<h1>20 - J - Explained well above.</h1>
<h1>30 - K - Each side of this square is 3cm long. If one vertex is at (2,0), another vertex must be at either (5,0), 3cm to the right, or (-1, 0), 3cm to the left.</h1>
<h1>33 - D - If you isolate y to get an easier form of this equation, you’ll get y = 2x - 4. You can then graph that and see that it goes through quadrants I, III, and IV.</h1>
<h1>34 - F - You can graph this and use the table of coordinate points to see where the equation goes through (1, x). If you look at the table, you can see that it goes through (1, 4). So, x must be 2.</h1>
<h1>37 - B - The question is asking for the perimeter of the outside of the figure. You know that one side of the square is 8cm long. Two sides of the square are included in the outside of the figure, so it must be 16 + something. So you’re choosing between B and C. I’m not sure if there is a technical way to choose between these. I chose B because it seems like 8 is around the diameter of the circle. So, 4 would be approximately the radius, and the circumference (perimeter of the circle) would be about 8pi.</h1>
<h1>38 - G - If you divide EGFH into triangles like the rest of the figure, you can see the ratio. EGFH has 2 triangles, and the shaded region has 6. The ratio of that would be 2:6, and reduced, 1:3.</h1>
<h1>42 - F - You can see this through right triangle trig, aka SOH-CAH-TOA.Tan = opposite/adjacent. You are looking for the side from the boat to the dock. You’re not looking for the hypotenuse, which SOH and CAH both include, so you would use tangent. Using the 52º angle, you go opposite that to the side you’re looking for and then adjacent to the 30 mile length side. (Sorry if that doesn’t make much sense)</h1>
<h1>44 - H - If you can reflect a line to another line and find it unchanged, it’s a line of symmetry. The lines of symmetry in that figure is basically all sides of both squares. So 8.</h1>
<h1>47 - C - I’m not 100% sure on this one. A bisector cuts an angle into to equal parts, as AE does to <BAC. I just figured <AEC would be a right angle.</h1>
<h1>51 - C - So, for every hour that Marcia spends making frames, she gives $3. Above in the little passage, it says it takes her 2 hours to make 1 small frame and 3 hours to make 1 large frame. In this week of December, she’s made 4 large frames and 2 small frames. So she has spent 12 hours on the large frames and 4 hours on the small frames for a total of 16 hours. Since she spent 16 hours making her frames, she will donate $48. Also in the passage part above, we’re given an equation for her profit, p = 30s + 70l, s for the small frames and l for the large frames. We know those values, so we can set up the equation</h1>
<p>p = 30(2) + 70(4) and get a profit of $340. The question is asking what percent of that week’s profit did Marcia donate. 14% of 340 is 47.6, which is the closest value to the $48 she actually donates. </p>
<h1>52 - I’m not sure about this one. I think you have to figure out what the graph is telling you and use those values in the profit equation, but it was confusing to me.</h1>
<h1>55 - C - You can pick values for x and y and test these. For example, if x is 2 and y is -5: (A) 2/-5 is not greater than 1. (B) |2|^2 = 4 is not greater than |-5|. © (2/3) - 5 is greater than (-5/3) - 5, so that works.</h1>
<h1>60 - F - If a number is 5 units from -3, then the difference between the number and -3 is 5. If the number is to the right of -3, x-(-3) = 5 and if the number is to the left of -3, then -3 - x = 5 or x-(-3) = -5. In either case the absolute value of x-(-3) is 5 so the solution set is |x-(-3)| = 5, or choice F.</h1>
