<p>@bluedevils it is 3/10</p>
<p>What was the scatterplot question? And have we figured out which one was the math experimental?</p>
<p>What do you guys think a -1 will be?</p>
<p>what was the answer to the k=x/y so solve for (x-y)/x in terms of k?</p>
<p>i remember putting the k+1 answer</p>
<p>does anyone remember a square with an area of 36 and a triangle on top and you had to find the possible point of one of the points? one point they gave you was (4,13). i think this one was in the experimental section. does anyone that didnt have experimental have this one?</p>
<p>@llaurenn
yeah i think that’s what i got too i put different random numbers twice and they all worked out</p>
<p>If k = (x/y), what is (x+y)/x in terms of k?</p>
<p>Answer is (k+1)/k</p>
<p>(k+1)/k
= ((x/y)+1)/(x/y)
= ((x/y)+(y/y))/(x/y)
= ((x+y)/y)/(x/y)
= ((x+y)/y)*(y/x)
= (x+y)/x</p>
<p>@flyingeagle
Think that’s experimental.</p>
<p>Do we know which math was the experimental yet? im hoping its the 17 median one because of course i thought it said mean…</p>
<p>llaurenn it wasn’t, i heard it was something about where p is undefined.</p>
<p>i spent like 30 minutes during the test figuring out which one it was and i kept recounting that i had 3 of each section… then i realized writing is only supposed to be 2 xD so i had writing experimental and also had the 17 median question</p>
<p>17 median was not experimental because I had it and I didn’t have experimental</p>
<p>What about that questions where it was like f(x)= 2x-5?
What’s the range if given -2<x<2?</p>
<p>I put -5<x<0. What about you guys?</p>
<p>Im sorry, but does anyone actually recall the full question to the 3/10 one? I don’t remember it exactly. Was there a chart?</p>
<p>@HeyImAlok,
it was f(x) = 2x^2 - 5
The answer is -5<=x<3</p>
<p>@Michael Thanks. How, though?</p>
<p>Can someone explain the three times the radius one?</p>
<p>The lowest number you can get is if you plug in 0 for x, and you get -5. The highest value you can get is when you plug in 2 for x, and u get 3.</p>
<p>^^^ Find the minimum value and the maximum value. Graph the function on a calculator and then determine the minimum value (in this case the y-value of the vertex), and determine the values for -2 and 2 (which are both 3). Or you could take the derivative, determine critical points, find maximum and minimum values.</p>
<p>2x^2 -5 is a parabola of upward concavity. Its minimum point is at x=0, and f(0) is -5. Inputting -2 and 2 into the function both yield 3. Therefore the range of f(x) (in the interval of [-2, 2]) is [-5, 3]. So -5 < f(x) < 3.</p>
<p>Edit: I don’t know how to type less than or equal to…</p>
<p>I think the question was if you triple the radius of a circle what is the increase in percent of the area of the circle.</p>
<p>@qqqqrt</p>
<p>you have Pir^2 and Pi(3r)^2 which equals 9Pir^2.</p>
<p>9Pir^2 is greater by 800% (8 * 100% = 8Pir^2)</p>
<p>@lemniscate @michael</p>
<p>The lowest number, 0 -> -5. Correct.
High value is NOT 2, the rule given is -2<x<2 where x is LESS THAN 2 and GREATER than -2. It’s <em>only</em> less than, not less than and equal to. Thus, highest plausible value would be 1. If you plug in 1, you get -1. I believe the correct answer is then -5<x<0, or A.</p>
<p>Am I missing something? Pretty sure highest value HAS to be less than 2, so it’s 1.</p>