6),(2,5),\ldots,(6,1)$ contribute $$\dfrac{1\cdot6+2\cdot5+3\cdot4+4\cdot3+5\cdot2+6\cdot1}{21^2}=\dfrac{56}{441}=\dfrac{8}{63}.$$

Thus, the correct answer is C.

Solution:

The total cost is $N(4B+5J)=253$ cents $=11\cdot23.$ Since $N \gt 1,$ $N\in\{11,23,253\}.$

If $N=253$ or $N=23,$ then $4B+5J$ equals $1$ or $11,$ impossible for positive integers.

So $N=11$ and $4B+5J=23,$ whose only positive solution is $B=2,$ $J=3.$ The jam costs $11\cdot3\cdot5=165$ cents $=\$1.65.$

Thus, the correct answer is D.

Solution:

Split the quadrilateral into two triangles of areas $R$ and $S,$ so the shaded area is $T=R+S.$

Comparing triangles that share an altitude, base ratios give $\tfrac{R}{3}=\tfrac{T+7}{10}$ and $\tfrac{S}{7}=\tfrac{T+3}{14}.$

Then $$T=R+S=3\cdot\tfrac{T+7}{10}+7\cdot\tfrac{T+3}{14},$$ so $10T=3(T+7)+5(T+3)=8T+36,$ giving $T=18.$

Thus, the correct answer is D.

Solution:

The hexagon is symmetric about $\overline{OP},$ so its area is twice that of trapezoid $AOPD.$

Draw $OF\parallel AD$ with $F$ on $\overline{PD}.$ Then $AOFD$ is a rectangle, so $DF=OA=2$ and $FP=PD-DF=4-2=2.$

Since the circles are externally tangent, $OP=2+4=6,$ so in right triangle $OFP,$ $OF=\sqrt{36-4}=4\sqrt2.$

Trapezoid $AOPD$ has parallel sides $OA=2$ and $PD=4$ with height $OF=4\sqrt2,$ giving area $\tfrac12(2+4)(4\sqrt2)=12\sqrt2.$ The hexagon area is $2\cdot12\sqrt2=24\sqrt2.$

Thus, the correct answer is B.

Solution:

Since a child is $9,$ the number is divisible by $9,$ so its digit sum $2(a+b)$ is a multiple of $9,$ which forces $a+b=9.$

There is also a $4$- or $8$-year-old, so the number is divisible by $4.$ Among numbers with two repeated digits summing to $9$ and divisible by $4,$ the number $5544$ is divisible by $1, 2, 3, 4, 6, 7, 8,$ and $9,$ but not by $5.$

So the eight ages can be $\{1,2,3,4,6,7,8,9\},$ and $5$ need not be among them. The age that is not necessarily a child's age is $5.$

Thus, the correct answer is B.