Maths, Addmaths and Statistics: Post your doubts here!

[ashiqbal]

thanks for making the correction...

no problem

 

[ashiqbal]

ashiqbal, syed1995....here is a question: The speed v m/s of a particle travelling from A to B, at time t s after leaving A, is given by v=10t-t^2. The particle starts from rest at A and comes to rest at B. Show that the particle has a speed of 5 m/s or greater for exactly 4√5 s.
Using 10t-t^2.=5, we get 5+2√5 and 5-2√5. The examiner report says to subtract one from the other. Why to do so?

is this from addmaths? we will make a graph, just like we do in every quadratic inequality. we have to find the range where it is from greater than 5. i might not be able to answer addmaths anymore for 4 days

 

[Haris Bin Zahid]

is this from addmaths? we will make a graph, just like we do in every quadratic inequality. we have to find the range where it is from greater than 5. i might not be able to answer addmaths anymore for 4 days
View attachment 11887

Thnx a lot....great man!
Can u plz elaborate a little bit!

 

[ashiqbal]

Thnx a lot....great man!
Can u plz elaborate a little bit!

bhayya time bohot tough hai!
farigh ho jaon ga tou online skype pai tution lai laina!

 

[asd]

asd: Thnx for all the help man! You surely are really knowledgeable and cooperative (MashaAllah).
Well....here is another question: The speed v m/s of a particle travelling from A to B, at time t s after leaving A, is given by v=10t-t^2. The particle starts from rest at A and comes to rest at B. Show that the particle has a speed of 5 m/s or greater for exactly 4√5 s.
Using 10t-t^2.=5, we get 5+2√5 and 5-2√5. The examiner report says to subtract one from the other. Why to do so?

Thanks, And oh sorry, I wasn't here. But i guess you got your answer ?

 

[HarisLatif]

i believe this is the wrong explanation.
they have asked for what? the smallest width that can ALWAYS hold 8 pencils, right? so we have to take the upper bound of the pencils's diameter. this is because if we take the lower bound of the pencils,6.5mm, the box could not hold 8 pencils if their actual diameter was more than 6.5 (which is totally possible) eg 7.5mm. what will happen then? the box could not hold 8 pencils. so in order for the box to hold the 8 pencils in ANY CASE, we will take upper bound of the diameter, ie 7.5mm
7.5mm*8 = 60 mm = 6 cm

My mind was yet unclear about the point kitecrystal told me ! anyways it's clear now THANKS

 

[Haris Bin Zahid]

Thanks, And oh sorry, I wasn't here. But i guess you got your answer ?

Yup, ashiqbal explained it. Thnx for ur consideration.
Will u be free after this maths paper?

 

[Haris Bin Zahid]

bhayya time bohot tough hai!
farigh ho jaon ga tou online skype pai tution lai laina!

Soooooorrrrrrrryyyyyyyy for disturbing. Really sorry. Thnx for taking out time for helping me out by sparing your precious time. I got the expanation. Really grateful.

 

[syed1995]

yea ashiqbal is right .. let our maths exams end .. then you will have all the time you need to ask addmaths questions!

wait 4 days till the 8th

 

[HarisLatif]

see s_10_qp22 q11 and please explain why in the marking schemes are widths given 2,2,5,5,10 instead of 1,1,2.5,2.5,5 because they have asked for 1 cm = 2 seconds in the question. so have they actually given the range in the ms or the width of each bar?
http://www.xtremepapers.com/papers/CIE/Cambridge International O Level/Mathematics D (Calculator Version) (4024)/4024_s10_qp_22.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International O Level/Mathematics D (Calculator Version) (4024)/4024_s10_ms_22.pdf

When 1 cm = 2 seconds then you can see the width will be the same as stated in the marking scheme. these are actually the width of each bar in the mark scheme.

 

[ashiqbal]

When 1 cm = 2 seconds then you can see the width will be the same as stated in the marking scheme. these are actually the width of each bar in the mark scheme.

hmm..ok

 

[UziB]

Some help in this would be appreciated
Q) The nth term of a sequence, S, is n³ + 2. The first four terms are 3, 10, 29 and 66.
The first four terms of another sequence, T, are 4, 12, 32 and 70. By comparing S and T, write down
a) the fifth term of T,
b) an expression, in terms of n, for the nth term of T.

 

[Spidey21]

whatsup

 

[Spidey21]

Some help in this would be appreciated
Q) The nth term of a sequence, S, is n³ + 2. The first four terms are 3, 10, 29 and 66.
The first four terms of another sequence, T, are 4, 12, 32 and 70. By comparing S and T, write down
a) the fifth term of T,
b) an expression, in terms of n, for the nth term of T.

a) S = 127
T = 131

b) n^3 +2+n (i think)

 

[UziB]

a) S = 127
T = 131

b) n^3 +2+n (i think)

Ah okay I got (a) part thanks, but shouldn't the fifth term of T be 132? If it's increasing by 1 each term from S, then the fifth term of T = fifth term of S + 5.
Got (b) part as well, and yes your answer is correct. Much appreciated, thanks again

 

[beeloooo]

http://clickpapers.net/past papers/math-/4024_s07_qp_1.pdf


q.14 last part !!
q.15 last part !!

help would be appreciated with explaination

 

[Spidey21]

Ah okay I got (a) part thanks, but shouldn't the fifth term of T be 132? If it's increasing by 1 each term from S, then the fifth term of T = fifth term of S + 5.
Got (b) part as well, and yes your answer is correct. Much appreciated, thanks again

it will be 132 i made a calculation error..

 

[Spidey21]

http://clickpapers.net/past papers/math-/4024_s07_qp_1.pdf


q.14 last part !!
q.15 last part !!

help would be appreciated with explaination

14 last part
see that there are two complete circles.
one with diameter 2x and the other with 4x
take the area of 4x and 2x and then
minus the area of 2x from 4x , make sure to keep the pi and there you go!!

15 last part
ratio it (similar triangles)
3:9
4:x
and then add 4 to x
hope it helps!

 

[beeloooo]

14 last part
see that there are two complete circles.
one with diameter 2x and the other with 4x
take the area of 4x and 2x and then
minus the area of 2x from 4x , make sure to keep the pi and there you go!!

15 last part
ratio it (similar triangles)
3:9
4:x
and then add 4 to x
hope it helps!

THANK YOU


btw whats your final answer than for the 14's last part ???

 

[Spidey21]

THANK YOU


btw whats your final answer than for the 14's last part ???

i have not done it..you want me to do it??