Author Topic: math doubts [NEW]  (Read 16070 times)

Offline simba

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Re: functions
« Reply #60 on: October 13, 2009, 05:35:52 pm »
each year a company gives a grant to charity. The amount given each year increases by 5% of its value in the preceding year. The grant in 2001 was $5000. Find :

(i) the grant given in 2011
(ii) the total amount of money given to the charity during the years 2001 to 2011 inclusive

Offline astarmathsandphysics

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Re: functions
« Reply #61 on: October 13, 2009, 06:24:14 pm »
I will answer when i get home in 3 hours

Offline astarmathsandphysics

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Re: functions
« Reply #62 on: October 13, 2009, 09:46:09 pm »
each year a company gives a grant to charity. The amount given each year increases by 5% of its value in the preceding year. The grant in 2001 was $5000. Find :

(i) the grant given in 2011
(ii) the total amount of money given to the charity during the years 2001 to 2011 inclusive
i)a_n=ar^{n-1} =5000*1.05^{10-1} = 7756.64
ii)S_n =\frac{a(1-r^n)}{1-r}=\frac {5000(1-1.05^10)}{1-1.05} =162889

Offline simba

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Re: functions
« Reply #63 on: October 13, 2009, 10:59:57 pm »
hw did u change da 5% to 1.05 ???

Offline Eamyzz

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Re: functions
« Reply #64 on: October 14, 2009, 01:03:44 am »
r=1+5/100 is hw they gt 1.05
anythng else? =)
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Offline simba

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Re: functions
« Reply #65 on: October 14, 2009, 11:28:09 am »
the second term of a geometric progression is 3 and the sum to infinity is 12
(i) find the first term of the progression



will someone solve it :)

Offline simba

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Re: functions
« Reply #66 on: October 14, 2009, 11:49:52 am »
q11 june 06 in part (i)  urgently plz

Offline slvri

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Re: functions
« Reply #67 on: October 14, 2009, 11:59:25 am »
the second term of a geometric progression is 3 and the sum to infinity is 12
(i) find the first term of the progression



will someone solve it :)

the nth term of a GP is given by an=arn-1
for the second term n=2
ar2-1=3
ar=3
a=3/r
the sum to infinity of a GP is given by S=a/(1-r)
a/(1-r)=12
a=12(1-r)
put a=3/r
3/r=12(1-r)
3=12r(1-r)
3=12r-12r2
12r2-12r+3=0
4r2-4r+1=0
(2r)2-2(2r)(1)+(1)2=0
(2r-1)2=0
2r-1=0
r=1/2
a=3/(1/2)=6
is this the correct answer?
i hate A level...........

Offline Eamyzz

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Re: functions
« Reply #68 on: October 14, 2009, 04:39:09 pm »
yes i gt da same answer tooo  :)
Live=Hope
Hope=Life
Life=Love
So,No life=No hope=DEATH
am alive =D

Offline simba

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Re: functions
« Reply #69 on: October 14, 2009, 04:44:15 pm »
guyz another question ....num.4 in nov.07

will someone help :???

Offline Ghost Of Highbury

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Re: functions
« Reply #70 on: October 14, 2009, 04:55:09 pm »
guyz another question ....num.4 in nov.07

will someone help :???

A level maths --> paper no?
divine intervention!

Offline simba

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Re: functions
« Reply #71 on: October 14, 2009, 04:56:56 pm »
its paper 1 ...pure 1

Offline badboy

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Re: functions
« Reply #72 on: October 14, 2009, 05:11:41 pm »
each year a company gives a grant to charity. The amount given each year increases by 5% of its value in the preceding year. The grant in 2001 was $5000. Find :

(i) the grant given in 2011
(ii) the total amount of money given to the charity during the years 2001 to 2011 inclusive
i)a_n=ar^{n-1} =5000*1.05^{10-1} = 7756.64
ii)S_n =\frac{a(1-r^n)}{1-r}=\frac {5000(1-1.05^10)}{1-1.05} =162889



ur answers r wrong i) 8144 ii)71034

Offline slvri

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Re: functions
« Reply #73 on: October 14, 2009, 05:43:36 pm »


ur answers r wrong i) 8144 ii)71034
(i)the grant given in 2011 should be the 11th term
the 11th term is given by a11=arn-1=5000*1.0511-1=5000*1.0510=$8144
(ii)total amount is the sum of the money from 2001 to 2011
so the sum from terms 1 to 11 is S11
=a(1-rn)/1-r
=5000(1-1.0511))/1-1.05
=5000(1-1.0511)/-0.05
=$71034
hope this helped :)
« Last Edit: October 19, 2009, 11:21:49 am by slvri »
i hate A level...........

Offline astarmathsandphysics

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Re: functions
« Reply #74 on: October 14, 2009, 07:15:36 pm »
Sod it you are righht. I shouldn't do these questions first thing in the morning.