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<li><p>The angles with P as a vertex that are also an angle of a triangle sum to 180. You can determine this by the fact that each one has a congruent angle on the opposite side of P that is not a part of any of the 3 triangles. Since the sum of the angles of all three triangles is 540, the sum of the angles that are an angle of a triangle and that do not have P as a vertex is 360. Of those angles, you know the measure of all except the one with a measure of x degrees, so you can set up an equation: other angles + x = 360. From that point, the problem should be quite easy to solve.</p></li>
<li><p>15 trays contain cups and 21 contain plates. If we assume that no trays have both a cup and a plate, then there must be 15+21 = 36 trays. This is not possible, since we already know that there are 25 trays. However, if we assume that 36-25 = 11 have both a cup and a plate, that leaves us with 21-11 = 10 with only plates and 15-11 = 4 with only cups. 10 trays with only plates + 4 trays with only cups + 11 trays with one of each = 25 total trays, so there are 11 trays with one of each.</p></li>
<li><p>We want to find RS<em>PS. The area of RST is 7, so RS</em>TS/2 = 7 and RS<em>TS = 14. Because we’re looking for RS</em>PS, we want to change the TS in RS<em>TS = 14 to PS. This is where we use the equations PT+TS = PS, obtained from the fact that PS is a segment with PT and TS as its only components, and PT = (2/5)</em>PS, which was given in the description. Substituting the second into the first to remove PT, the unneeded length, we obtain (2/5)<em>PS + TS = PS, or (3/5)</em>PS = TS. Because of this, we can substitute (3/5)<em>PS for TS in RS</em>TS = 14, giving us RS<em>(3/5)</em>PS = 14, which is equivalent to RS<em>PS = 14</em>(5/3) = 70/3. </p></li>
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<p>Are you sure it’s 18, or did I make a mistake somewhere?</p>
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<li><p>AB = AC and E and D are midpoints, so ED is parallel to BC and it is halfway between A and BC. As you draw lines parallel to BC starting at BC and moving toward A, their length decreases linearly from the length of BC to 0 as distance from BC increases. Because E and D are midpoints of the sides on which they are located, BC is halfway between A and BC, so its length will be half of BC’s. 4 = (1/2)BC gives the solution, which is BC = 8.</p></li>
<li><p>This one, strangely enough, seems to be the easiest of all the questions you missed. The first student had the entire length of rope, which we will represent by the unit length 1. He cut it in half, leaving him two sections, each of length 1/2. He kept one of these and gave the other to the next student, who cut his rope of length 1/2 into sections of length 1/4 and 1/4, kept one, and gave one away. The students who received the rope of length 1/4 cut it into two halves, each of length 1/8… If you notice the pattern, each piece of rope that the students keep has a length which is a fraction with a denominator that is a power of 2. Of the answer choices, only 1/16 has a denominator that is a power of 2; it is equal to 1/(2^4), so it is a length that would occur in the sequence.</p></li>
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