<p>lol ye... bad conceptual error... add vectors requires vertex and arrow connections :</p>
<p>.1 standard deviation on my end as well... the symbol one i just remember i got 3.</p>
<p>for the symbol one was answer choice C = 3??</p>
<p>If C was 3, that's what I got.</p>
<p>^Eh, you prolly can, but with law of cosines you need to remember to use 130. Personally I can't remember physics to save my life and vectors are super foreign, so I just made a triangle and cos'd it. lol.</p>
<p>no but is c 3 my question?</p>
<p>how did u get that answer</p>
<p>please someone was the symbols question answer choice C...plz neone no?</p>
<p>wait for the vector one...weren't the magnitudes 4 and 8? i got sqrt(4^2+8^2-2(4)(8)cos(130)) and got 11 as my answer. did anyone else do this....or did i completely misread the problem? :/</p>
<p>
[quote]
why cant you just find the components of the vectors, add them together, and use the Pythagorean theorem?
[/quote]
I assumed that since it gave you an angle, you'd have to use law of cosines since you don't know the side length of the angle it gave you.
You know two side lengths but not their angles, and you know an angle of the unknown length(the magnitude of the two vectors). So you had to have done the law of cosines to find it.</p>
<p>But, silly me! I was in radians mode!!! I didn't see my answer, so I just did the pythagorean theorem even though I didn't see the answer.</p>
<p>umm it was almost a last minute guess.......so yea i m not sure so thats y i m asking ig ne if u got c it was question 26 the first one in the second column...try 2 remeber</p>
<p>im fairly sure it was 6 and 8 because i remember bubbling in 6.2, which is only attainable with 6 and 8 :)</p>
<p>i dont remember...some people say its 8 and 4 and some say its 6 and 4</p>
<p>I'm 100% sure it was 8 and 4 because I kept doing the problem over and over.</p>
<p>well 6.2 is wrong, i think, lol, an addition of a 6 vector going up at 50 degrees from a horizontal 8 vector can't give a 6.2 vector sum... srsly</p>
<p>The answer choices were 6.2 something and 12? I personally got 11.95; don't know where I went wrong.</p>
<p>Was the period of the weird trig graph 2pi?</p>
<p>haha ye... i got that problem wrong. see my posts above ^.^ XD</p>
<p><em>edit</em> and the trig one indeed had 2 pi period.</p>
<p>awesome; at least I know I got one right</p>
<p>hey u no for the answer with the triagnle hypot was 1.... u al got 1 - cosx rite..? ok...i think i mie have done sin instead of cos by mistake...plz tell me there was no option which said 1 - sin(x)</p>
<p>Yes, it was 2pi. Probability is a bizznitch.</p>
<p>I think there was a 1 - sin^2 x but not 1-sin x? This is the polar one, right?</p>