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Offline al noor

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AS MATH !
« on: November 13, 2009, 06:27:23 pm »
Hello,
I attached a file ,,, could some1 please help me by writting the answers as i want to check them..
Thankyou in advance :D
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Offline cooldude

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Re: AS MATH !
« Reply #1 on: November 13, 2009, 09:13:05 pm »
Quote from: al noor on November 13, 2009, 06:27:23 pm
Hello,
I attached a file ,,, could some1 please help me by writting the answers as i want to check them..
Thankyou in advance :D

since nobodys told u until now, so i thought i wud and i also saw some easy qs :D , sorry but i cant answer all of them since im still doing igcse maths. neway ur answers--

a) i) V=15000-(2000*3)
=>£9000
ii)10000=15000-2000t
t=(10000-15000)/-2000
t=7 yrs
iii) this is because after 4 yrs the gradient does not remain constant and after 4 yrs the gradient does not remain -2000

b) i) 5000=18275e-0.24t
=>5000/18275=e-0.24t
=>t=5.4004 yrs

ii) % error=(error/actual value)*100
error=1700-[18275*e-0.24*10]
error=£42.1294
=> % error=(42.1294/1700)*100
=> % error=2.4782%

c) i) i don't know
ii) £/yr
iii) the gradient represents the rate of decrease of the value of the car over time
d) i don't know

hope i helped

« Last Edit: November 13, 2009, 09:16:04 pm by cooldude »
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Offline astarmathsandphysics

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Re: AS MATH !
« Reply #2 on: November 13, 2009, 09:28:13 pm »
d)Not ksqrt(t) because the value increases from the graph. The other one is better V=18275*e-0.24*t
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Offline al noor

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Re: AS MATH !
« Reply #3 on: November 14, 2009, 10:58:09 am »
Thankyouuu :P
Quote from: cooldude on November 13, 2009, 09:13:05 pm
since nobodys told u until now, so i thought i wud and i also saw some easy qs :D , sorry but i cant answer all of them since im still doing igcse maths. neway ur answers--

a) i) V=15000-(2000*3)
=>£9000
ii)10000=15000-2000t
t=(10000-15000)/-2000
t=7 yrs
iii) this is because after 4 yrs the gradient does not remain constant and after 4 yrs the gradient does not remain -2000

b) i) 5000=18275e-0.24t
=>5000/18275=e-0.24t
=>t=5.4004 yrs

ii) % error=(error/actual value)*100
error=1700-[18275*e-0.24*10]
error=£42.1294
=> % error=(42.1294/1700)*100
=> % error=2.4782%

c) i) i don't know
ii) £/yr
iii) the gradient represents the rate of decrease of the value of the car over time
d) i don't know

hope i helped


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Offline al noor

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Re: AS MATH !
« Reply #4 on: November 14, 2009, 10:58:53 am »
Thankyoouu soo much :D
Quote from: astarmathsandphysics on November 13, 2009, 09:28:13 pm
d)Not ksqrt(t) because the value increases from the graph. The other one is better V=18275*e-0.24*t
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Offline al noor

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Re: AS MATH !
« Reply #5 on: November 14, 2009, 11:09:00 am »
Quote from: astarmathsandphysics on November 13, 2009, 09:28:13 pm
d)Not ksqrt(t) because the value increases from the graph. The other one is better V=18275*e-0.24*t

ummm  astar can u help me with c(i) and (d) as for (d) i didnt understand wat u wrote in the last post?
Thankyou in advance:D
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Offline MaNi_DaDuDe

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Re: AS MATH !
« Reply #6 on: November 14, 2009, 11:13:19 am »
Quote from: al noor on November 14, 2009, 11:12:44 am


Avoid double posting please.

Thanks.
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