Fri Jul 23, 2004 11:36 am
wow, thanks m.Bison, your explaination helped me a lot!
Sat Jul 24, 2004 3:22 pm
untouchable dragon wrote:wow, thanks m.Bison, your explaination helped me a lot!
No problem. I have an exam coming up Monday on integration techniques (including IBP), so it was good practice for me anyway.
Edit: Heh, while studying for my test this morning, I found out I mixed up the formula for IBP.
It should really be ∫udv = uv - ∫vdu. Though, it didn't make a difference in this case, as e^x is its own antiderivative; I got the right answer using the wrong method (hence why I didn't catch the error sooner). Sorry about that.
Fri Jul 30, 2004 4:00 pm
Integration... Its hard!
Cant some one help me integrate tan^2 2x?
I Know the integratal of tan^2x = tanx - x + c, but I cannot figure out how to include that 2...
Cant some one help me integrate tan^2 2x?
I Know the integratal of tan^2x = tanx - x + c, but I cannot figure out how to include that 2...
Fri Jul 30, 2004 4:03 pm
M. Bison wrote:untouchable dragon wrote:wow, thanks m.Bison, your explaination helped me a lot!
No problem. I have an exam coming up Monday on integration techniques (including IBP), so it was good practice for me anyway.
Edit: Heh, while studying for my test this morning, I found out I mixed up the formula for IBP.
It should really be ∫udv = uv - ∫vdu. Though, it didn't make a difference in this case, as e^x is its own antiderivative; I got the right answer using the wrong method (hence why I didn't catch the error sooner). Sorry about that.
hehe! It doesn't matter, I knew the formula, I just didn't know the process XD How did you do in your test?
Fri Jul 30, 2004 5:08 pm
Masquerade wrote:Integration... Its hard!
Cant some one help me integrate tan^2 2x?
I Know the integratal of tan^2x = tanx - x + c, but I cannot figure out how to include that 2...
Anytime you see composite functions inside an integral (that is, f(g(x))), try looking for ways to apply u-substitution. In this case, g(x) = 2x, and f(g(x)) = (tan(2x))^2.
∫(tan2x)^2 dx
Let u = 2x, du = 2dx.
Solve for dx and you'll find that dx = du/2.
Replace the x's with u's and you'll get:
∫(tanu)^2 du/2
Bring out the constant 1/2 (from the du/2).
1/2 ∫(tanu)^2 du = 1/2(tanu - u).
Replace the u's with what you substituted them for (2x).
1/2(tan2x - 2x). Distribute and add the constant of integration:
(tan2x)/2 - x + C.
hehe! It doesn't matter, I knew the formula, I just didn't know the process XD How did you do in your test?
It went well. Math is most definitely my strong point, academically.
Unfortunately, the same can't really be said about anything else heh.
Sat Jul 31, 2004 12:45 pm
M. Bison wrote:Masquerade wrote:Integration... Its hard!
Cant some one help me integrate tan^2 2x?
I Know the integratal of tan^2x = tanx - x + c, but I cannot figure out how to include that 2...
Anytime you see composite functions inside an integral (that is, f(g(x))), try looking for ways to apply u-substitution. In this case, g(x) = 2x, and f(g(x)) = (tan(2x))^2.
∫(tan2x)^2 dx
Let u = 2x, du = 2dx.
Solve for dx and you'll find that dx = du/2.
Replace the x's with u's and you'll get:
∫(tanu)^2 du/2
Bring out the constant 1/2 (from the du/2).
1/2 ∫(tanu)^2 du = 1/2(tanu - u).
Replace the u's with what you substituted them for (2x).
1/2(tan2x - 2x). Distribute and add the constant of integration:
(tan2x)/2 - x + C.
Ohh! Thanks M.Bison! I have a bunch of other questions just like these! I'l be able to work them out now ^^;;
Sun Aug 01, 2004 6:15 am
I'm doing quadratic graphs here, does anyone know how to find the equation of the line of symmetry of the curve?
Sun Aug 01, 2004 6:49 am
Okay, for simplicity here is your quadratic eqn: y = ax^2 + bx + c
Step one is to find the vertex of the equation. Do that by using this equation x = -b/2a.
Once you get this value, plug that value into your original quadratic to get your y value. You now have your vertex.
Now since your equation was in the form y = ax^2 + bx + c, the only point that is really important is that x value, since if you try to graph it, you will notice that the curve will open up or down...
so your answer wouuld be the x = (whatever value you had).
hopefully that helped
Step one is to find the vertex of the equation. Do that by using this equation x = -b/2a.
Once you get this value, plug that value into your original quadratic to get your y value. You now have your vertex.
Now since your equation was in the form y = ax^2 + bx + c, the only point that is really important is that x value, since if you try to graph it, you will notice that the curve will open up or down...
so your answer wouuld be the x = (whatever value you had).
hopefully that helped
Mon Aug 02, 2004 9:56 pm
I am obviously missing something with this math problem...
(2y+5)-(y+3)+(7y-3y)=17
I came up with 9/5...and I am apparantly wrong. What in the world am I doing wrong?
(2y+5)-(y+3)+(7y-3y)=17
I came up with 9/5...and I am apparantly wrong. What in the world am I doing wrong?
Mon Aug 02, 2004 10:56 pm
No problem, Masquerade. Integration can be fun once you get used to it (seriously). 
(2y+5)-(y+3)+(7y-3y)=17
First, distribute that negative sign to get (2y+5)-y-3+(7y-3y)=17. Now, because everything else is being added, you can remove the parenthesees.
2y + 5 - y - 3 + 7y - 3y = 17
Combine terms.
(2y - y + 7y - 3y) + (5 - 3) = 17
5y + 2 = 17
Subtract 2 from each side.
5y = 15
Divide each side by 5.
y = 3
By the way, it seems your mistake comes from not properly distributing the negative sign.
I'm guessing you had simplified -(y + 3) to -y + 3? Instead, it should be -y - 3.
-(a + b)
You can think of the - as a "-1":
-(a + b) = -1(a + b)
Therefore, you multiply everything by (or distribute the) -1.
(-1)a + (-1)b = -a - b
Drama Queen wrote:I am obviously missing something with this math problem...
(2y+5)-(y+3)+(7y-3y)=17
I came up with 9/5...and I am apparantly wrong. What in the world am I doing wrong?
(2y+5)-(y+3)+(7y-3y)=17
First, distribute that negative sign to get (2y+5)-y-3+(7y-3y)=17. Now, because everything else is being added, you can remove the parenthesees.
2y + 5 - y - 3 + 7y - 3y = 17
Combine terms.
(2y - y + 7y - 3y) + (5 - 3) = 17
5y + 2 = 17
Subtract 2 from each side.
5y = 15
Divide each side by 5.
y = 3
By the way, it seems your mistake comes from not properly distributing the negative sign.
I'm guessing you had simplified -(y + 3) to -y + 3? Instead, it should be -y - 3.
-(a + b)
You can think of the - as a "-1":
-(a + b) = -1(a + b)
Therefore, you multiply everything by (or distribute the) -1.
(-1)a + (-1)b = -a - b
Tue Aug 03, 2004 12:53 pm
M. Bison wrote:Drama Queen wrote:I am obviously missing something with this math problem...
(2y+5)-(y+3)+(7y-3y)=17
I came up with 9/5...and I am apparantly wrong. What in the world am I doing wrong?
(2y+5)-(y+3)+(7y-3y)=17
First, distribute that negative sign to get (2y+5)-y-3+(7y-3y)=17. Now, because everything else is being added, you can remove the parenthesees.
2y + 5 - y - 3 + 7y - 3y = 17
Combine terms.
(2y - y + 7y - 3y) + (5 - 3) = 17
5y + 2 = 17
Subtract 2 from each side.
5y = 15
Divide each side by 5.
y = 3
By the way, it seems your mistake comes from not properly distributing the negative sign.
I'm guessing you had simplified -(y + 3) to -y + 3? Instead, it should be -y - 3.
-(a + b)
You can think of the - as a "-1":
-(a + b) = -1(a + b)
Therefore, you multiply everything by (or distribute the) -1.
(-1)a + (-1)b = -a - b
Wow, I can't believe I didn't catch that. Thank you so much, you rock. :D
Tue Aug 03, 2004 1:36 pm
ScottNak wrote:Okay, for simplicity here is your quadratic eqn: y = ax^2 + bx + c
Step one is to find the vertex of the equation. Do that by using this equation x = -b/2a.
Once you get this value, plug that value into your original quadratic to get your y value. You now have your vertex.
Now since your equation was in the form y = ax^2 + bx + c, the only point that is really important is that x value, since if you try to graph it, you will notice that the curve will open up or down...
so your answer wouuld be the x = (whatever value you had).
hopefully that helped
Aha, that sure did.
Thanks for the help.
Sun Aug 08, 2004 9:24 am
This may sound really dumb, since you guys are all doing algebra, and I am doing financial maths.
I don't know the formula, for weekly costs. I've worked out like say I use something, and it lasts for a whole month what would be my weekly cost.
Like say I buy an item that cost $2.00 eg a packet of chips, and it lasts me for 2 weeks, how much would it cost me weekly? I don't think it is like $2.00 divide 2 = $1.00, since the monthly formula isn't just like dividing by four, to work out the weekly cost. But it could be, I don't really know.
Since if it lasts me monthly it wouldn’t be $0.50. The monthly formula isn’t like amount of cost divide 4 = $0.50. Since you end up with 48 weeks in a year overall. The monthly formula is; the amount the good costs x 12 divide by 52 = answer. Eg; Something costs me $3.00, and it lasts me for a whole month. It would be like; $3.00 x 12 divide 52 = $0.69, and $0.69 is therefore my weekly cost if an item lasts me for one month.
Does anyone know what I mean? Oh also how do you work out days? Like say I buy some chocolate, and it costs me $4.00 and it lasts me three days, how much would it cost me weekly?
Please help me! I will be forever grateful.
I don't know the formula, for weekly costs. I've worked out like say I use something, and it lasts for a whole month what would be my weekly cost.
Like say I buy an item that cost $2.00 eg a packet of chips, and it lasts me for 2 weeks, how much would it cost me weekly? I don't think it is like $2.00 divide 2 = $1.00, since the monthly formula isn't just like dividing by four, to work out the weekly cost. But it could be, I don't really know.
Since if it lasts me monthly it wouldn’t be $0.50. The monthly formula isn’t like amount of cost divide 4 = $0.50. Since you end up with 48 weeks in a year overall. The monthly formula is; the amount the good costs x 12 divide by 52 = answer. Eg; Something costs me $3.00, and it lasts me for a whole month. It would be like; $3.00 x 12 divide 52 = $0.69, and $0.69 is therefore my weekly cost if an item lasts me for one month.
Does anyone know what I mean? Oh also how do you work out days? Like say I buy some chocolate, and it costs me $4.00 and it lasts me three days, how much would it cost me weekly?
Please help me! I will be forever grateful.
Sun Aug 08, 2004 11:57 am
First of all, let's assume a week has seven days. Soo... a bag of chips costs two dollars, and it lasts you two weeks.
2 dollars = 2 weeks
divide both sides by 2
1 dollar = 1 week
So yeah, it would be one dollar a week.
Chocolate that lasts you three days?
4 dollars = 3 days
divide both sides by 3
4/3 dollars ($1.33) = 1 day
multiply both sides by 7
28/3 dollars ($9.33) = 1 week
That's pretty much it... just divide both sides by an equal constant.
2 dollars = 2 weeks
divide both sides by 2
1 dollar = 1 week
So yeah, it would be one dollar a week.
Chocolate that lasts you three days?
4 dollars = 3 days
divide both sides by 3
4/3 dollars ($1.33) = 1 day
multiply both sides by 7
28/3 dollars ($9.33) = 1 week
That's pretty much it... just divide both sides by an equal constant.
Thu Aug 19, 2004 11:09 pm
All right y'all, I really need some help in answering 3 questions over the characters, Damon and Pythias:
--What were their first and last names
--What were some of their contributions
--When were they born and when did they die
All help is very, very much appreciated. I am having a bad day, and I can't find anything that will answer these questions.
--What were their first and last names
--What were some of their contributions
--When were they born and when did they die
All help is very, very much appreciated. I am having a bad day, and I can't find anything that will answer these questions.