Oh hello
This's just strange. We did a small project "Making ice-cream" You know, put milk and sugar and vanilla into a zimploc, place the pack into a bigger one stuffed with ice and salt. We've luckily gone through the questions he asked. But the last question just doesn't make sense to me...
Okay, thanks Dextercioby. Viet dao, you meant that I compute incorrectly? Hmmm let me do it again.
Or, we can 'break out' the absolute by devide into two situation: x-1 >= 0 or x-1 <0 it's long, but not so confusing.
Hello,
Well, I have the equation:
x^2-2x-m|x-1|+m^2=0 (1) They ask me what m is so that the equation has solution(s)
Well, I let t=|1+x| and solve the problem with t (no more x) and I have the last equation: t^2-mt+m^2-1=0 (2)
in order that (1) has solution(s), (2) must have solution(s) t...
Ah, by the way, We have:
If \Delta =0, the subtance is not 'full' so; it has no \pi_{cc} linkage, no 'cycle.
If \Delta =1, the subtance has 1 \pi_{cc} linkage OR 1 'cycle'
If \Delta =2, the subtace has 2 \pi_{cc} linkages OR 1 \pi_{cc} linkage + 1 'cycle' OR 2 'cycles'
So, what is this...
Well, hello!
I was informed a new formula. As a non-native, I can't give out the definitons, I hope that you can help me call some.
Well, here is the formula:
\Delta=\frac{2+\Sigma n_i (V_i-2)}{2}
Well, as far as I can say, \Delta here is equal to the total of the linkage C=C+ number of...
Ohh, sorry, I'm so clumsy, I didn't mean to draw a straight line connecting M and N, there was nothing connecting them at all. yes, they meant to cross the resistances.
And that's helped. Thanks!
BTW, I got difficulty when I need to go down for the next line, what's the syntax in LaTex for...
Well, I'm quite idle in Lunar New year's holidays so I decide to do some exercises on Physics, I got stuck when solving the problem, then I tried with the suggestion on my text book, but still, I don't understand the problem, please help me out! btw, sorry, something has gotten on LaTex or I'm...
Uhmm... alright!
Just imagine that sinx = \alpha
Then the thing above is equal to:
\lim_{x\rightarrow0}\frac{ln(1+\alpha)}{\alpha} (*)
Then, this limit is a consequence limit of e:
\lim_{x\rightarrow\infty}(1+\frac{1}{x})^x = e
So. (*) =1
Hmmm... So far, I've done it. But I think tis way is better:
\lim_{x\rightarrow0} \frac{ln(1+sinx)}{x}
=\lim_{x\rightarrow0} \frac{ln(1+sinx)}{sinx} . \frac{sinx}{x}
=1.1 = 1
Is it right?
Well, hello!
I have difficuties solving a problem.
Well, the first 3 ones are done successfully:
a. Calculate out the lim (x -> inf) of (1+2/x)^x
But assuming that 2/x = 2/t, I got the result to be e^2
b. Do the same thing with the lim (x ->0) of (1+3x)^(1/x)
Well, yeah "do the same...