<p>Seriously lookingup8 reread my earlier post</p>
<p>No guys number 59 with digging the hole was (v/lw) - d. It was answer choice E, not D. I confirmed it with many others taking it that put the same answer and my friend who has an extremely good memory.</p>
<p>Oh and Ethan the answer is 7/cos 56. If you found the central angle, which was 56 degrees, the other radius was adjacent to it and to find the hypotenuse you have cos 57 = 7/x. Then you solve for x to get 7/cos 57. 28 was not the central angle… it was an inscribed angle. And the inscribed angle didn’t form a right triangle, but the central angle did.</p>
<p>I said 7/cos58. I went back and checked that problem a lot and am pretty sure I got it right. The triangle was isosceles because the legs were 2 radii. That meant the center angle was 124, meaning the adjacent angle was 56. You had to use 56 since it was in the bigger triangle and it was law of cosines. Adjacent over hypotenuse which in this case was AO, the longest side. Hope that helps.</p>
<p>7/cos(56). Stop arguing over it. I’m 100% sure that’s the answer.</p>
<p>for the one with the multiplying polynomial at the end:
isnt the answer somewhat along the lines of x(x2 + 6x) + 5x because x and (x2+6x) counts as 1 multiplication and 5x counts as the 2nd?</p>
<p>yes halcyoncxd,
that’s the answer</p>
<p>No it was x(x(x+7) + 5) - something else. There wasn’t a 5x in the answer choice I picked. I know that for sure…</p>
<p>and it said in the question that a constant times a binomial is one multiplication, but the computer program never said that a constant multiplied by x is one multiplication, it wasn’t designed that way… otherwise you could said that a constant like 1 or 2 is a multiplication because it’s being multiplied by x^0, which is 1.</p>
<p>^by the definition you stated x(x(x+7))+5x-1 would work as well if the computer doesn’t count the multiplication of x^0 and -1, then a multiplication of x and 5 would be under the same principle. The multiplication mainly encompasses the parentheses as “multiplications”.</p>
<p>Sillyup seems to be 100% confident about everything. Think you got a 36. Are you also confident what the curve will look like cuz I need to know.</p>
<p>Do you g?uys remember the answer choice for that question</p>
<p>Sillyup can youbsraw out how you got the cos56? I’m still pretty sure its 7/sin28</p>
<p>@booberrypi78
That is true, but that just further proves my point… Unless you were just adding something that I forgot.</p>
<p>I thought that I had cos56 also. I think that one of the triangles was solvable, and that the angle which was on the same line (supplementary angle) was 124. And because the triangles shared a side and you were trying to find the hypotenuse it should have been cos56. That being said I could be totally wrong because I was rushing like hell and ran out of time.</p>
<p>@matthew5
The 5x being outside of the parantheses</p>
<p><a href=“https://docs.google.com/file/d/0Bz7re510txaibVBWc0ZUcGZBaHc/edit[/url]”>https://docs.google.com/file/d/0Bz7re510txaibVBWc0ZUcGZBaHc/edit</a></p>
<p>How is the curve looking guy?</p>
<p>purpledinos, it wasn’t a right angle… ?</p>
<p>did we reach a consensus on the 2a+2b question yet?</p>