Mathematics: Post your doubts here!

[Rutzaba]

lny - ln70 = e^-3t - 1

Can someone rearrange this to make y the subject? If you would show me steps that would be helpful

by ln proprty if the are being substracted it mean they are eing divided

so take ln common... ln(y/70) =e^-3t - 1

then wen we will remove the ln on the other side the whole thing wud become the power of e

y/70 =e^ (e^-3t - 1)
y=70 e^ (e^-3t - 1)

 

[Minato112]

Please help with my question too people

Can U plz repost it?

 

[peter lucantoni]

I NEED SOME HELP! In the question 6(ii) ,where is the 1.282 come from? thank you so much!



please !

 

[hassanbmj]

Mechanics problem:
Q: A particle is moving along a straight line with constant acceleration. In an interval of T sec ot moves D meters; in the next interval of 3T sec it moves 9D meters. How far does it move in a further interval of T seconds ?
Answer 5D.
Your help will be much appreciated.​

 

[Rutzaba]

Mechanics problem:​

Q: A particle is moving along a straight line with constant acceleration. In an interval of T sec ot moves D meters; in the next interval of 3T sec it moves 9D meters. How far does it move in a further interval of T seconds ?​

Answer 5D. ​

Your help will be much appreciated.​

dude i got a perfect eye sight...wat i aint got is... mechanics xD

 

[syed1995]

dude i got a perfect eye sight...wat i aint got is... mechanics xD

same xD

 

[Dug]

Mechanics problem:​

Q: A particle is moving along a straight line with constant acceleration. In an interval of T sec ot moves D meters; in the next interval of 3T sec it moves 9D meters. How far does it move in a further interval of T seconds ?​

Answer 5D. ​

Your help will be much appreciated.​

s = ut + ½ at²

At t = T, s = D
D = uT + ½ aT²
uT = D - ½ aT² --- i

At t = T + 3T, s = D + 9D
10D = 4uT + 8aT² --- ii

Put (i) in (ii)
10D = 4D - 2aT² + 8aT²
aT² = D --- iii

Put this back in (i)
uT = D - ½ D
uT = ½D --- iv

At t = 5T,
s = 5uT + (25/2) aT² --- v

Plug (iii) and (iv) into (v)
s = (5/2)D + (25/2)D
s = 15D

Distance covered in the final interval = Total distance - Distance covered in the first 4T seconds
= 15D - 10D
= 5D

 

[Sandhya Mahat]

Help me solving this guys: Solve the equation:
Mod(x-1) + Mod(x+1) -1 = Mod(x)

 

[princess787]

can anyone help me in 9709_w12_qp_31 question 3! please! and if you have solved it please upload the whole paper! i solved the papers... bt d MS is not clear! blank spaces everywhere! loads of doubt!! neeed to understand the method! if nt the whole paper! please help me in question 3! thanx! :*

 

[yubakkk]

pLz help this question. its from book.

For a biased cubical dice the probability of any particular score between 1 and 6 (inclusive) being obtained is inversely proportional to that score. Find the probability of scoring a 1.

 

[Scafalon40]

Q5

a. ii)

5 _ _ _

No. of arrangements = 6³ = 216

b. ii)

n(B) = 6

n(G) = 8

If the cousins are in, 3 boys are already selected. The number of ways to select the remaining team is 11C2.

If the cousins are out, we need to select 5 students for the team which is simply 11C5.

No. of ways = 11C2 + 11C5 = 517

wait could you elaborate please? why did you take 6^3?

 

[Dug]

wait could you elaborate please? why did you take 6^3?

Because repetition is allowed in that question. It would be 6*5*4 if repetition was not allowed.

 

[ZainH]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w08_qp_1.pdf

The last question's last part , I'm getting one of the signs wrong for some reason. They changed the equation when taking the inverse? :\
If someone could do it with steps included I'd love them forever .

 

[Scafalon40]

Because repetition is allowed in that question. It would be 6*5*4 if repetition was not allowed.

Sorry but I still don't understand
I've never encountered a question like this before

 

[Minato112]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w08_qp_1.pdf

The last question's last part , I'm getting one of the signs wrong for some reason. They changed the equation when taking the inverse? :\
If someone could do it with steps included I'd love them forever .

h(x) = 6x−x^2
= - (x^2 - 6x)
= - [ (x-3)^2 - 9]
= 9 - (x-3)^2

To find h*inverse*
Let y = h(x)
y = 9 - (x-3)^2
(x-3)^2 = 9 - y
x - 3 = (+or -) *root* 9 - y
x = (+or -) [*root* 9 - y ] + 3

Since the domain is positive, we take the positive root.
Therefore h*inverse* = [*root* 9 - x ] + 3

 

[ZainH]

h(x) = 6x−x^2
= - (x^2 - 6x)
= - [ (x-3)^2 - 9]
= 9 - (x-3)^2

To find h*inverse*
Let y = h(x)
y = 9 - (x-3)^2
(x-3)^2 = 9 - y
x - 3 = (+or -) *root* 9 - y
x = (+or -) [*root* 9 - y ] + 3

Since the domain is positive, we take the positive root.
Therefore h*inverse* = [*root* 9 - x ] + 3

The part I highlighted, how come you took 9 to the other side but changed the sign of 'y' ?

 

[Dug]

Sorry but I still don't understand
I've never encountered a question like this before

Read this

 

[Minato112]

The part I highlighted, how come you took 9 to the other side but changed the sign of 'y' ?

I didnt move the 9. I switched - (x-3)^2 with y. Got it now?

 

[ZainH]

I didnt move the 9. I switched - (x-3)^2 with y. Got it now?

Not exactly, why are you equating the equation while finding the inverse?
This is what I did.

To find H inverse
Let y = h(x)

h(x)= 9-(x-3)^2
h(x)-9= (x-3)^2
*root*h(x)-9= x-3
[*root*h(x)-9]+3 = x

H inverse = [*root*h(x)-9]+3 where as yours and the MS's is "9-x" instead of "x-9"


EDIT: I undestand what you did, but why is my method wrong?

 

[princess787]

can anyone help me in 9709_w12_qp_31 question 3 and 10 ii) ! please!