The math help thread

Halp anyone? ;_;

Mean Value Theorem states that there must exist some value c in [4, 8] for which f'© equals the mean value of f'(x) over [4, 8], i.e. the slope from 4 to 8, rise over run: f(8)-f(4) / 8-4.

So, for any value of c:
-5 ≤ f'© ≤ 3
-5 ≤ f(8)-f(4) / 4 ≤ 3
-20 ≤ f(8)-f(4) ≤ 12

RIGHT.
f'© = (f(b) - f(a)) / (b - a)

SOOO...
-5 < (f(8) - f(4)) / (8 - 4) < 3.
Then just multiply both sides by the denominator.

As long as I can do that question on the exam, I'm good.
Thanks!

EDIT: FFS the Webwork closed at 10:30. Every online section has ALWAYS closed at 11:30. FFFFFFFFFFFFFFFFFFFFFFf.

RIGHT.
f'© = (f(b) - f(a)) / (b - a)

SOOO...
-5 < (f(8) - f(4)) / (8 - 4) < 3.
Then just multiply both sides by the denominator.

As long as I can do that question on the exam, I'm good.
Thanks!

EDIT: FFS the Webwork closed at 10:30. Every online section has ALWAYS closed at 11:30. FFFFFFFFFFFFFFFFFFFFFFf.

That looks kinda like Lagrange something. (didn't learn maths in English)
My bad, its more common name is indeed Mean Value Theorem. Darn 3rd world country's math.

^ Lagrange error bound is something else, which has to do with approximating integrals using Simpson's Rule if I recall.

2^(n^2)-2^n=5


Solve for n algebraically.

Couldn't you just make the 5 a base of 2? I dunno, but that's the first thing I thought of.

2^x=5
x log 2 = log 5
x = log5 / log2

2^(n^2) - 2^n = 2^(log5/log2)

And since the bases are the same..

n^2 - n = log5/log2 and solve?

I'm under the impression you can't take a log of two terms being added or subtracted and break them up.
Tried your way out on my calculator, and came up with n = 1 +- zi, where z is some number I didn't want to copy down.
I know this has real solutions, via graphing.

Yeah. I just graphed both of the equations and they came out different.. <_< Dunno why though. I thought I just substituted 5 with 2^(log5/log2) but I guess not.

ln ( 5 + 7) != ln 5 + ln 7
:[
I remember my teaching talking about this with some imaginary trig (sin x = 2), using hyperbolic trig functions. Ended up with the same thing, I think, except with base e. Ended up doing a quadratic formula on it, although I can't remember the details....

2^(n^2)-2^n=5


Solve for n algebraically.

Where


Noah

Putting in the two solutions from that into the equation, I get -2.144 and 14.87. Not really that close to 0. o.0

Putting in the two solutions from that into the equation, I get -2.144 and 14.87. Not really that close to 0. o.0

Then you have probably misinterpreted. I get 2.10 and -1.10 as results.

Keep in mind that



which approximates to 10.29.

Noah

Graphed 2^(n^2)-2^n-5=0 and found the positive zero with the graphing calculator's zero thing. 1.75. o.0
Seems a bit off from 2.10.

Are you sure you can take a log of each term on one side like that?
ln(5+7)-(ln(5)+ln(7))=-1.07. That doesn't seem equal to me.

Yeah, you can't do that.

Having said that I don't know what you can do, except factor out a 2^n on the left side which leaves you at a dead end as far as I know.

let a=2^n
thus the equation a^2-a=5
solve a (do it urself i'm too lazy)
then n= log a/ log 2

let a=2^n
thus the equation a^2-a=5
solve a (do it urself i'm too lazy)
then n= log a/ log 2

It's 2^(n^2), not (2^n)^2.

Unless it is (2^n)^2, in which case that does work.

Graphed 2^(n^2)-2^n-5=0 and found the positive zero with the graphing calculator's zero thing. 1.75. o.0
Seems a bit off from 2.10.

Are you sure you can take a log of each term on one side like that?
ln(5+7)-(ln(5)+ln(7))=-1.07. That doesn't seem equal to me.

This is correct, you cannot apply logarithms like I did. My, my, I seem to go old.

let a=2^n
thus the equation a^2-a=5
solve a (do it urself i'm too lazy)
then n= log a/ log 2

You cannot do it that way. If a = 2^n, then 2^(n^2) = a^n. (By the way: Are you 100% sure it's 2^(n^2) and not (2^n)^2? )

[spoiler=Scrabbles, most won't likely help any]
I'll see what I'll be able to do for tomorrow, as it's gettin late over here[/spoiler]

Noah

*facepalm*
There's a difference between 2^(n^2) and (2^n)^2?

Need some help on this one problem.

A winch at the top of a 12 meter building pulls a pipe of the same length to a vertical position, as shown in figure 29. The winch pulls in rope at a rate of -0.2 meters per second. Find the rate of vertical change and the rate of horizontal change at the end of the pipe when y = 6.

Here's a picture

http://i33.tinypic.com/n2xut2.jpg


I got root21/30 for dy/dt and -root7/30 for dx/dt.
Is this correct? I didn't feel like using law of cosines, so I dropped two perpendiculars, one to the building and one to the floor. I got that for when y = 6, s = root84 and x is root108. Can anyone confirm for me?

I found some answers online that say dy/dt is 1/5 and dx/dt -root3 / 15