<p>what was the answer to the question that said if x is greater than y which one is positive
and it gave three choices…?</p>
<p>i put k+4, i tried k-4 and it didnt seem right</p>
<p>@ozfried2693 I believe it was 4 people taking both.</p>
<p>What was the “88 median temp” question? I forget what it asked.</p>
<p>@ arsenalcrazy</p>
<p>try using the numbers r = 2 t = 10 and p = 20
You’ll see that these numbers both satisfy the equations and give you the answer as well.</p>
<p>The answer was k-4</p>
<p>x^2 - 12 = k
you need to subtract 4 from both sides.
x^2 - 16 = k - 4 = (x+4)(x-4)</p>
<p>I know the answer on the k+4 or k-4 was choice E, which I’m fairly certain was k-4.</p>
<p>You have to subtract 4 from both sides so that -12 equals -16 and then K becomes k - 4</p>
<p>If you subtract 4 from one side you have to do the exact same thing to the other side.</p>
<p>I vote k-4.</p>
<p>^That’s what I put. I went over that question about 3 times to make sure.</p>
<p>How do you think the curve will be on this math?</p>
<p>it’s k - 4, just substitute any number for x
I used 6, so 6^2 - 12 = 24
(6+4)(6-4) = 20 and since k = 24 it’s k - 4.</p>
<p>Yeah I believe it’s k-4 cause if x^2-12=k, x^2-12-4 [12 & 4 make 16]=k-4.</p>
<p>Anyone remember the physics/chemistry venn diagram? Was it 2 or 4?</p>
<p>I would say
-1 = 770 or 760
-2 = 740
-3 = 710
-4 = 690
etc.</p>
<p>Physics/Chem VD was experimental.</p>
<p>haha I hate the math curve.</p>
<p>I got k-4. If you say x is 4, then 16-12=4. Then (4-4)(4+3)=0</p>
<p>To get from 4 to 0, you subtract 4.</p>
<p>^
Pretty harsh considering math was pretty easy for the average high schooler this time and CR (vocab) was harder.</p>
<p>The curve is going to be harsh.</p>
<p>hopefully I got an 800 there are only two I am unsure of. Does anyone know the answer to that one with two parallel lines that were bisected, than it asked for an angle?</p>
<p>Okay, what is the median temp question? I got 93…that doesn’t sound right. Maybe I read the question incorrectly, because I remember it being easy.</p>