Recent content by Barchie

Fine so if the losses from ventilation never outweighed the heat being provided by the candle would the room eventually one day reach the temperature of the flame? (if the candle burned forever)

Hey All, Maybe a silly question, but... Is there a limit to how much a certain element can heat a body of air, say a room or an oven? Maybe more clear with an example. If i had a candle lit in the corner of my room for ever, would the temperature of the room continuously rise (slowly)...

Update - I have got my answer the long winded way, it took me about 15/20 finding the factors and doing synthetic division for 1,-1,2,-2,3,-3 ... then i found (x-3) to be a factor. after that i was left with x^4+45x^2+324 - to which luckily i found first try 9 and 36 to be factors that multiply...

Perhaps i need to brush up on exactly what it is, but i have always been under the impression i was using the rational root theorem to solve these types of problems, however the method i would usually undertake would be to 1. Get a list of numbers for set p which would be factors of the...

Homework Statement Consider the polynomial p(x) = x^5+45x^3+324x-3x^4 - 135x^2 - 972 . Given that p has some integer roots on the real and imaginary axes, factorise p into linear and quadratic factors with real coefficients. Enter your answer a set of factors in the form { x-1, x+4...

ok no worries. have resolved this issue. for absolute value |x^2-9| i factor to |(x+3)(x-3)| for x < 3 | -(x+3)(x-3) which evaluates for 3- to: (-x-3)(x-3)/x-3 | cancels to: -x-3 lim x-> 3- = -6 for the 3+ (x+3)(x-3)/(x-3) cancels to: x+3 lim x-> 3+ = 6 and if the limits...

Homework Statement For f(x) = |x^2-9|/x-3 and a=3, discuss the limiting behavior of f(x) as x->a+, as x->a- and as x->a The Attempt at a Solution Am i right to say that the limit for x-> 3 is undefined? I am also a little confused with the a+ and the a- for a+ i would say that 3+^2-9/3+-3 =...

Thanks dude :-) Got the answer now! Appreciate the response.

ima douche.. been looking at this one for way too long now! thanks a heap for your help matey :)

So if K = to .5 instead of 1 then popping it back into c= 1/-k makes c = -2 if c = -2 k =.5 Then putting it back into the first equation to check still doesn't get me the right answer: x = 1/k(t-c) x = 1/.5(2--2) x = 1/2 I must have missed something :-(

aha... .5 = 1/2k+1 0.5(2k+1) = 1 k+.5 = 1 k = 1-.5 k = 0.5 But if i sub this into the original formula x = 1/k(t-c) x = 1/.5(2--1) x = 1/.5*3 unfortunately still not getting it :( This still doesn't work

My bad, think i got a little excited. ok... My attempt again: Equation#1: 1 = 1/k(0-c) Equation#2: .5 = 1/k(2-c) Rearrange Equation 1 to get: c = 1/-k Sub into Equation#2 to get: 0.5 = 1/k(2-c) 0.5 = 1/2k-kc 0.5 = 1/2k-k(1/-k) 0.5 = 1/2k+1 k = 1 Therefor c = 1/-1 = -1 So i figured i...

Ok i think I am onto what you are saying, are you able to let me know if I am on the right track. 2 Equations: 1: 1 = 1/k(0-c) 2: .5 = 1/k(2-c) Rearrange the first one to equal: 1 = 1/-c -c = 1 c = -1 Substitute c into eq#2 to give: .5 = 1/k(2+1) .5 = 1/(2k + k) .5(2k+k) =...

Ok :-) With that, 1 = 1/k(0-c) I get: 1 = 1/-kc and, am unsure how to get one of the 2 variables i don't know. If i re-arrange i always have 2 unknowns stopping me from coming to an answer.

Awesome :-) Managed to get the answer. Unfortunately i am having an issue now with the second question, where i am asked to find k and C, given that the initial concentration of x is 1.0 and it is half consumed in 2 seconds. Im a little bit lost here, so really detailed explanations would be...