Given the rather large number of homework help threads here in WCT, I'm merging them all into one big thread to consolidate everything.

Need help with your homework, this is the place to ask.

eta (04/22/08) Do note that this doesn't mean posting your homework/project/test and having us do it for you. Feel free to use examples if it'll help you get a better grasp on what you're not understanding, but do not use this thread as a means to cheat your way to better marks. Think of this like a tutor, we help you understand so you can learn to do things on your own, and doing everything for you does nothing of the sort.

January 2012 Update: Sticky!

Okay, I know the question I'm about to ask will seem simple and stupid but it is just one of those days. I don't have time to ask my math teacher because there is a test tomorrow, so please help.

First, let's get this out of the way:
1 / [2(x + 1)] = 1 / (2x + 2)
Right?! If I am already wrong, please tell me.

Okay, then I have to integrate the equation.
∫1 / [2(x + 1)]dx = (1/2) ∫[ 1 / (x +1)]dx = (1/2) ln (x + 1)
Now I'm pretty sure this was done correctly. Correct me if I'm wrong.

This is how I integrated the other form.
∫[1 / (2x + 2)]dx = (1/2) ∫(1/u)2dx = (1/2) ln u = (1/2) ln (2x + 2)
u = 2x + 2 u'= 2
Now what did I do wrong?

I know it must be very stupid mistake because I'm already calculus BC and way past this material. And yes, I double-checked deriving both of my solutions and came back with the original answer. Who can spot out my error? Please Help!!!!

Well, you may not believe this but both equations are correct.
Indefinite integrals are defined only up to a constant. We can manipulate the second equation a little to get the first:
(1/2)ln(2x+2)=(1/2)ln(2(x+1))=(1/2)(ln(x+1)+ln(2))=(1/2)ln(x+1)+ln(2)/2
Which differs from the first equation by a constant.

Oh wow, you're right! So I'm not insane. Thank you so much. Now I can sleep through the night knowing that I don't have a huge problem with basic calculus. It's just so frustrating when you can't make things fit.

well you might have to follow tariff laws, quota laws, like international laws, and i know there is some kind of organization that oversees these kind of actions. maybe something to do with united nations. i was looking through my econ book, but couldnt find much.

maybe anti-trust laws (also known as competition policies)

Prove that g(x, a, 1) =< f(x) for all x and a > 1 and 1 >= lambda =< sqroot(2a-1), where f(x) = (4e^(-a)a^(lambda+a)x^(lambda-1))/(GAMMA(a)(a^lambda + x^lambda)^2).

Sorry for the formatting, but I didn't know how else to write it.

Prove that g(x, a, 1) =< f(x) for all x and a > 1 and 1 >= lambda =< sqroot(2a-1), where f(x) = (4e^(-a)a^(lambda+a)x^(lambda-1))/(GAMMA(a)(a^lambda + x^lambda)^2).

Sorry for the formatting, but I didn't know how else to write it.

Please help!

I'd love to help, but I'm not sure what the definition of g(x,a,1) is.
Also, are x and a real or complex?

I am currently trying to complete a physics lab report. The ultimate goal is to calculate gravity experimentally, via data collected from a simple pendulum with that has simple harmonic movement. (I think that's how you translate it anyway...)

I have the following data:

-Length of the pendulum
-Time it takes the pendulum to complete 20 oscilations (with an angle of less than 4º, represented by "t")
-Period, represented by T
-Period squared, T^2

And I'm supposed to calculate g (gravity) in cm/seconds^2. I'm kind of lost... what should I do?

"All these people having fun are playing for a gigantic 40000$ award, while I'm here sweating, players asking me stupid layer questions we all hate, and I'm thirsty.

I tried that, but I get a weird number (25.something cm/second squared) and I'm supposed to be getting the actual acceleration of gravity... I don't know where I'm going wrong. I mean, I calculated the period like this:

20 oscilations every t seconds, so t/20 = Period

I square that value and then divide the length of the pendulum by it, and get that weird answer...

"All these people having fun are playing for a gigantic 40000$ award, while I'm here sweating, players asking me stupid layer questions we all hate, and I'm thirsty.

I tried that, but I get a weird number (25.something cm/second squared) and I'm supposed to be getting the actual acceleration of gravity... I don't know where I'm going wrong. I mean, I calculated the period like this:

20 oscilations every t seconds, so t/20 = Period

I square that value and then divide the length of the pendulum by it, and get that weird answer...

"All these people having fun are playing for a gigantic 40000$ award, while I'm here sweating, players asking me stupid layer questions we all hate, and I'm thirsty.

"All these people having fun are playing for a gigantic 40000$ award, while I'm here sweating, players asking me stupid layer questions we all hate, and I'm thirsty.

"All these people having fun are playing for a gigantic 40000$ award, while I'm here sweating, players asking me stupid layer questions we all hate, and I'm thirsty.

Ok so I have been sick for a while and I haven't had the chance to find out how to solve these problems but I guess that with an example of each I can solve the rest of the problems I have anyway here goes:

I need to make a graph and show at least 1 period for:
y=cos(x + Pi/4)
y=3 cos x + 3
y= -2 cos(x - Pi/2)

0.719375 for 13cm
0.881375 for 20cm
1.210375 for 36cm
1.622375 for 65cm
1.869625 for 89cm
1.889225 for 93cm
2.416375 for 150cm
2.571125 for 173cm

Kiljae8n: the first one is simple. To whatever value of X you have (in RADIANS, be careful not to use degrees) add Pi/4. In other words, what would usually happen yo Y when X= Pi/4 happens when X = 0 instead.
Second one: Take your original cos graph and add 3 to every Y (move it upwards three units). Afterwards, multiply every Y by 3 (in other words, "stretch" the graph so that it has a greater amplitude)
Third one: combine 1 and 2. NOTE: do NOT raise or lower the graph.

"All these people having fun are playing for a gigantic 40000$ award, while I'm here sweating, players asking me stupid layer questions we all hate, and I'm thirsty.

Okay, I was sick, and then gave blood at blood drive, so I missed some classes, and I'm having a bit of trouble with some chemistry problems:

1) 10.00 grams of hydrocarbon are burned in insufficient oxygen. Not all of the carbon is converted to CO2. 12.86 g of water form along with 22.25g of CO2. The only substance present after the reaction is complete is CO. a) what is the empirical formula of the hydrocrabon? b) how many gram sof carbon monoxide form?

With this one, I can't seem to figure out where to go from, since neither the amounts of oxygen or CO are given.

The other one I'm having trouble with is:
3) Street heroin, C21H23O5N . HCl was diluted with mannitol, C6H14O6. If 1.000 grain of this mixture is completely combusted in pure oxygen gas, 72.50 mL of gaseous carbon dioxide (d = 1.89 g/L) are formed. What is the percentage of heroin in the mixture?

1) 10.00 grams of hydrocarbon are burned in insufficient oxygen. Not all of the carbon is converted to CO2. 12.86 g of water form along with 22.25g of CO2. The only substance present after the reaction is complete is CO. a) what is the empirical formula of the hydrocrabon? b) how many gram sof carbon monoxide form?

For this one, consider that you have the mass of hydrocarbon. This mass is made up of only hydrogen and carbon. If you can figure out how much is hydrogen, then you can easily calculate the mass of carbon, the hydrogen/carbon ratio and thus the empirical formula of the hydrocarbon.

The other one I'm having trouble with is:
3) Street heroin, C21H23O5N . HCl was diluted with mannitol, C6H14O6. If 1.000 grain of this mixture is completely combusted in pure oxygen gas, 72.50 mL of gaseous carbon dioxide (d = 1.89 g/L) are formed. What is the percentage of heroin in the mixture?

For this one, I'm not sure what 1.000 grain means, but I will assume you know the combined mass of the mixture. I would do this one first by writing out your two combustion equations. Then you can use your products side to figure out the amount of carbon in the mixture. Go from there.

I'm having huge trouble on this problem, it's difficult as **** and no matter what I do I get the wrong answer:

2+2

I've tried multiplying 2*2, which gives me something like 5, and I've subtracted 2-2, which gave me 1, and I just can't figure it out. I have a vague idea of what to do, but I just need some help.

Help has come in the form of a bit of basic algebra. I feel that it'll shed some light on your problem here.

Basically:

S + T = W

See? Not too hard, but I'll break it down a bit further. The S in this case stands for 'spam' and the T stands for 'light trolling'. And the W? That stands for 'Warning'. I love math.

Need help with your homework, this is the place to ask.

eta (04/22/08) Do note that this doesn't mean posting your homework/project/test and having us do it for you. Feel free to use examples if it'll help you get a better grasp on what you're not understanding, but do not use this thread as a means to cheat your way to better marks. Think of this like a tutor, we help you understand so you can learn to do things on your own, and doing everything for you does nothing of the sort.

January 2012 Update: Sticky!

Okay, I know the question I'm about to ask will seem simple and stupid but it is just one of those days. I don't have time to ask my math teacher because there is a test tomorrow, so please help.

First, let's get this out of the way:

1 / [2(x + 1)] = 1 / (2x + 2)

Right?! If I am already wrong, please tell me.

Okay, then I have to integrate the equation.

∫1 / [2(x + 1)]dx = (1/2) ∫[ 1 / (x +1)]dx = (1/2) ln (x + 1)

Now I'm pretty sure this was done correctly. Correct me if I'm wrong.

This is how I integrated the other form.

∫[1 / (2x + 2)]dx = (1/2) ∫(1/u)

~~2dx~~= (1/2) ln u = (1/2) ln (2x + 2)u = 2x + 2 u'= 2

Now what did I do wrong?

I know it must be very stupid mistake because I'm already calculus BC and way past this material. And yes, I double-checked deriving both of my solutions and came back with the original answer. Who can spot out my error? Please Help!!!!

Indefinite integrals are defined only up to a constant. We can manipulate the second equation a little to get the first:

(1/2)ln(2x+2)=(1/2)ln(2(x+1))=(1/2)(ln(x+1)+ln(2))=(1/2)ln(x+1)+ln(2)/2

Which differs from the first equation by a constant.

Most lipids in the average human diet are:

A) unsaturated fatty acids.

B) saturated fatty acids.

C) glycerophospholipids.

D) triacylglycerols.

Any ideas which one it is? I've been doing research and am getting conflicting info.

I've wrote that you must follow that countries laws, but I cant find anything else. Help!!

maybe anti-trust laws (also known as competition policies)

haves/wantslist:Thanks Zaphod of High~Light Studios for my sweet SIG

!

Prove that g(x, a, 1) =< f(x) for all x and a > 1 and 1 >= lambda =< sqroot(2a-1), where f(x) = (4e^(-a)a^(lambda+a)x^(lambda-1))/(GAMMA(a)(a^lambda + x^lambda)^2).

Sorry for the formatting, but I didn't know how else to write it.

Please help!

and tax laws too (not just local but the foreign countries tax laws that you're doing business too)

a trip to your local custom office will get you most of the answers that you need

cookie wizards of the the simic

The extendo siggy thingy currently dead

I'd love to help, but I'm not sure what the definition of g(x,a,1) is.

Also, are x and a real or complex?

GAMMA is the gamma distribution.

x and a are real.

I am currently trying to complete a physics lab report. The ultimate goal is to calculate gravity experimentally, via data collected from a simple pendulum with that has simple harmonic movement. (I think that's how you translate it anyway...)

I have the following data:

-Length of the pendulum

-Time it takes the pendulum to complete 20 oscilations (with an angle of less than 4º, represented by "t")

-Period, represented by T

-Period squared, T^2

And I'm supposed to calculate g (gravity) in cm/seconds^2. I'm kind of lost... what should I do?

20 oscilations every t seconds, so t/20 = Period

I square that value and then divide the length of the pendulum by it, and get that weird answer...

20 oscilations every t seconds, so t/20 = Period

I square that value and then divide the length of the pendulum by it, and get that weird answer...

That gives me g = around 0.039 cm/t^2.

I don't know how else to solve for g using the data I have from the lab... should I just turn it in like this?

I need to make a graph and show at least 1 period for:

y=cos(x + Pi/4)

y=3 cos x + 3

y= -2 cos(x - Pi/2)

0.719375 for 13cm

0.881375 for 20cm

1.210375 for 36cm

1.622375 for 65cm

1.869625 for 89cm

1.889225 for 93cm

2.416375 for 150cm

2.571125 for 173cm

Kiljae8n: the first one is simple. To whatever value of X you have (in RADIANS, be careful not to use degrees) add Pi/4. In other words, what would usually happen yo Y when X= Pi/4 happens when X = 0 instead.

Second one: Take your original cos graph and add 3 to every Y (move it upwards three units). Afterwards, multiply every Y by 3 (in other words, "stretch" the graph so that it has a greater amplitude)

Third one: combine 1 and 2. NOTE: do NOT raise or lower the graph.

1) 10.00 grams of hydrocarbon are burned in insufficient oxygen. Not all of the carbon is converted to CO2. 12.86 g of water form along with 22.25g of CO2. The only substance present after the reaction is complete is CO. a) what is the empirical formula of the hydrocrabon? b) how many gram sof carbon monoxide form?

With this one, I can't seem to figure out where to go from, since neither the amounts of oxygen or CO are given.

The other one I'm having trouble with is:

3) Street heroin, C21H23O5N . HCl was diluted with mannitol, C6H14O6. If 1.000 grain of this mixture is completely combusted in pure oxygen gas, 72.50 mL of gaseous carbon dioxide (d = 1.89 g/L) are formed. What is the percentage of heroin in the mixture?

Thanks in advance for any help.

For this one, consider that you have the mass of hydrocarbon. This mass is made up of only hydrogen and carbon. If you can figure out how much is hydrogen, then you can easily calculate the mass of carbon, the hydrogen/carbon ratio and thus the empirical formula of the hydrocarbon.

For this one, I'm not sure what 1.000 grain means, but I will assume you know the combined mass of the mixture. I would do this one first by writing out your two combustion equations. Then you can use your products side to figure out the amount of carbon in the mixture. Go from there.

2+2

I've tried multiplying 2*2, which gives me something like 5, and I've subtracted 2-2, which gave me 1, and I just can't figure it out. I have a vague idea of what to do, but I just need some help.

Help has come in the form of a bit of basic algebra. I feel that it'll shed some light on your problem here.

Basically:

S + T = W

See? Not too hard, but I'll break it down a bit further. The S in this case stands for 'spam' and the T stands for 'light trolling'. And the W? That stands for 'Warning'. I love math.