Sarahisme

lol, anyone at all?

lol, cheers DQW!

so what do peoples think?

ok i've got another method....here we go We want to find a [math] \delta > 0 [/math] (depending on [math] \epsilon [/math]) so [math] |x-4| < \delta \implies |h(x)-h(4)| < \epsilon [/math] [math] i.e. |\sqrt(x) - 2| < \epsilon [/math] Note [math] |\sqrt(x) - 2| = \frac{|x - 4|}{|\sqrt(x) + 2|} [/math] Choose [math] \delta [/math] so that [math] \delta\frac{1}{|\sqrt(x) +2|} < \epsilon [/math] Note [math] |x-4| < \delta \implies |x| < 4 + \delta [/math] Choose [math]\delta[/math] to be < 1 then [math] |x| < 5 [/math] Then choose [math] \delta = min(1,\epsilon(\sqrt(5) + 2)) [/math] Then [math] |x-4| < \delta \implies |x|<=5, \ Since \ \delta<=1 [/math] Then [math] |\sqrt(x) - 2| = \frac{|x - 4|}{|\sqrt(x) +2|} [/math] [math] <\delta\frac{1}{\sqrt(5) + 2} <= \epsilon, \ since \ \delta <=\epsilon(\sqrt(5) +2) [/math] Q.E.D

...although i dunno how part 2 is going to work...

ok so far this is what i have gotten: [math] |\sqrt(x)-\sqrt(4)|=|\sqrt(x)-\sqrt(4)|\times\frac{|\sqrt(x)+\sqrt(4)|}{|\sqrt(x)+\sqrt(4)|}=\frac{x-4}{|\sqrt(x)+\sqrt(4)|} [/math] Let: [math] \frac{3}{2}<x<\frac{5}{2} [/math] Then [math] \frac{x-4}{|\sqrt(x)+\sqrt(4)|}\leq|x-4|\times\frac{1}{|\sqrt(\frac{3}{2})+2|}=|x-4|\times\frac{1}{\sqrt(\frac{3}{2})+2} [/math] So [math] |\sqrt(x)-\sqrt(4)|\leq|x-4|\times\frac{1}{\sqrt(\frac{3}{2})+2} [/math] Let [math] |x-4|\times\frac{1}{\sqrt(\frac{3}{2})+2}<\epsilon [/math] [math] \therefore |x-4| < (\sqrt(\frac{3}{2})+2)\epsilon [/math] So choose [math] \delta = (\sqrt(\frac{3}{2})+2)\epsilon [/math] [math] \therefore |x-4| < \delta \implies |h(x) - h(4)| < \epsilon [/math] well, yep, hows that?

hey, its been awhile since i could do these, could i have a little bit of help, if there is someone willing out there? Cheers Sarah

oh so you have to have all this array stuff...?

ahh never mind, win some, lose some , its all good, thanks for the offer MetaFrizzics

oh right, oops , thanks

i take that as an, everything is perfectly correct??

cool! thanks DQW!

...

ok here is question and my answers...: [math] \lambda=2L=2\times0.580m=1.16m [/math] [math] \mu=\frac{M}{L_(wire)}=\frac{0.00225kg}{1.00m}=0.00225(kg/m) [/math] [math] v=\sqrt{\frac{F}{\mu}}=\sqrt{\frac{29.4N}{0.00225}}=114.3(m/s) [/math] [math] f=\frac{v}{\lambda}=\frac{114.3(m/s)}{1.16m}=98.54 \ Hz=98.5 \ Hz [/math] how'd i do? -Sarah

just wanted to make sure i am doing this correctly.... ok my answers are: (1) [math] \lambda_n=\frac{2L}{n}, \ n=1,2,3,... [/math] (2) [math] v=\sqrt{\frac{F}{\mu}} [/math] [math] \mu=\frac{M}{L} \ \ \ \ (mass \ per \ unit \ length) [/math] [math] f=\frac{v}{\lambda}=\frac{n\sqrt(\frac{F}{\mu})}{2L}=\frac{n\sqrt(F)}{2L\sqrt(\mu)} [/math] ok yep thats it (hopefully the latex stuff works, i cannot see it on my computer at the moment, so just tell me if it shows up incorrectly (or not at all) ) Cheers Sarah

right ok, yep that makes sense i think, thanks Yggdrasil

umm here is the question "Here V is a nonzero finite-dimensional vector space: If dim V = p and if S is a linearly dependent subset of V, then S contains more than p vectors. True or False?" Now i am not sure , i could argue either way, although the answers says that it is false. i think it is false because it doesnt say that S is a basis, and so S can have less than p vectors and still be a subset of V. on another note, can a z-dimensional vector space V have a basis with less than z vectors in it? (or is this a related note? )

oh right, yep of course, sorry yeah i get it now thanks matt

what this about?

hmmm...in the hint it says to use the comparison theorem. i don't see why we would really need to, but anyway

oops sorry yes thats obviously true, i just keep 'typing my mind' as it were, ignore that last post

this set is 3-dimensional, right? because there would need to be -1, t and t^{2}, right?

oh ok, yep, i see.

for part (b) again use the equation for adding relativistic velocites together [math] u=\frac{u^{'}+v^{'}}{1+\frac{u^{'}v^{'}}{c^{2}}} [/math] So u = the speed of one ship relative to the other. [math] u=\frac{30000+30000}{1+\frac{(30000)^{2}}{c^{2}}} =\frac{60000}{1\times10^{-8}} =59999.9994(m/s) =6.0000\times10^{4} [/math] and the final answer should be too 5 sig. figures, as the information given in the question (for part b) is to 5 sig. figs. so what did people think of all that? i am pretty damn sure its right!

for part (a) use the equation for adding relativistic velocites together [math] u=\frac{u^{'}+v^{'}}{1+\frac{u^{'}v^{'}}{c^{2}}} [/math] So u = the speed of one ship relative to the other. [math] u=\frac{0.6c+0.6c}{1+\frac{(0.6c)^{2}}{c^{2}}} =\frac{1.2c}{1.36} =0.88c = 0.9c [/math] oh by the way, how do you make an equation go to the next line using latex? eg. u=42x = ... =.... etc.