Kilbaha!

[Whatlol]

I am 100% confident in my workings above.

[Slumdawg]

I am 100% confident in my workings above.

Fair enough. Make sure you work it out with your teacher. I'll see with mine on monday if I remember.

God who knew one question could cause so much controversy. Does anybody else have any ideas?

[8039]

Does anybody else have any ideas?

f*$%# this question in the a$# is my idea

It goes against everything that's worked for me thus far ;(

[Slumdawg]

I'm more annoyed that it wasted so much of my god damn time.

Never answering another question again on vn till exams are OVER.

The solutions could be wrong. I just kept working and trying different stuff till I got an answer and the one I got was B which is what Kilbaha had which is why I thought it was right, hence I tried to explain what I did. I do get the logic behind the other answers though, it makes sense but the calc still seems to spit out B still.

Somebody pm me when they get to the bottom of this. I need to go and finish a WHOLE lot of work.

Kilbaha and methods can get stuffed.

[kenhung123]

Thanks for your all your hard work and help!
I guess we can blame Kilbaha, if they are wrong.

[Martoman]

Right. Kilbaha are wrong. To prove they are wrong use herons formula on the triangle you form. the points which form the triangle of interest are:

(11/5,11/5) (11/7,11/7) and (11/6,0)

side lengths between first two points: 0.888934,
Between 11/5 point and 11/6 = 2.230346
between 11/7 and /6 = 1.593104

herons formula that bad boy in wolfram alpha and you get answer A. Try it out here. http://www.wolframalpha.com/input/?i=Area+of+triangle+herons+formula&a=

Answer A as everyone else got. Kenhung your original working is fabulous and what I used to get my answer. You break it up into two separate integrals.

[8039]

Right. Kilbaha are wrong. To prove they are wrong use herons formula on the triangle you form. the points which form the triangle of interest are:

(11/5,11/5) (11/7,11/7) and (11/6,0)

side lengths between first two points: 0.888934,
Between 11/5 point and 11/6 = 2.230346
between 11/7 and /6 = 1.593104

herons formula that bad boy in wolfram alpha and you get answer A. Try it out here. http://www.wolframalpha.com/input/?i=Area+of+triangle+herons+formula&a=

Answer A as everyone else got. Kenhung your original working is fabulous and what I used to get my answer. You break it up into two separate integrals.

 :smitten:

[kenhung123]

Reuploaded the full version of the insight question. Just having trouble with e and f. If anyone can help that would be greatly appreciated

[Jdog]

yeah for that area question, i broke it up into two parts,

a part before the cusp of the modulus and te one after it.

i got a as well.

[8039]

Reuploaded the full version of the insight question. Just having trouble with e and f. If anyone can help that would be greatly appreciated

I'm gona work on this question in full when I get back from school. But maybe they want the distance from marker A and the wave to be equal to 0? maybe somehow make it that the maximum of the equation is far enough to reach marker A? I'll get back to this when I read it properly... if it does have something to do with the maximum then it would make sense to make t=0 since cos(0)=1 aka the maximum. Is the value of S a negative value by any chance?

[kenhung123]

Yea, I am quite lost because I'm not sure how they produce a maximum a t=0 when they didn't say anything about initial distance is maximum.

[8039]

Yea, I am quite lost because I'm not sure how they produce a maximum a t=0 when they didn't say anything about initial distance is maximum.

Would it by any chance be 19? If I'm right then you just assume it's maximum since that the longest distance from the waters edge to the marker A. We want it to be right on the marker so maximum should be equal to 0. t=0 just makes everything inside the cos equal to 0 which is the maximum value of cos. For a minimum cos is equal to -1.

fail

[kenhung123]

Its 3 :S weird question..

[kenhung123]

May as well upload solutions

[Blakhitman]

Where does it say t=0?