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#1 Re: Help Me ! » **_ Parametrics _** » 2005-08-25 14:05:22

use the focus directrix form of the eqn of a parabola:

x² = 4py (or y² = 4px for the horizontal case); note that p = the distance from the vertex to the focus = the distance from the vertex to the directrix

now, mathsyperson has already done most of the work!

take x² = 16y = 4(4y), so we see p = 4.

this means our parabola (with vertex (0,0) ) has focus (0, 4) and directrix y = -4

#2 Re: Help Me ! » I was doing alg 2 hw, and stumbled upon Geometry problem Help please » 2005-08-25 13:47:44

sounds like the answer's supposed to be:

A = pi (y+5)²/4 = (pi/4) (y² + 10y + 25)

#3 Re: Help Me ! » Need help » 2005-08-25 13:39:18

2sinxcosx + cos^2x = 1

=>

2sinxcosx - sin^2x = 0 (bc of the pythagorean identity)

=>

(2cosx - sinx)sinx = 0

=>

2cosx=sinx or sinx = 0

=>

tan x = 2 or sin x = 0

so x = arctan 2 or x = n(pi), where n is any integer.

#4 Re: Help Me ! » Hep with Trig. please :) » 2005-08-25 13:31:05

ahgua, do not worry about dividing by cos x in the (correct) sol'n given by wcy; sin x/cos x is defined as tan x, so we do not need to worry about division by zero, the dfn of tan x already forbids it!

another way to show x=0 is a sol'n was already given by wcy in that soln; the sin x = 0 half of the soln...

#5 Re: Help Me ! » logarthimic operations : order of precedance? » 2005-08-25 13:20:16

right, I missed that ambiguity!

correct, mikau: it doesn't matter which comes first, not even the exponent rule... the only ordering you have to obey is PEMDAS, as MathsIsFun said.

#6 Re: Help Me ! » limits - try this one! » 2005-08-25 13:17:05

darn... I was trying to be too fancy.... replace all the question marks in that above post with the symbol 'infinity' (or '+infinity', if you prefer)

#7 Re: Help Me ! » limits - try this one! » 2005-08-25 13:15:52

First, recall the theorem ( a(n) -> 0 ) => ( (a(n))^(1/n) -> 0 ) , where n -> ∞ in both limits. (the proof of this is pretty straightforward, but I can do it if you like)

Then note the well known fact that (x^n)/(n!) -> 0 as n -> ∞; put this together with the above theorem and get:

x/((n!)^(1/n)) -> 0 as n -> ∞.

Since x is constant in the limit, we must have:

(n!)^(1/n) -> ∞ as n -> ∞, and it's positive infinity since n! > 1 for all n in N.

[replaced ? with ∞ for you - mathsisfun]

#8 Re: Help Me ! » logarthimic operations : order of precedance? » 2005-08-25 11:39:43

if you use them correctly, there is no preferred order; you should be able to work any equation using any approach (i.e. order) you wish. just make sure that you always follow the order of operation rules for arithmetic!

if you would post a specific question where you ran into trouble, I would be more than happy to explain what I'm talking about using your situation as an example...

oh and btw the second one should be: (log m) - (log n) = log (m/n)

#9 Re: Help Me ! » Good Will Hunting needed! - Can anyone solve this problem » 2005-08-25 07:13:10

another thing.... does 'x' in the eq'n signify multiplication? or is it a variable? you have to understand what kylekatarn is trying to say in his post: it is virtually impossible to solve one eq'n in 5 unknowns! you have to tell us which is known, and which is unknown (like wcy said, we have to know whether or not that 'e' you've used is the natural exponential!); i.e., do we solve for a? or some other variable?
I'm getting to think this isn't a serious topic... just like kylekatarn said! especially since you didn't simplify the (8+18-20) part of it before you posted....

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