<p>what was the answer to the last problem on one of the math problems with 3 equations with letters</p>
<p>Hyakku, I did that, too.</p>
<p>I did tan105/tan(whatever angle it was) = 10/x.
Hopefully, that was the right equation...</p>
<p>skater - z = 2</p>
<p>beads sure was 76</p>
<p>lesos, yeah i got the same :)</p>
<p>x^2/3 the answer y.</p>
<p>Good, there's still hope! =)</p>
<p>for the parabola and rectangle area question is it 27? because i got (3,0) and (0,9) for the points of the rectanle, after factoring the quadratic equation. the factored form was (2x-3)(x-3)</p>
<p>Lesos:I think the equation would use sin rather than tan. It is called the sine law.</p>
<p>It crossed the x axis at the FIRST intercept, which was 1.5. 3 was the second time it crossed. So you had to do 1.5 x 9, not 3 x9.</p>
<p>Predictions for a score if I did... </p>
<p>2 wrong, 1 omit
or
2 omit, 1 wrong? </p>
<p>Can't remember if I ommitted the square question or not, I know I omitted xy= x + y.</p>
<p>yea i agree mike</p>
<p>xy=x+y</p>
<p>x=1.2
y=14</p>
<p>sorry, but 16.8 does not = 15.2</p>
<p>(arbitrarily chosen numbers)</p>
<p>external anglle - sum of two opposit
mid points x,y, z I+II+III?</p>
<p>gk23: the point was (1.5,0) not (3,0). U used the wrong factor.</p>
<p>Yea it was the law of sines, I was kinda surprised trig was up there, didn't really know how I would've solved it without trig.</p>
<p>So I'm pretty sure the first or second one was expiremental. If everyone got the popcorn, I can't really remember what the expiremental section had I cant remember the second one. Anyone know?</p>
<p>aww that was such a retarded mistake i made with the rectangle area thing ;(</p>
<p>I DIDNT GET POPCORN! So thats experimental, I had two writing ones. :p</p>
<p>
[quote]
Lesos:I think the equation would use sin rather than tan. It is called the sine law.
[/quote]
Ahhhh, you're right.
Yeah... I thought that the SAT didn't test trig. Isn't that one of the differences between the SAT and ACT- the ACT tests trig, and the SAT doesn't?</p>
<p>
[quote]
256 = the area of the square with the circle of 16pi
[/quote]
</p>
<p>I thought that at first, but then I over-analyzed it and assumed it wanted the area of the square that the circle hadn't taken up, and I came out to 56 or 59 or something.</p>
<p>gahhhhh</p>