<p>i remember there were three lines. i also think only one was labelled. </p>
<p>can you also refresh my memory. there were three lines drawn and five points. all of the points were above the lines. where exactly was 'a'? i remember putting the one that was second from the left.</p>
<p>weren't there only two lines? k & l</p>
<p>Really...? I'm also pretty sure it was reflected twice. Ohh well, we'll just have to wait and see.</p>
<p>are we referring to the same question?</p>
<p>i'm referring to the one in the first math section, it wasn't even near the last questions, so it was not intended to be difficult.</p>
<p>There were two reflections, I'm positive.</p>
<p>i don't know :|
i know i reflected a point p twice, once over each line</p>
<p>i put A...were there any other hard math ?s</p>
<p>i usually have a photographic memory when it comes to hard questions, but i'm near blank on this one. i didnt consider it to be very difficult...</p>
<p>all i can remember is "if point p is reflected across line m (or l???), then which of the following would be the new point?"</p>
<p>really? i had only two lines, with two reflections
i hope they're different problems.</p>
<p>lol, it's possible we're not even talking about the same question. The problem I'm talking about had line l and line m intersecting each other. Point P had to be reflected across line l, then line m.</p>
<p>i think it was actually 60 because if it was 30 that means the bottom right angle of the parallogram would be 90 (because the other piece of it was 60) and it obviously wasnt a 90 angle it was a 120 degree angle so it had to be 60</p>
<p>unless there was a misprint, i'm pretty sure there were three lines. the two intersecting ones were below the one u had to reflect across.</p>
<p>can anyone remember the exact wording of the question???</p>
<p>slope -3/4?</p>
<p>to make sure we're talking about the same reflection question:
were all the points near the top (above the lines)?
and was 'a' the lefmost point?</p>
<p>Im pretty positive the angle they were asking for was 60 degrees. They were asking for the angle opposite the congruent line of the triangle on the right. The angle opposite the congruent line in the triangle on the left was 60 and we know the left and right triangles are congruent so i think the answer to that one was 60 degrees.</p>
<p>OPPPPPPPPP, I think your reasoning is almost right, but it's slightly off: wouldn't the right angle be at the CENTER of the parallelogram, therefore making it a special case of a parallelogram (a "kite," right?). you would have angles of 60, 60, 120, and 120 which add up to 360, and if you drew the triangles in it would be two equilateral triangles adjacent to each other ( like
/\/ if I could connect the top/bottom). Draw the diagonals, and they would bisect the angles and produce four right triangles.</p>
<p>yea true it was a 60 60 120 120 parallelogram. but the angle they were asking for was right next to one of the 60 degree angles of the equilateral triangle and that angle was definitely 120 degrees in total.</p>
<p>which triangle one is this???</p>
<p>is it the one with the three triangles and it asked for the measure of the angle in the top right.....(and the answer was 30 degrees)</p>
<p>i thought it wanted the bottom left....****</p>
<p>Yeah, I got 30 degrees. I thought it wanted the top right...</p>