Author Topic: math doubts [NEW]  (Read 16136 times)

Offline slvri

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Re: math doubts [NEW]
« Reply #120 on: November 14, 2009, 07:03:24 pm »
ok.....on tenth bday 1000 is put and on 18th bday it bcomes 1000(1.06)8 cuz for 8 years interest is put on it(on 11th to 18th bdays)
on eleventh bday 1000 is put and on 18th bday it becomes 1000(1.06)7 bcuz for 7 years interest is put on it(12th to 18th bdays)
and so on till on 18th bady 1000 is put but it stays 1000 bcuz thats the end of the time period
this forms a geomtric progression 1000+1000(1.06)+1000(1.06)2+.....+1000(1.06)8
a=1000, r=1.06, n=9(NOT 8!)
S9=1000(1.069-1)/(1.06-1)=$11491.32
is this correct?
« Last Edit: November 14, 2009, 07:05:20 pm by slvri »
i hate A level...........

Offline slvri

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Re: math doubts [NEW]
« Reply #121 on: November 14, 2009, 07:04:51 pm »
oh sory........then -63 is the minima cuz -63 is the value of the function when the value of x is 2
i hate A level...........

zara

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Re: math doubts [NEW]
« Reply #122 on: November 14, 2009, 07:06:55 pm »
oh sory........then -63 is the minima cuz -63 is the value of the function when the value of x is 2
ayte! so if it was +63 then we say its the maxima?

Offline slvri

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Re: math doubts [NEW]
« Reply #123 on: November 14, 2009, 07:12:02 pm »
no....if theyve said itz a minimum point then 63 will still be a minima......
its not necessary that a minima is always negative....it can be positve tooo..
and a maxima can be negative
i hate A level...........

zara

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Re: math doubts [NEW]
« Reply #124 on: November 14, 2009, 07:15:20 pm »
no....if theyve said itz a minimum point then 63 will still be a minima......
its not necessary that a minima is always negative....it can be positve tooo..
and a maxima can be negative
ohkay...thnkx..

Offline astarmathsandphysics

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Re: math doubts [NEW]
« Reply #125 on: November 14, 2009, 08:08:23 pm »
Jane's bank balance put $1000 into a savings bank account for her on each birthday from her 10th to her 18th. The account pays interest at 6% for each complete year that the money is invested. How much money is in the account on the day after her 18th birthday+
?It will be <br />1000+1000*1.06+10000*1.06^2+....+1000*1.06^8
it is a geometric sequence with first term 1000 and common raio 1.06 and 9 terms
S_9=\frac {a(1-r^n)}{1-r}=\frac {1000(1-1.06^9)}{1-1.06}=11491.32

Offline astarmathsandphysics

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Re: math doubts [NEW]
« Reply #126 on: November 14, 2009, 08:14:17 pm »
If coordinates are (2,-63); is the y coordinate considered for deciding the maxima or minima?
n if its -ve then its minima n if +ve then maxima?
It is not y that has to be negative for a min. \frac {d^2 y}{dx^2} has to be positive
and  \frac {d^2 y}{dx^2} has to be negative for a max

explained here
http://www.astarmathsandphysics.com/a_level_maths_notes/C1/a_level_maths_notes_c1_maxima_and_minima_the_second_differential_criterion.html

zara

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Re: math doubts [NEW]
« Reply #127 on: November 15, 2009, 06:16:46 pm »
yea i got it astar Thanks..

Offline astarmathsandphysics

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Re: math doubts [NEW]
« Reply #128 on: November 20, 2009, 08:22:31 pm »
The function f is defined by f : 2x^2-12x + 13 for 0 <=x <=A, where A is a constant.
(i) Express f(x) in the form a(x + b)2 + c, where a, b and c are constants. [3]
(ii) State the value of A for which the graph of y = f(x) has a line of symmetry. [1]
(iii) When A has this value, find the range of f.

i)2x^2-12x + 13=a(x + b)^2 + c=ax^2+2abx+ab^2+c
2x"2=ax^2 so a=2
-12x=2abx=4bx so b=-3
13=ab^2+c=2*(-3)^2+c so c=-5
f(x)=2(x-3)^2-5
ii)A=6
iii)max value of f(x) is f(6)=13
range of f is [-5,13]

zara

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Re: math doubts [NEW]
« Reply #129 on: December 10, 2009, 03:35:12 pm »
Find the sum of the infinite series 1/103+1/106+1/109+... expressing ur answer as a fraction in its lowest terms.

Hence express the infinite recurring decimal 0.108108108...as a fraction in its lowest terms.

Offline vanibharutham

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Re: math doubts [NEW]
« Reply #130 on: December 10, 2009, 03:41:57 pm »
Find the sum of the infinite series 1/103+1/106+1/109+... expressing ur answer as a fraction in its lowest terms.

Hence express the infinite recurring decimal 0.108108108...as a fraction in its lowest terms.

We find that r = 0.001, and a = 0.001

S to infinity = a / 1 - r

=====> 0.001 / 1 - 0.001
=> 1 / 999

Since

1/ 999 = 0.001001001001001 etc....
2/999 must equal 0.002002002002 etc

Hence 108/999 must equal 0.108108108108108

In its simplest form 108/999 = 4/37

:) 
A genius is 1% intelligence, 99% effort.

zara

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Re: math doubts [NEW]
« Reply #131 on: December 10, 2009, 03:46:13 pm »
thank you mate!

zara

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Re: math doubts [NEW]
« Reply #132 on: December 10, 2009, 03:52:37 pm »
Charles borrow $6000 for a new car. Compound interest is charged on the loan at a rate of 2% per month. Charles has to pay off the loan with 24 equal monthly payments. Calculate the value of each monthly payment.

Offline @d!_†oX!©

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Re: math doubts [NEW]
« Reply #133 on: December 10, 2009, 04:52:54 pm »
I am not sure if it's right or not...i just gave it a try...please check the answer if you have one and reconfirm....
A = P ( 1 + r/100 )n
6000=P(1+2/100)24
thus, P= $3730.328928
and because the monthly payments are equal, thus each monthly payment= 3730.328928/24 = $155.43

 :) :)
AAL IZZ WELL!!! ;)

zara

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Re: math doubts [NEW]
« Reply #134 on: December 20, 2009, 03:36:31 pm »
if anyone uses Pure mathematics 1 by Hugh Neill & Douglas Quadling can u plz turn over to page 253 Ex.16D Q4.
i got rong answer prolly cause one of my limit is rong and thts whr im confused, can summon plz do n lemme kno?