<ol>
<li>First we’ll find all possible values of p and t. The parabola intersects l at (p,5) so we have 5 = -p^2 + 9, so p = ±2. Also, the parabola intersects l at (t, -7), so -7 = -t^2 + 9, t = ±4.</li>
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<p>The slope of line l is (5-(-7))/(p-t) = 12/(p-t). We want p-t to be negative, but close to 0 (so that slope is minimized). This is optimal when p = 2, t = 4, so the minimum slope is 12/(2-4) = -6.</p>
<ol>
<li><p>The buyer saves $1.88 on the two bottles. Per bottle, this is $0.94.</p></li>
<li><p>A little tricky to explain, but realize that if the large cube has side length n, then the unpainted cubes will form a cube of side length n-2. There are 27 unpainted cubes, so (n-2)^3 = 27, n = 5. So there are 5^3 = 125 cubes total.</p></li>
<li><p>(2+4+…+100) - (1+3+…+99) = (2-1) + (4-3) + … + (100-99) = 50</p></li>
<li><p>I and II are not true (consider a rectangle that is not a square). III must be true, since the sum of the interior angles is 360°. Hence the answer is D.</p></li>
<li><p>I have a feeling this is incorrect. Since (x+y)/2 is the average of x and y, it will also be the midpoint of x and y, so y corresponds to point E.</p></li>
<li><p>Setting k = 0 implies x = 7 or x = 2. So A is correct.</p></li>
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<p>For 3, I’ve seen a lot of problems regarding painting cubes (on SAT’s and elsewhere, including MATHCOUNTS and AMC). But yes, that is correct. Basically, given an n x n x n cube, only the outer cubes have paint on them. So the unpainted cubes are the ones that are in the inner (n-2) x (n-2) x (n-2) cube.</p>
<p>For 7, pretty much. You just have to notice k = 0 gives 3(x-7)(x-2) = 0 → x = 7, 2. Alternatively, you can see that x = 7 and x = 2 are such that the LHS and RHS of the equation are equal. Plugging in x = 7, we see that 0 = k.</p>
