IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: al noor on November 13, 2009, 06:27:23 pm

Title: AS MATH !
Post by: 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

Title: Re: AS MATH !
Post by: cooldude 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

Title: Re: AS MATH !
Post by: 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

Title: Re: AS MATH !
Post by: al noor 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

Title: Re: AS MATH !
Post by: al noor 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

Title: Re: AS MATH !
Post by: al noor 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

Title: Re: AS MATH !
Post by: MaNi_DaDuDe 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.