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Author Topic: Mathematics Help Thread  (Read 229020 times)

MagmaMcFry

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Re: Mathematics Help Thread
« Reply #945 on: February 25, 2013, 01:59:46 pm »

I have figured out a rough estimate of God's minimum power in watts, based on the size of the Revelation city and the mention that God has enough glory to brighten the city up like day, and a few simplifications (mainly that the city just happens to be a perfectly flat plane :P).
Wouldn't that just be city area times sunlight irradiance?
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MonkeyHead

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Re: Mathematics Help Thread
« Reply #946 on: February 25, 2013, 03:11:10 pm »

Treat it as a perfect black body (which God would surely be able to create) and apply Stefans Law.

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Re: Mathematics Help Thread
« Reply #947 on: February 25, 2013, 05:05:30 pm »

Hot black bodies are not the most efficient method of illumination.

And even with pure visible light illumination who will be taking away all the waste heat (unless all objects in that city are 100% white and light is "recycled")?
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RedWarrior0

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Re: Mathematics Help Thread
« Reply #948 on: March 12, 2013, 01:57:34 pm »

I'm trying to figure out the volume of a cone using a triple integral, but I seem to be failing in my limits of integration. It's a right circular cone with the tip at the origin and base of y^2 + z^2 = 5 at x=5, and somehow I'm getting 1000/3...
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MagmaMcFry

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Re: Mathematics Help Thread
« Reply #949 on: March 12, 2013, 02:46:42 pm »

I'm trying to figure out the volume of a cone using a triple integral, but I seem to be failing in my limits of integration. It's a right circular cone with the tip at the origin and base of y^2 + z^2 = 5 at x=5, and somehow I'm getting 1000/3...
intX(0, 5,
intY(-x, x,
intZ(-sqrt(x²-y²),sqrt(x²-y²),
  1
))),

or alternatively
intX(0, 5,
intTheta(0, 2pi,
intR(0, X,
   r
))),

or even more alternatively
intX(0, 5,
  pi*x²
),

or simply 1/3 * 5 * (pi*5²).
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WealthyRadish

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Re: Mathematics Help Thread
« Reply #950 on: March 12, 2013, 05:04:11 pm »

Stop making me miss calc III! D:

Lack of math has been killing me lately.
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Guardian G.I.

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Re: Mathematics Help Thread
« Reply #951 on: March 28, 2013, 12:45:27 pm »

I need to solve an equation:


Apparently, the algebraic identity can't be applied here.

(my knowledge of English mathematical terminology is minimal; if I sound vague, sorry about that!)
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Jim Groovester

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Re: Mathematics Help Thread
« Reply #952 on: March 28, 2013, 12:49:01 pm »

Is that capital X the same variable as the x in the exponent?
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Guardian G.I.

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Re: Mathematics Help Thread
« Reply #953 on: March 28, 2013, 12:50:03 pm »

Is that capital X the same variable as the x in the exponent?
Yes.
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MagmaMcFry

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Re: Mathematics Help Thread
« Reply #954 on: March 28, 2013, 12:50:48 pm »

I need to solve an equation:


Apparently, the algebraic identity can't be applied here.

(my knowledge of English mathematical terminology is minimal; if I sound vague, sorry about that!)

Your equation transforms to log_10(x) = log_x(10), which translates to ln(x)/ln(10) = ln(10)/ln(x) or ln(x)² = ln(10)², so ln(x) = +-ln(10); therefore x = 10 or x = 1/10.
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MonkeyHead

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Re: Mathematics Help Thread
« Reply #955 on: March 28, 2013, 12:56:41 pm »

log10 x = logx 10

1/log10 x - log10 x = 0

Simplify to give logx = 1 or logx = -1

Therefore x=10 x=1/10


Damn, Ninja'ed.

Guardian G.I.

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Re: Mathematics Help Thread
« Reply #956 on: March 28, 2013, 12:58:21 pm »

I need to solve an equation:


Apparently, the algebraic identity can't be applied here.

(my knowledge of English mathematical terminology is minimal; if I sound vague, sorry about that!)

Your equation transforms to log_10(x) = log_x(10), which translates to ln(x)/ln(10) = ln(10)/ln(x) or ln(x)² = ln(10)², so ln(x) = +-ln(10); therefore x = 10 or x = 1/10.
ln(x) means natural logarithm log_e(x), am I right?

Also, I don't understand, how does x^(lg(x)) transform into log_10(x)? Figured that out by myself.
« Last Edit: March 28, 2013, 01:07:03 pm by Guardian G.I. »
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MonkeyHead

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Re: Mathematics Help Thread
« Reply #957 on: March 28, 2013, 12:59:26 pm »

Yep.

Guardian G.I.

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Re: Mathematics Help Thread
« Reply #958 on: March 28, 2013, 01:15:26 pm »

log10 x = logx 10

1/log10 x - log10 x = 0

I don't understand, how did you transform log10 x = logx 10 into 1/log10 x - log10 x = 0?
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da_nang

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Re: Mathematics Help Thread
« Reply #959 on: March 28, 2013, 01:25:35 pm »

log10 x = logx 10

1/log10 x - log10 x = 0

I don't understand, how did you transform log10 x = logx 10 into 1/log10 x - log10 x = 0?
Change the base from x to 10: logx(10) = lg(10)/lg(x) = 1/lg(x). Thus the equation becomes lg(x) = 1/lg(x). After that it's merely algebra.
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