BF2S Forums Homework Help Thread

So um, can anyone explain what this poem means to them? Like, what you think it means?

If you say in a poem "grass is green,"
They all ask, "What did you mean?"

"That nature is ignorant," you reply,
"And on a deeper level - youth must die."

If you say in a poem "grass is red",
They understand what you have said.

ok math:

i should be able to do this but i cant think straight right now:

Randy is building a fence at the side of his warehouse.  He has 120 m of fencing and plans to use the side of the warehouse as one side of the rectangular fenced area.  What are the dimensions of the maximum area Randy can enclose.

heres the formula:

A=(120-21)(y)

please help!!

=Karma-Kills= wrote:

A bit of AS Level maths you say? Ok here you go.

The whole question reads:



"a) Given that

(2 + x)^5 + (2 - x)^5 = A + Bx^2 + Cx^4,

Find the value of the constants A, B and C."



Ive done that, A = 64, B = 160, C = 20

But then it says



"b) Using the substitution y = x^2 and your answer to part a) solve

(2 + x)^5 + (2 - x)^5 = 349."



I dont understand whtat the by substitutioin bit means...

I.E. (2 + x)^5 + (2 - x)^5 = 64 + 160 x^2 + 20x^4 (from a)

So does it want you to solve 64 + 160y + 20(x^2)(y)

I tried this and got ~ +- 30, which is way off.

*Confused*

PS the answer is given in the back of the book if you want.

Karma for the working solution!

whats the answer? so if i got it right i will post my working

=Karma-Kills= wrote:

Ok the answer is

+/- Square root of 3/2

Ya get it?

Thanks for trying anyways

cool, I got that xD

20x^4+160x^2-285=0          I just took -285 from both sides.
x^2(20x^2+160)-285=0       Factorise

Now u replace each x^2 with Y

Y(20Y+160)-285
20Y^2 + 160Y -285       Now u have a quadratic equation involving Y
so Y= 1.5 or -9.5 ( I used my calculator to find out this)

Subsitute 1.5 back into one of the equations either 20x^3+160x^2=285 or y=x^2
I used the y=x^2 cuz its easier
1.5=x^2
so X=  +- square root of 1.5

U can't use the -9.5 cuz u can't square root -ve number

Last edited by rabee2789b (2006-12-22 07:25:09)

Mong0ose wrote:

rabee2789b wrote:

U can't use the -9.5 cuz u can't square root -ve number

Thats not technically true, the square root of a negative number is a complex number

i.e. (-1)^0.5 = i (or j; depending on where you are taught and the convention you use)

These complex numbers are used when converting equations from the time domain into dimensionless equations, can be very useful

Good work on solving the problem above though

Yes, but hes doing AS Maths, hes not meant to know how to square root -ve number. If hes doing Further maths I would agree abut wut ur saying ^^

Last edited by rabee2789b (2006-12-22 07:44:54)

rabee2789b - thanks.

Yeh i have heard about real and fake numbers and all that... but havent really been taught it (i did Further Maths for GCSE, but only doing normal A level now).

Anyways thanks a lot, i see how you got it and it all makes sense.

All i needed to do was replace the remaining x^2 with a y... *slaps forehead*

Easy when you know how eh?!

Last edited by =Karma-Kills= (2006-12-22 10:57:26)

Got some more!

*Feels dumb for posting so much in this thread*

Logs now...

"Calculate the value of y for which 2 log (to the base 3) y - log (to the base 3) (y + 4) = 2"

That reads, "2 log to the base 3 of y - log to the base 3 of (y + 4) = 2 if that helps....


This is my attempt:

2 log (to the base 3) y - log (to the base 3) (y + 4) = 2

log (to the base 3) y - 1/2 log (to the base 3) (y + 4) = 1

Then the tricky bit, im pretty sure you have to use the Laws of Logs to combine the 2 minuses into 1 divide (e.g something like "log (to the base 3) y / 2(Y+4) )

But im not sure how to excatly.

I can tell you the answer is 12...

Any help (and solutions!) gets +1!

Oh and please clarify what would y / (y + 4) equal? Something like 4y^-1?

Thanks guys

say "{your teacher's name} is gay" and watch your class love you.

Wehay.

I answered my own question.

Any one who wonders, the key is

log (to the base 3) y - 1/2 log (to the base 3) (y + 4) = 1

Take the half back up as a power and then use Law of Logs to divide... easy peasy

Archer wrote:

2 + 2 = ?

wtf mang this question is boterin me fr liek n hour lolol can some1 plz help me

teh teachr is so gey

4

bump ^^^^^

Man I remember doing trigonometric identities/proofs two years ago... they were really annoying, proofs and identities are possibly the hardest thing you'll ever do in math.

Can't help you though, I forgot all my identities. Give me a list of the identities and I can try.