OK, my regular uploading site, fileden, isn't working, so this is a band-aid and hopefully I can come up with a better link tomorrow. I'd prefer a direct link, but this will have to do.
Your # 18 makes no sense. You make a -sin^2x*cos^2+sin^2=sin^2(1-cos^2). I don't know how you did that, I'm not saying its wrong, but you definitely did more than one step and i have no idea what those steps were.
Well, after a long time, i figured out that other question, but for 87, i'm pretty sure it should be y+2 not y-2 because the center is (0,-2) and it's y-k, so y--2=y+2. So methinks it should be y+2.
For number 83 we found that you standard equation is off and that instead of (y-4)^2=4(x+3), h should equal 2 becuase the center is at (2,4) and not (3,4)..You accidentally put the focus into the standard equation
okay this one, i'm not to sure about, but i think for 103, the magnitude is closer to 8.5, not three, but i'm not real confident with those problems, but I think it should be 8.5, because you would have <3i,8j> (right? I think...) and the magnitude of that vector is like 8.5
Okay. So in 91 you say that a=2 so isnt your graph wrong because you show that the vertices are at (2,5) and (2,-3) but in your answer on the right you say the vertices are (2,3) and (2,-1) cause the center is (2,1) and you add and subtract a which is 2 and you get (2,3) and (2,-1)
Number 7 should be
ReplyDeletesin(x)cos(x)
the bottom cancels out
Your # 18 makes no sense. You make a -sin^2x*cos^2+sin^2=sin^2(1-cos^2). I don't know how you did that, I'm not saying its wrong, but you definitely did more than one step and i have no idea what those steps were.
ReplyDeleteWell, after a long time, i figured out that other question, but for 87, i'm pretty sure it should be y+2 not y-2 because the center is (0,-2) and it's y-k, so y--2=y+2. So methinks it should be y+2.
ReplyDeleteFor number 83 we found that you standard equation is off and that instead of (y-4)^2=4(x+3), h should equal 2 becuase the center is at (2,4) and not (3,4)..You accidentally put the focus into the standard equation
ReplyDelete#7- Not sure what you mean...
ReplyDelete#87- You are correct. My mistake.
#83- You are correct. My mistake.
Keep them coming: EC for mistake finders!
Hey pierce and me(harry) both get it right? cause yeah we ate at Panda then found the error at the church aha.
ReplyDeleteYeah, that's fine. Good job. Panda still sounds good.
ReplyDeleteNumber 7 is correct in the key. I screwed up a little...
ReplyDeletefor 92 in your "equation" answer, i believe you have an extra 'x' attached to the -12 on the right hand side.
ReplyDeleteNumber 94, the second vertex should be (3, -7) i believe, because a=5, not 6.
ReplyDeleteokay this one, i'm not to sure about, but i think for 103, the magnitude is closer to 8.5, not three, but i'm not real confident with those problems, but I think it should be 8.5, because you would have <3i,8j> (right? I think...) and the magnitude of that vector is like 8.5
ReplyDelete#92-Yup extra x is right. Not sure why I would put that on there...weird
ReplyDelete#94-Oh yes this was brought to my attention by Ashley R earlier today I think. Meant to let everyone know...
#103-Typo in the problem. The second coefficient should be 3. Sorry.
FYI I'm going to bed. With any luck I'll be in @ 6:45 tomorrow morning if any of you have more questions.
ReplyDeleteOkay. So in 91 you say that a=2 so isnt your graph wrong because you show that the vertices are at
ReplyDelete(2,5) and (2,-3) but in your answer on the right you say the vertices are (2,3) and (2,-1) cause the center is (2,1) and you add and subtract a which is 2 and you get (2,3) and (2,-1)
#82
ReplyDeleteyou've got a typo in your solution.
should be
y^2 = 8x
Yes, the graph appears to be pretty crude.
ReplyDelete