The Homework Help Thread

[Mr. Wobbels 664]

My failing at stoichiometry or the high concentration?

 

No the concentration. That'd burn a hole through someone's face fairly easily.

[Aureity 3,055]

No the concentration. That'd burn a hole through someone's face fairly easily.

 

Ah wait, one sec. I'm getting like less than 1M now that I redid it with pv = nRT.

 

(P 728.9)(V 1.2) = n(R 62.4)(T 294)

 

According to this, n should be 0.0478...?

[Mr. Wobbels 664]

Ah wait, one sec. I'm getting like less than 1M now that I redid it with pv = nRT.

 

(P 728.9)(V 1.2) = n(R 62.4)(T 294)

 

According to this, n should be 0.0478...?

 

Yeah just fixed it. Apparently PV/nT=R Just doesn't like me.

 

(.984 atm)(1.2 L)=n(.082057 atm*L/mol*K)(294.15 K)

n = .0489 mol H₂

.0978 mol HCl / .8 L = .122 M HCl

 

Looks better. The Ideal Gas Law is a lot more accurate apparently.

[Ragtime 60]

inb4 dimensional analysis

[Aureity 3,055]

inb4 dimensional analysis

 

Too late.

[Nothing392 61]

I have a question for math homework. It actually wasn't assigned but it was one of the next problems on the page I am really curious as to how to do it. I'm only in 8th grade, so it will probably be pretty easy for the rest of you.

 

So it's a rational function:

 

 

I have to graph it, but I'm not really sure what to do. I know it has a vertical asymptote at 2 and -2 and the horizontal asymptote is 1. Beyond that, I have no idea how to make a table for this.

[Aureity 3,055]

I have a question for math homework. It actually wasn't assigned but it was one of the next problems on the page I am really curious as to how to do it. I'm only in 8th grade, so it will probably be pretty easy for the rest of you.

 

So it's a rational function:

 

 

I have to graph it, but I'm not really sure what to do. I know it has a vertical asymptote at 2 and -2 and the horizontal asymptote is 1. Beyond that, I have no idea how to make a table for this.

 

 

 

Well, that turns into f(x) = (x - 1)(x + 1) / (x - 2)(x + 2) if you factor it. Which really isn't that helpful, but I do it anyway just to check for holes in the graph (which you'll learn about later).

 

The horizontal Asymotote, since the numerator's X degree = the denominator's y-degree, is y=1 (from the coefficients, 1/1 = 1)

Vertical Asymptotes, as you can tell by the denominator, are x=-2 and x=2.

 

Since there are no holes, graph those three lines on your paper. You should have something that looks like bars now. Now, basically fill the top-left and top-right, and fill the one bottom-center with curved lines that get closer and closer to the asymptote as x or y approaches infinity/negative infinity, but doesn't ever actually touch the asymptotes. It should looks like some sort of weird trinity thing when you're done.

 

As for the table, just plug in random stuff for X.

For example, if X = 0; then Y = -1/-4 = 0.25. If X = 1; then Y = 0. Just plug in random X values and find their Y-values.

Edited April 13, 2012 by Aureity

[Why-Gnome-Ear-Fifty 400]

WOLFRAM ALPHA XD

 

Need to revive thread for my own benefit, call me selfish XD

 

I need to find the what values of p make the series "sum from 2 to infinity 1/(n^p*lnn)" convergent. I've tried the comparison test, but I can't get the original series to relate to 1/n^p and the limit comparison test doesn't seem to work due to the fact that it's not biconditional. Any help?

[Talemage 47]

Right, first off, I kind of figured that there be different homework threads and not just one thread(which is dead now, by the way) where everyone posts different homework of different subjects.

 

If any mod thinks that this shouldn't exist, then you may remove it, but keep in mind that you will be causing several people to fail in their homework in the future But really, though.

 

Anyway, to start, I have some a little a lot of things to be solved. It isn't homework, but it needs to be done. They're activities that they expect us to do in our free time. "Self Learning Kits" of the "Personalized Education Program," where every grade gets the same handouts.

 

I have this every week, so you might expect more stuff every Monday. It would be nice if you showed how you got the answer, and not just post the answer and not explain how you got it. I'm not trying to be lazy, I'm also trying to answer it on my own(difficult, as ponies are hard to remove from any brony's mind[i'd imagine]).

 

(Brace yourselves, lots of algebraic terms are coming!)

 

Evaluate each expression:

 

3x/5y1 for x= -2, y= 2

- I figure that I need to do something like replace x and y with the given numbers (-2 and 2), but I haven't the slightest on what to do after that.

 

x+y/3 + x+y/3 or x= 15, y= 9

 

 

Perform the indicated operation:

 

3ab + 7ab - 12ab =

- I'm thinking the answer is -2ab. Since all the variables are the same (ab), all I really have to do is solve it as if it weren't there. I don't know, that's what the book says. Not to solve it like the variables aren't there, but to ignore it if it's the same variable for each term, then add it after solving what's left of the question.

 

(x^3 + 5x^2 - 3) - (1 - 2x - 3x^3)

- Methinks the answer is -2 + 5x^2 - 5x^3. If I'm wrong, correct me.

 

Add (x^3 + 5x^2 - 3) to the difference when (-3x^2 + x + 2) is subtracted from (1 - 2x - 3X^3)

- -2x^3+8x^2-3x-4 ?

 

Perform and simplify. Use the special products formulas.

Since the book says nothing about the special products formulas, and my internet is gone as I wrote this, I'd be thankful if you told me what the hay it is.

 

(3k - 1)^2=

 

(3x - 5y)(3x + 5y)

- Assuming I have to multiply both in parentheses once everything in them has been solved.

 

(2h - h)^2=

 

(2x + y + 5)^2=

 

(h/4 -5)(h/4 +5)

 

 

Factor the following completely:

No idea how to do this, will consult the almighty and knowledgeable math book for help. I did do one, however.

 

2(pi)r - 4(pi)^2=

- I think it's 2(pi) (r-2[pi]), though I don't know why, though. You're welcome to explain what I just did which I have no clue as to how I even got to that end in the first place.

 

p(q+r) -x(q+r)=

(q+r)(p-x) ?

 

m^3 + 64=

- This doesn't have any parentheses, so I'm guessing it's easier harder.

 

x^2 - 25=

 

x^2 + 8x + 15=

 

 

Now for part two. Yep, part two because we were supposed to get the items above LAST WEEK, but because they weren't complete yet and they were finished today, we got them today. So we got both parts today. Today because our professor just remembered today.

 

Oh, by the way, we'll be having this every week. Not every other week, not every ten days. Every. Single. Week. Until March, 2013. Good luck coping with all those math stuff

 

Solve each of the following(oh, how I love that sentence~)

I feel like this thread will be closed because it's too long

 

2-m/5 = 3 m= ?

 

5x - 10 = 50 x= ?

Yes, I think it looks a bit easy, but I just want to be sure.

 

5a+5/6a-5 = 1 a= ?

 

5x/3 - 3x/2 = 4 x= ?

 

7(2x - 5) = 3 (4x - 1) +6 x= ?

 

 

And, last for the rest of the week:

 

Find the solution set and sketch the corresponding graph of each of the following inequalities.

I remember a faint part of this from last year. I think the solution set is the set of numbers that, if applied to the given term, would make it correct. That's all I think I can remember, and that's all I can describe.

 

Oh, and this involves < and all that stuff. Also, it involves something like 'not lower than,' which could also refer to a number higher than or the number itself. Look at me, a near fourteen year-old telling a community whose majority is somewhere between the ages of 16 to 20 or higher what something means.

 

I'll be using <= for 'less than or equal to,' and => for 'greater than or equal to.'

 

14 - 6x <= 26

 

-5 < 2x - 1

 

2x - 3 <15

 

8 - 2(x+1) => 2

 

2x - 5 < -3

 

 

That's all. And I'm not tired of typing. I've been writing nearly nonstop today at school(high school, what we call second year).

 

And don't worry, you won't have to solve that on your own. Since you'll probably be awake when I'm asleep, and the majority of you probably don't go to school anymore(correct me if i was wrong there), you'll be able to do more stuff than I will. I'll still be trying to solve this, though. One way or another. At the most, I'll be able to solve the first sets of items today. At the least, nothing at all.

 

This will be changed every time I think I solved one correctly, with my answer in bold.

 

(everything I said above). Um.. that is, if you don't mind... and if you're free.

 

Also, I had to wake up early today just to post this topic. No internet even when I went to sleep.

Edited June 27, 2012 by Midnight Glimmer

[Ezio Auditore 702]

 

Uhh all joking aside I'm not too good at math so I can't help you here. I don't remember anything about algebra even if it was 2 years ago. If you need geomatry help let me know

Edited June 26, 2012 by Ezio Auditore

[MVC NVMXD JVRXD 2,406]

If I have nothing to do later, I might help you out.

[Victoria and Co. 782]

Some answers:

 

3x/5y1 (that's a fraction, not divide) for x= -2, y= 2

-3/5 or -.6

x+y/3 + x+y/3 (both are fractions) for x= 15, y= 9

16

3ab + 7ab - 12ab =

-2ab

(x^3 + 5x^2 - 3) - (1 - 2x - 3x^3)

4x^3+5x^2+2x-4

Add (x^3 + 5x^2 - 3) to the difference when (-3x^2 + x + 2) is subtracted from (1 - 2x - 3X^3)

-2x^3+8x^2-3x-4

(3k - 1)^2=

9k^2-6k+1

(3x - 5y)(3x + 5y)

9x^2-25y^2

(2h - h)^2=

h^2

 

 

Edited June 26, 2012 by Vinyl Pon3 Scratch

[Dreadmallon 353]

Must I repeat my earlier status?

 

"Dear Algebra, stop asking us to find your x. She's not coming back."

[Aureity 3,055]

Feel free to drop me a PM if you'd like any help from me.

[The Awesome One 1,315]

This isn't Algebra of my level so I don't know it, Sorry. I can give you advice though. Knowing maths comes from memorizing all the methods and formulas and practicing and practicing and not stopping practicing. It's pretty easy once you get the hang of it.

[Victoria and Co. 782]

I'm pro at math, I already posted some answers, so if you want the easy way out, feel free to just copy them.

Now, if you want to learn how to do it, just shoot me a PM and I'll be happy to explain.

[Key Gear 6,663]

Well, my brain is too fried to render you direct assistance, but I can make a recommendation for the future. Have you ever heard of Wolfram Alpha? It's a really cool site that is very helpful for all kinds of math problems. Basically, you can put a formula in and receive an answer out. It's no substitute for knowing how to work the problems yourself, but it's a great way to find out if you're on the right track.

 

To use the site, just paste one of your problems in. After some computation, the solution should appear (along with a whole bunch of other stuff). For example, check this out.

Edited June 26, 2012 by Scootacool

[Tom 891]

Well, my brain is too fried to render you direct assistance, but I can make a recommendation for the future. Have you ever heard of Wolfram Alpha?

 

This is very helpful, but it won't always say how it got the answer.

 

Alas when I get the free time I can work these out for you myself, but it's getting late now.

 

Also:

 

 

3x/5y1 (that's a fraction, not divide) for x= -2, y= 2

-3/5

x+y/3 + x+y/3 (both are fractions) for x= 15, y= 9

16

3ab + 7ab - 12ab =

-2ab

(x^3 + 5x^2 - 3) - (1 - 2x - 3x^3)

4x^3+5x^2+2x-4

Add (x^3 + 5x^2 - 3) to the difference when (-3x^2 + x + 2) is subtracted from (1 - 2x - 3X^3)

-2x^3+8x^2-3x-4

(3k - 1)^2=

9k^2-6k+1

(3x - 5y)(3x + 5y)

9x^2-25y^2

(2h - h)^2=

h^2

 

 

 

All these are correct. Edited June 26, 2012 by Tom

[Victoria and Co. 782]

All these are correct.

 

Like a sir.

[PlsDeleteAccount 157]

I did this less than 3 months ago...

 

and I can't remember how to do it again.

 

Bollocks.

[Talemage 47]

Oh, joy.

I just found out that I was told to be fractions are actually supposed to be divided. Why don't they get things straight at the start!?

 

 

 

I'm trying.

 

Well, I still have an incredibly short a good one day left. We're supposed to submit it on Friday.

 

Here's what I've tried:

 

p(q+r)-x(q+r)

(q+r)(p-x)

m^3+64

(m+4)(m^2-4m+16)

x^2-25

(x-5)(x+5)

x^2-8x+15

(x+5)(x+3)

(2x+y+5)^2

4x^2+4xy+20x+y^2+10y+25

 

I see that my answer for some of the ones are already correct, so I'll leave those out.

 

Still need

3x/5y-1 for x= -2 and y= 2

 

(h/4 -5)(h/4 +5)

 

And all of the ones from part two.

 

Again, I'd appreciate it if you could give your way on how you solved it, so I won't just stay dumb and copy it. Vinyl Pon3 Scratch, you're kind of an exception because I found out how you solved most of it.

 

As much as I love solving stuff, I also need a required time to do it. They don't expect me to answer two whole pages of algebra in five days.

[A Blithering Div 247]

Find the solution set and sketch the corresponding graph of each of the following inequalities.

I remember a faint part of this from last year. I think the solution set is the set of numbers that, if applied to the given term, would make it correct. That's all I think I can remember, and that's all I can describe.

 

Oh, and this involves < and all that stuff. Also, it involves something like 'not lower than,' which could also refer to a number higher than or the number itself. Look at me, a near fourteen year-old telling a community whose majority is somewhere between the ages of 16 to 20 or higher what something means.

 

I'll be using <= for 'less than or equal to,' and => for 'greater than or equal to.'

 

14 - 6x <= 26

 

-5 < 2x - 1

 

2x - 3 <15

 

8 - 2(x+1) => 2

 

2x - 5 < -3

 

 

That's all. And I'm not tired of typing. I've been writing nearly nonstop today at school(high school, what we call second year).

 

And don't worry, you won't have to solve that on your own. Since you'll probably be awake when I'm asleep, and the majority of you probably don't go to school anymore(correct me if i was wrong there), you'll be able to do more stuff than I will. I'll still be trying to solve this, though. One way or another. At the most, I'll be able to solve the first sets of items today. At the least, nothing at all.

 

This will be changed every time I think I solved one correctly, with my answer in bold.

 

(everything I said above). Um.. that is, if you don't mind... and if you're free.

 

Also, I had to wake up early today just to post this topic. No internet even when I went to sleep.

 

All of those are slope-intercept i.e. y=mx+b. So 14 - 6x <= 26 is really 26 => -6x + 14. First we must solve for x: like so:

 

26 => -6x + 14

-14 - 14 leaving us with

12 => -6x

/-6 /-6

-2 <= x Make sure to switch the sign in front of the number and the less than/greater than sign when the number being divided by is negative.

 

That means the slope is -6/2. Your first point on the graph should be on (-2, 26). Now I don't know if you have to make a table of values because the wording sucks but you can at least graph it now as you have the slope. And because y is greater than or equal to the equation, you shade in the area above the line. I'm pretty sure that's how you do it and I'll leave you to solve the rest.

 

edit:

3x/5y-1 for x= -2 and y= 2

 

That's basic substitution. Then its PEMDAS i.e. order of operations: Parenthesis, Exponents, Multiplication, Division, Addition and lastly Subtraction.

Edited June 27, 2012 by A Blithering Div

[Talemage 47]

Seems that I'm the only user here who uses this.

 

Find the quotient.

-15a^10/3a^4=

 

(a^2/b)^-4=

 

1/3^-3= Is this 27?

 

No, this isn't everything, I just answered everything else. Everything you do not see here that is in my homework is done and I find easy.

I just looked at the notes I took about product rules, power rules, product of a power and power of a quotient.

 

On another note, I'll guess that Mal is one of the people following this

[Whiteshade 1,426]

-5a^6 I believe for the first

For coefficients divide normally (-15/3) but for exponents if the same base then subtract (10-4)

 

a^-8/b^-4 for the second

An exponent of an exponent means multiply, also brackets so the exponent applies to both a and b. Lastly, when multiplying a positive by a negative, the product is negative as well

 

For the 3rd, yes its 27

No brackets so exponent only applies for the 3, and the exponent is -3, so its 1/27, and 1 divided 1/27 is basically 1*27/1 which is obviously 27

Edited July 7, 2012 by Whiteshade

[Talemage 47]

-5a^6 I believe for the first

For coefficients divide normally (-15/3) but for exponents if the same base then subtract (10-4)

 

a^-8/b^-4 for the second

An exponent of an exponent means multiply, also brackets so the exponent applies to both a and b. Lastly, when multiplying a positive by a negative, the product is negative as well

 

For the 3rd, yes its 27

No brackets so exponent only applies for the 3, and the exponent is -3, so its 1/27, and 1 divided 1/27 is basically 1*27/1 which is obviously 27

 

 

Thanks. I figured out how you got the answers with what you said.