<p>It was B. I got it wrong too.</p>
<p>^^ I’m pretty sure I put B. Because you could find the distance between point C and I believe it was point B. and the distance was 16<em>sqrt5 if I’m not mistaken. So you could plug in the points in the answer choices along with point B into the distance formula to see if they matched 16</em>sqrt5. I’m pretty sure that answer choice B, when plugged into the distance formula, came out to be 16*sqrt5.</p>
<p>i am going to kill myself y’all</p>
<p>It’s okay; don’t be sad :(. I’ve thought that I did badly on a couple of tests and ended up doing pretty well. Just wait and see, and your scores might end up being better than you expected.</p>
<p>Yeah true, I thought i did horrible on the september one but somehow got a 31 which is decent… thanks :)</p>
<p>■■■, got 3 perfect 36’s on sciences the 3 days before. I rushed like 2 passages in 5 minutes. Hoping for like a 31. ._.</p>
<p>I don’t think 16 was even an answer choice for the last question. It was like A) 2 B) 3 C) 4, I put A…</p>
<p>Does anyone remember the inequality union one? Interest three-part problem?</p>
<p>Inequality/Union was the last answer choice.</p>
<p>Interest three-part problem?</p>
<p>Thats already posted in the beginning/middle of the thread.</p>
<p>16 was an answer choice and it was E.</p>
<p>Consolidated Answers: ( PLEASE ADD or CORRECT)</p>
<ol>
<li>60 degrees=Angle AOC </li>
<li>16 ( Answer to #60)</li>
<li>Quadrants I and IV ( for he triangle shifted)</li>
<li>2- sqr 5 ( for the ax^2+bx+c)</li>
<li>72 degrees ( for the angle in the pentagon</li>
<li>a^2 + b^2 <c^2</li>
<li>3pi ( for quadrilateral and 4 circles-looking for shaded region</li>
<li>null set </li>
<li>14 ( spinner- how many times they would do 4 times)</li>
<li>6 (spinner- if it spins 540 degrees where will it land)</li>
<li>60% spiiner probability</li>
<li>(20,40) coordinates of point</li>
<li>8K ( what was this question???)</li>
<li>7 (arithmetic sequence greater than 200)</li>
<li>0<y<3 ( does anyone remember the other answer choices?)</li>
</ol>
<p>Just for the record, I am from Ohio and got form 72A. :)</p>
<p>Hi smart people… so, this may seem like a silly question, but why are the vertices of the triangle in quadrants 1 and 4? </p>
<p>Doesn’t “x+4” ultimately cause the figure to shift to the left 4?
For some shape (perhaps parabolas) I recall learning that the sign indicated a shift in the opposite direction?</p>
<p>I agree with all of the other answers in the list above</p>
<p>@soxfan did you have a graphing calculator? If you did, you could have just graphed it in there.</p>
<p>Th coordinate were shifted positively, so it goes right 4. If the addition is inside the function, such as (x+4)^2, then it would be shifted left 4</p>
<p>Muffin Man: I did, but I didn’t think twice about my decision 
FutureDoc: Thanks for clarification</p>
<p>Yesterday was defintely not my finest 5 hours of testing</p>
<p>This was my justification for the circumference of the 4 circles. I believe the answer was 6pi. If all of e circles have the same radius measurements within that particular quadrilateral then all of the angles of the quadrilateral must be equal with angle measures of 90 degrees each. Obviously 90 degrees is 1/4 of the 360 degrees in the circle. Therefore, the circumferences of each circle will be 3/4 of the entire circumference. The entire circumference of 1 circle was 2pi (radius was 1). 2pi times 3/4 is 6pi/4 (the circumference for just one circle). Multiply that by 4 (24pi/4) to get 6pi as the circumference for the shaded regions of the 4 circles.</p>
<p>Actually I can’t seem to remember if the question asked for circumference or area. If the question asked for area then it would be 3pi.</p>