Recent content by Rodrae

Could you explain why it is parallel and not series parallel. Is Z3 and Z4 is series? or not? Why? Can you give me ways how to determine if a circuit is parallel or series by just looking on it? Is the flow of current has something to do with determining the kind of circuit even there is no...

I KEEP TELLING INTO HIM ( NOT MENTIONING THE NAME ) THAT THIS IS A PARALLEL CIRCUIT BUT HE ALWAYS KEEP TELLING ME THAT THIS IS A SERIES-PARALLEL... SOMEONE TELL ME. WHAT CIRCUIT IS IT? IS FLOW OF CURRENT DETERMINE THE KIND OF CIRCUIT? PLEASE ALSO INDICATE PROOF THAT THIS IS A _______...

I keep telling into him ( not mentioning the name ) that this is a parallel circuit but he always keep telling me that this is a series-parallel... Someone tell me. What circuit is it? Is flow of current determine the kind of circuit? Please also indicate proof that this is a _______...

Just a simple explanation. Calculus might increase our ability to analyze things and IQ. It also test our own limits:smile:

You might do that.. Sometimes there are things that is written in a book, etc. that is really confusing to understand. Maybe a simple explanation might be helpful when you are just starting calculus rather than those with lots of formula. Like those that I red before about the law of derivatives...

ok. Thanks a lot

So therefore you mean that ... for example y= \sqrt{(cos180)(sin270)} = \sqrt{(-1)(-1)} = \sqrt{(-1)^2} and we can't say (-1)2(1/2) = (-1)2/2 = (-1)1 = -1 and you say that it is absolute value so it will =1 and will never be equal to -1? and\sqrt{x^2} is just like |x| so...

y = \sqrt{x^2- 2x + 1} The y' is always the slope right? Then if we simplify the equation y =\sqrt{(x-1)^2 }----or---y = \sqrt{(1-x)^2} y = x - 1 y = 1-x Checking:---------------Checking: \sqrt{(x-1)^2}-------------\sqrt{(1-x)^2}...

Ok. Thanks a lot :smile:

In 1/x y can be negative if x is negative and goes closer and closer but will never reach the graph y=0 and x=0 but in 1/|x| y will never be a negative no because |x| is always positive the graph goes closer and closer but will never reach to x=0 and y=0

1/|x| is discontinuity at zero and non removable... because we couldn't simplify the equation to remove x w/c is 0 The graph will move closer but never reach x=0

ok thanks a lot

/x/ = y is non removable discontinuity x but is /x/^2 = y is a removable discontinuity or not? and why? x^2