Post Your Problem Here (2 Viewers)

[velox]

hmm thats not 2u...or is it? Do the limit to infinity.

 

[mojako]

I NEED HELP QUICKLY!!!!!!!!!!!!PLZZZZZZ

john borrowed 2000 in 1st jan 1980. he agreed to pay back 300 on ist jan in each succeeding year and to add 6% percent per annum interest on the amount owing during the year completed
find amount still owing immediately afta 1st jan 1985?
FIND THE NUMBER OF YEARS NECESSARY TO FREE HIIMSLEF OFF DEBT?

since nobody has answered this and this (http://www.boredofstudies.org/community/showpost.php?p=1054668&postcount=18) I'll attempt it

Let A_k be the amount still owing after k years.
A_0 = 2000
A_1 = A_0*1.06 - 300 (* is for times)
A_2 = A_1*1.06 - 300
continuing and rearranging,
A_k = 2000*1.06^k - ( 300 + 300*1.06 + 300*1.06^2 + ... + 300*1.06^(k-1) ) (^ is for power)
(if you don't know how it gets to this expression, just remember it for other loan repayment questions)
Now "300 + 300*1.06 + 300*1.06^2 + ... + 300*1.06^(k-1)" is a GP with a=300, r=1.06. n=k
Sum of GP = a(r^n-1)/(r-1) = 300(1.06^k-1)/(1.06-1) = 5000(1.06^k-1)

A_k
= 2000*1.06^k - 5000(1.06^k-1)
= -3000*1.06^k + 5000

Now, when it's immediately after 1st Jan 1985 it's been exactly 5 years. k=5.
A_5
= -3000*1.06^5 + 5000
= 985.32 dollars

When he's free from debt, A_k=0
0 = -3000*1.06^k + 5000
5000 = 3000*1.06^k
1.06^k = 5/3
k = ln(5/3) / ln(1.06)
(to help you remember this formula, try something like 2^k=8, in which k=ln8/ln2 not ln2/ln8)
k = 8.76...
It would take 9 years to free himself off debt.
(note: if for example k=8.12..., it would also take 9 years and not 8 years)


Now I'll find the asymptote of y=x^2 / 2x+3.
dividing by x^2, y = 1 / (2/x+3/x^2)
as x->infinity, 2/x->0 and 3/x^2->0
so y -> 1/0, y->infinity
doesn't work unfortunately

So let's try another method.
dividing by x, y = x / (2 + 3/x)
as x->infinity, 3/x->0
y->x/2
So the asymptote is y=x/2


Now y=2x+3 / x^2+4
y = (2 + 3/x) / (x + 4/x)
as x->infinity, y->2/x->0
so the asymptote is y=0 which is the x-axis
note: y=f(x)/g(x) won't have oblique asymptote when the degree of g(x) is greater than the degree of f(x)

 

[BoganBoy]

hi all:
i have this seeming easy question, but cant figure out how to do it. here it goes:


A^2*B^3 = 1323. A and B are integers. Find A and B.

beside trial and error, is there any other ways? thanks. btw, im onlly starting year 11, so dont use too complicated methods. thanks.

 

[Slidey]

List all prime factors of 1323, excluding 2 (as the number is not even).

Test 3, 5, 7, 11, 13, 17, 19, 23, 31, 37, et cetera.

Then, test each one again. If it works twice, note that. If it works thrice, note that. Call it X Now find 1323/X^3 or 1323/X^3, and repeat the procedure, not using the number X this time.

3 works. 3 works again. And again...

So 3^3 is a factor of 1323.

1323/3^3 = 49. 49=7^2, therefore (A,B)=(7,3)

 

[Trev]

urgh how can u b bothered with so much trial and error!

 

[Slidey]

^ Considering the first trial worked, it din't require any bother. What required bother was writing the explanation out.

 

[BoganBoy]

wow, thanks slide_rule. u rock so much. when do u do ur HSC? ill shall be looking for ur name in the top achiever's list.

 

[Trev]

his name is funny lol

 

[Slidey]

I do it in 2005.

Don't expect too much from me, but thanks for the faith.

 

[Captain pi]

I do it in 2005.

Don't expect too much from me, but thanks for the faith.

2005?

Birthday:
31 December, 1969

...

HSC Graduation:
N/A.

Explain, please. Many of your other posts seem - to me - as if you are taking a higher level course (Tertiary) than you profess; I don't know, maybe that just reveals the scope of my mathematics education.

 

[Trev]

he is actually in yr 12 like us mr pi., he just knows alot about uni stuff bcoz he researchs alot bout it, is that bout rite slide? (dont correct me again!)

 

[Slidey]

On the contrary Captain Pi, I assure you I graduate in 2005; I merely did not wish to express this fact in my profile.

Yes, that's correct, Trev. I am interested in mathematics, science, linguistics, philosophy, education and psychology. It is an unstoppable urge of mine to learn more about these subjects.

 

[littleboy]

can you guys help we out with this question?

Prove that:

a)[(sin^2 a - cos^2 a)(1 - sin a cos a)] / cos a (sec a - cosec a)(sin^3 a + cos^3 a)
= sin a

b)(1+ cosec^2 A tan^2 C) / (1 + cosec^2 B tan^2 C)
= (1+cot^2 A sin^2 C) / (1 + cot^2 B sin^2 C)

c) If tan a +sin a = x

and if tan a - sin a = y,

prove that x^4 + y^4 = 2xy(8 +xy)



thanks

 

[Trev]

for a) (i changed degree of a to degree of x coz im used to it)
[(sin^2 a - cos^2 a)(1 - sin a cos a)] / cos a (sec a - cosec a)(sin^3 a + cos^3 a)
= [(sinx - cosx)(sinx + cosx)(1-sinxcosx)]/cosx(1/cosx - 1/sinx)(sinx+cosx)(sin²x - sinxcosx + cos²x)

= [(sinx - cosx)(1-sinxcosx)]/[cosx({sinx-cosx}/sinxcosx)(sin²x - sinxcosx + 1 - sin²x)

= [(sinx - cosx)(1-sinxcosx)]/[cosx({sinx-cosx}/sinxcosx)(1-sinxcosx)

= [(sinx - cosx)]/[({sinx-cosx}/sinx)

=sinx yay!

 

[Trev]

can any1 get the othas? i tried...... and failed lol

 

[mojako]

for (b):
Not sure if the question is correct... if it is then I can't do it

for (c):
I used binomial expansion to expand the LHS, and manipulated / compared LHS and RHS at the same time, and then it turns out that "tan^2 @ - sin^2 @" must be the same as "tan^2 @ * sin^2 @" for the question to be true.
tan^2 @ * sin^2 @ = tan^2 @ * (1 - cos^2 @) = tan^2 @ - sin^2 @
now, you should write the proof properly, going strictly from LHS to RHS (or the other way around).

 

[Trev]

for (b):
Not sure if the question is correct... if it is then I can't do it

for (c):
I used binomial expansion to expand the LHS, and manipulated / compared LHS and RHS at the same time, and then it turns out that "tan^2 @ - sin^2 @" must be the same as "tan^2 @ * sin^2 @" for the question to be true.
tan^2 @ * sin^2 @ = tan^2 @ * (1 - cos^2 @) = tan^2 @ - sin^2 @
now, you should write the proof properly, going strictly from LHS to RHS (or the other way around).

tan^2 @ * sin^2 @ = tan^2 @ * (1 - cos^2 @) = tan^2 @ - sin^2 @
........shown:
tan²xsin²x = tan²x - sin²x
by LHS...
= (sin²x/cos²x)sin²x
= sin²x(1-cos²x)/cos²x
= sin²x/cos²x - sin²xcos²x/cos²x
= tan²x - sin²x

 

[mojako]

tan^2 @ * sin^2 @ = tan^2 @ * (1 - cos^2 @) = tan^2 @ - sin^2 @
........shown:
tan²xsin²x = tan²x - sin²x
by LHS...
= (sin²x/cos²x)sin²x
= sin²x(1-cos²x)/cos²x
= sin²x/cos²x - sin²xcos²x/cos²x
= tan²x - sin²x

didn't I just did that when I wrote "tan^2 @ * sin^2 @ = tan^2 @ * (1 - cos^2 @) = tan^2 @ - sin^2 @"

so did u work out (b)?

 

[Trev]

didn't I just did that when I wrote "tan^2 @ * sin^2 @ = tan^2 @ * (1 - cos^2 @) = tan^2 @ - sin^2 @"

so did u work out (b)?

awww yeh lol and my way took lots longer
didnt work out b) tho
mojako hav u done hsc yet? if so wat math marks did u get

 

[FinalFantasy]

Prove:
(1+ cosec² A tan² C) / (1 + cosec² B tan² C)
= (1+cot² A sin² C) / (1 + cot² B sin² C)

LHS=(1+ cosec² A tan² C) / (1 + cosec² B tan² C)
divide sec² C on top and bottom
=(cos² C+cosec²Asin²C)\(cos²C+cosec²Bsin²C)
=(1-sin²C+cosec²Asin²C)\(1-sin²C+cosec²Bsin²C)
=(1+sin²C(cosec²A-1))\(1+sin²C(cosec²B-1)) ----->(1)
Now I'll Prove cosec² x-1=cot² x
We know 1=cos²x+sin²x, divide sin²x on both sides we get
cosec²x=cot²x+1
therefore
cosec² x-1=cot² x
Now back to (1)
=(1+cot²Asin²C)\(1+cot²Bsin²C)=RHS