<p>it asked for the area, not the circumference. I am certain.</p>
<p>It definitely asked for area: Area of the Four circles was 4pi minus the other areas to get 3pi</p>
<p>This may have already been discussed, but for question 59, what was the exact question/parameters given? I remember the equation was y=3(a^x) and it asked you to find the range of y, but what was the exact question? Did it give specific requirements for a (ie, greater than 1, non-negative, etc.?) and/or x, and what was the correct answer?</p>
<p>I trust myself enough to solve according to area and not circumference. I just couldnt remember what I put or what the question was. I interpreted it a the way I previously described except 3/4 of the area would be 3pi/4. Multiply that by 4 to get 12pi/4= 3pi</p>
<p>Musiciife73, the asymptote never got to 0 and a first value of 3 means the range w for any graph of that type would be 0<y<3</p>
<p>Is there a consensus answer on the question that some people are saying the answer is +/- 8/9? I remember getting something like +/- 2sqrt(2)/5.</p>
<p>Great, that’s what I think I put (I ran out of time and guessed). Alos, @Math284, I got +/- [2sqrt(2)]/5. I remember double checking after finding my answer, unless I read the question wrong. I actually don’t even remember which question that was. Was it the sin(theta) question?</p>
<p>Yup, that would be it.</p>
<p>no it was def 2root2/5 but I only put negative for that. I wasn’t sure if it was both negative and positive.</p>
<p>It was both positive and negative 2 root 2/5. Sin functions can be both positive in thetas of 90 to 270. Sin takes positive in 90 to 180 and negative in 180 to 270.</p>
<p>what was the exact question for #59 y=3(a^x)?
what did it say a and x were?</p>
<p>Also, what was the exact question for the sin quadrant question? I know the answer was + or - 2 root 2/5, but what exaclty was the question. I know it said it was between 90 and 270 degrees, but what was the rest of it? What else did they give us?</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>I am 99% sure #15 was 3<y because regardless of which number you plugged in for y since the problem was like 3(a)^x or something like that (I don’t remember the exact problem), the graph would ALWAYS cross the y-axis at (0,3) and continue increasing to infinity. It asked for when x is 0 or greater, so the answer would be (3, infinity) or something like that since the range would start at 3 every time.</p>
<p>It was this:</p>
<p>y= 3(a^x) where x is NEGATIVE and a>1.</p>
<p>Ah, perhaps I did have a different test form then. Ohioans.</p>
<p>Am I the only one that disagrees with consolidated answer 4? why would it have to be 2-sqrt(5)? I thought you didn’t have enough information to answer the question.</p>
<p>I agree. X was negative and a was 3 on my form.</p>
<p>60% spiiner probability</p>
<p>What was the question?</p>
<p>Yea, I got y>3 also. I plugged in a few numbers in my calculater and y increased to infinity each time…</p>
<p>I don’t remember x having to be negative, tho a>1.</p>
<p>I’ve been thinking about this a lot, and I’m convinced that there was enough information. The key piece was that a, b, and c are rational. There are a lot of ways to convince yourself that the only number y such that 2+sqrt5 * y and 2+sqrt5 + y are both rational (which is necessary and sufficient for y to be the right answer) is 2-sqrt5, but you can also look at what the quadratic equation would give and see that the only way to get the sqrt5 is to get both it both as a positive and negative.</p>
<p>Source: I’m pretty good at math.</p>
<p>let’s say you had a quadratic equation, whenever you simplify it out(hoping you don’t have to use the quadratic formula) to solve x, you could have something like (x+2)^2=5. So x+2=+/- root 5. Then x=-2 +/- root 5. See there would be 2 solutions, with the -2 staying the same and having +/-</p>