Pink Poogle Toy Forum

What I'm confused about is why you used 25 when the number was .25? Did you multiply it by 10 to get it to 25? And I'm not sure I get this log stuff, I've never seen it used before.

Thanks for helping me though.

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Another question which has got me puzzled,

Determine the future value of each investment/loan,

$6750 borrowed for 23 weeks at 9%/a, compounded monthly.

The equation for future value is: A = P(1+i)^n

A = Future Value
P = Present Value
i = Rate (.09 in this case) divided by the number of periods (12 in this case)
n = How long the money is borrowed for (In this case, it's 23 but our teacher wants us to turn it into a term in years, like 4 months would be 1/3 of a year). This would than be multiplied by the number of periods which is 12 in this case.

Therefore, can anyone help me convert 23 weeks into something which would describe the number as something of a year?

I know that sounds really confusing but if you read my example (Bolded), I'd think you'd be able to sorta understand what I'm saying.

23 weeks would be 23/52ths of a year. (Ugh, I hate interest questions. :P)

A = 6750 (1 + (0.09/12))^(23/52 * 12)
A = 7023.08

Something like that...if I read your problem correctly...

That's it. Thank you very much!

Because 1600=5^2 x 2^6, or 25x2^6.

Imparfait is generally used for events that have happened continually. For example:

Mark ate the dog.
(Mark a mangé le chien)

uses the passé composé because Mark went up, and ate the dog. He didn't eat the dog for several days. However

Mark was playing with the dog
(Mark jouait avec le chien)

uses the imperfect because he was playing with the dog for several hours. THat's basically what it says. It's like the difference between "was ____ing" and "_______ed" in English.

Plus: You more or less always use the imperfect for être and avoir.

In the future, there is an absoultely brilliant site which you can use here

Ah, that makes sense now. Thanks guys!

You are awesome. I always have this problem, and can never remember when to use which one. Then the teacher says "Write a 400 word journal entry on ____, using both the passe compose and the imparfait tenses", and I just go O_O.

I'm having trouble with integration :/

Differentiate 3sec^3x (sec to the power of 3) with respect to x, and hence find ∫ [18 sec^3 x.tan x] dx

Another problem I am having is this one:

Find the definite integrals using x= 6 ln2, x= 6 ln4, as lower and upper limits to solve ∫ e^x/3 dx

I know that ∫ e^x/3 dx = 3 e^x/3, but I am having trouble solving the 'limits' part.

Thanks in advance

Thanks a lot dude. :D *bookmarks page*

To differentiate 3(sec(x))^3, you use the power rule with the chain rule for derivatives.

(3(sec(x))^3)' = 9(sec(x))^2 * (sec(x))' = 9(sec(x))^2 * sec(x)tan(x) = 9(sec(x))^3*tan(x).

Because differentiation cancels the effect of integration, we know that:

∫9(sec(x))^3*tan(x) dx = 3(sec(x))^3 + C

Thus, to get ∫18(sec(x))^3*tan(x), multiply both sides by 2.

∫18(sec(x))^3*tan(x) dx = 6(sec(x))^3 + C

Your second question is simply plugging in numbers by the Fundamental Theorem of Calculus.

You know the antiderivative is 3e^(x/3), so evaluate it from 6ln2 to 6ln4.

3e^(6ln(4)/3) - 3e^(6ln(2)/3).

Factor out a 3 and divide out the 3 in the exponent:
3(e^(2ln(4)) - e^(2ln(2)))

Remember e^x and ln(x) are inverse functions. Thus, e^(2ln(4)) causes the e and ln(4) to cancel to: 4^2, or 16. Similarly, e^(2ln(2)) = 2^2, or 4.

16 - 4 = 12. Don't forget the 3 we factored out earlier; 12 * 3 = 36.

Thank you M. Bison! You are always very helpful. I actually understood everything, you're great!

No problem. Just glad I could help.

Ok, I have a Visual Arts project due tomorrow. Which I haven't started

I need to find an Aboriginal artwork painted by one of ten people or something and write about it.
I have my print. Its called "Heaven and Earth" and it's painted by Sally Morgan.
Now, Aboriginal artworks are quite hard to find on the internet, and I need to know the Title, Date painted, Media and Size.
I have the title so I need:
Date painted
Media
Size
Thanks to anyone who can help but if not then dont worry.
It is now approx 5:38 EST and I should probably be going at about 9:00 EST so3 1/2 hours from now don't bother. If you don't want to help then thats fine, but it would be greatly appreciated.
EDIT: I got some info on This web site but not really enough to use. I assume the size there would be the image size as opposed to the paper size, and also from this I know that it was published in 1990.

Ok, we're doing review in Chemistry this week and I cant figure out/remember/find how to do this question in my old text book. If someone with some knowledge of high school chemistry could explain what Im supposed to do I think I can solve it on my own.

1330 mL of gaseous hydrogen were at STP. The pressure was reduced at a constant temperature. The new volume was 1520mL. What was the new pressure in mm Hg?

Boyle's Law states that, assuming constant temperature, initial P*V must equal final P*V.
That is, P1 * V1 = P2 * V2.

Initial pressure, P1 = 760.0 mmHg (STP means pressure is 1 atm = 760 mmHg).
Initial volume, V1 = 1330 mL.
Final pressure, P2 = ?
Final volume, V2 = 1520 mL

Solve Boyle's Law for your unknown, P2, and plug in the known values.

P2 = (P1*V1)/V2
P2 = (760.0 mmHg * 1330 mL) / (1520 mL)

Notice the units mL cancel, leaving you with P2 = <b>665 mmHg</b>.