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Old February 12th, 2009, 06:08 AM   #31
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Default Re: HOMEWORK HELP THREAD!
I already did that.

With a diagram.
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Old February 12th, 2009, 06:13 AM   #32
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Default Re: HOMEWORK HELP THREAD!
Quote:
Originally Posted by Ken View Post
I already did that.

With a diagram.
typing on a psp takes forever
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Old February 12th, 2009, 06:13 AM   #33
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Default Re: HOMEWORK HELP THREAD!
Quote:
Originally Posted by Ken View Post
I don't know why you're having trouble with that. The answer is 50%.


You should have learned these box-like diagrams.
Punnett squares? Yeah suck at that too, sort of.
alright thanks.
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Old February 12th, 2009, 09:45 AM   #34
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Default Re: HOMEWORK HELP THREAD!
How do you suck at them, it's like:

draw a square
divide it into four smaller squares
write the first genotype down the left side
write the second genotype across the top
multiply
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Old February 13th, 2009, 01:34 AM   #35
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Default Re: HOMEWORK HELP THREAD!
okay ken this one's for you

take the indefinite integral of: x(x+4)^1/2

the answer is (2/5)(x+4)^(5/2) - (8/3)(x+4)^(3/2) + c

these are usually pretty easy but I dunno how the fuck the book got that answer.
so far all I have is that
u = x+4
du = 1

I tried having u = (x+4)^(1/2) then du be (1/2)(x+4)^(-1/2) but that doesn't seem to work either.

[edit] okay I got a little further.
u = x + 4
x = u - 4

so I have (u-4)(u)^(1/2)

stuck again!
Last edited by stephaknee; February 13th, 2009 at 01:44 AM.
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Old February 13th, 2009, 01:38 AM   #36
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Default Re: HOMEWORK HELP THREAD!
Wtf I already have an exact thread like this, with the exact same title!!! That's not fair.
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Old February 13th, 2009, 01:55 AM   #37
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Default Re: HOMEWORK HELP THREAD!
Quote:
Originally Posted by Gatorade View Post
Wtf I already have an exact thread like this, with the exact same title!!! That's not fair.
Dogen was inspired by me and stole my idea for a homework thread which needs to be stickied.
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Old February 13th, 2009, 01:58 AM   #38
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Default Re: HOMEWORK HELP THREAD!
Quote:
Originally Posted by Ziz View Post
Dogen was inspired by me and stole my idea for a homework thread which needs to be stickied.
So did you have a homework thread before me and dogen? I made mine like a month ago... no, more like 2 weeks ago. I don't remember exactly when...
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Old February 13th, 2009, 01:59 AM   #39
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Default Re: HOMEWORK HELP THREAD!
Quote:
Originally Posted by Gatorade View Post
So did you have a homework thread before me and dogen? I made mine like a month ago... no, more like 2 weeks ago. I don't remember exactly when...
Who cares answer my hw question.
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Old February 13th, 2009, 02:02 AM   #40
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Quote:
Originally Posted by Ecstasy View Post
Who cares answer my hw question.
I care and I don't know what an indefinite integral is.
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Old February 13th, 2009, 02:19 AM   #41
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Default Re: HOMEWORK HELP THREAD!
Quote:
Originally Posted by Ecstasy View Post
okay ken this one's for you

take the indefinite integral of: x(x+4)^1/2

the answer is (2/5)(x+4)^(5/2) - (8/3)(x+4)^(3/2) + c

these are usually pretty easy but I dunno how the fuck the book got that answer.
so far all I have is that
u = x+4
du = 1

I tried having u = (x+4)^(1/2) then du be (1/2)(x+4)^(-1/2) but that doesn't seem to work either.

[edit] okay I got a little further.
u = x + 4
x = u - 4

so I have (u-4)(u)^(1/2)

stuck again!
Ahaha. I am in the same position at my calc class. But the problems I get screwed on are friggen harder. Its easier to use Integration by Parts but it looks like you are asked using Integration by Substitution. Which is stupid. It should be solved using Parts. Substitution is a bitch to use in this case. It could be a substitution and a parts.

integral (u - 4 )(u)^1/2

w = u - 4
dv = u^1/2
dw = dx
v = 2/3 u^3/2

(u - 4)( 2/3 u^3/2) - integral ( 2/3 u^3/2 dx)

which gives us:

2/3 u ^5/2 - 8/3 u^3/2 - 4/15 u^5/2

which is

2/3 (x + 4)^5/2 - 8/3 (x + 4)^3/2 - 4/15 (x + 4)^5/2

I'm on the right track. It is a substitution and parts problem. I think the issue is early on, picking what u is. I'm gonna give it a few more whirls.
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Old February 13th, 2009, 02:29 AM   #42
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Default Re: HOMEWORK HELP THREAD!
Alright I have to do a Chemistry lab report and I am really confused on one of the questions. We dissolved solid NaOH in HCl and we recorded the initial and final temperature of the solution and calculated the change in temperature (T final- T initial) and it was 9.3 degrees Celsius. The question states:

Calculate the experimental enthalpy change of this reaction in kJ per mole of NaOH.

What do I do to find this?
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Old February 13th, 2009, 02:41 AM   #43
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Default Re: HOMEWORK HELP THREAD!
Quote:
Originally Posted by Ulti View Post
Ahaha. I am in the same position at my calc class. But the problems I get screwed on are friggen harder. Its easier to use Integration by Parts but it looks like you are asked using Integration by Substitution. Which is stupid. It should be solved using Parts. Substitution is a bitch to use in this case. It could be a substitution and a parts.

integral (u - 4 )(u)^1/2

w = u - 4
dv = u^1/2
dw = dx
v = 2/3 u^3/2

(u - 4)( 2/3 u^3/2) - integral ( 2/3 u^3/2 dx)

which gives us:

2/3 u ^5/2 - 8/3 u^3/2 - 4/15 u^5/2

which is

2/3 (x + 4)^5/2 - 8/3 (x + 4)^3/2 - 4/15 (x + 4)^5/2

I'm on the right track. It is a substitution and parts problem. I think the issue is early on, picking what u is. I'm gonna give it a few more whirls.
Thanks, I figured it out (you lead me in the right direction!) Answer, in case you were wondering:

u= x+ 4
x = u-4

(u-4)(u)^1/2

g=u^(1/2)

(u-4)(g) = (gu) -4(g)
= (U^1/2)(u) - 4(u)^1/2
=(u)^(3/2) - 4(u)^(1/2)
= 2/5u^(5/2) - 8/3u^3/2
=(2/5)(x+4)^(5/2) - (8/3)(x+4)^(3/2)
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Old February 13th, 2009, 02:48 AM   #44
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Hm. That is a pretty new way to solve the problem for me. I think I am missing a step <<' Oh well. Tutor will crank out all my issues. Glad I could help you :3
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Old February 13th, 2009, 03:09 AM   #45
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Default Re: HOMEWORK HELP THREAD!
In all honesty, Gatorade, we need 2 threads. Look at how busy it gets.
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