smurfy:

Here goes...

Let A = 2x and P = x
Using A = P[1 + (r/100)]^n,

2x = x[1 + (4/100)]^n
2 = (1 + 0.04)^n
2 = (1.04)^n

Take log of both sides:

log 2 = log (1.04)^n
nlog(1.04) = log 2
n = log 2/log(1.04)
n = 17.67
n = 18 years (approx)

jauntee02: i'l give only the first person to solve this (within 12 hours) a recharge card.

1..The current distance between
Buenos Aires and Cape Town at
latitude 34° S is 6890 km, and
the sea floor spreading rate at
the South Atlantic mid-ocean
ridge (i.e. between the African
and South American plate) is 35
mm yr-1. Assuming that both
cities lie near to the edge of their
respective continental margins,
calculate to the nearest million
years how long it has been since
these two continents separated.
*****Hint: it may help if you first
calculate how far the plates
would move in 1 million years.
2..A palaeontologist has
discovered a new fossil
Pterosaur species (i.e. flying
reptile) in sedimentary strata
near Buenos Aires, and also near
Cape Town. The strata are dated
as Early Cretaceous (120 Ma) in
age. Using a reconstruction of
the fossil skeleton, the
palaeontologist has suggested
that the animal probably ate fish,
and could have flown long
distances across the sea at an
average of 60 km h-1. To the
nearest hour, determine how
long, it would have taken the
Pterosaur to fly across the Early
Cretaceous Atlantic Ocean.

Laplacian:
1.)...the seperation rate,
r=35mm yr-1=35*10^-3 m yr-1
or
r=35*10^-3*10^6/10^6 m yr-1
or
r=35 km/million-yrs....
So,
the time T, taken for a seperation of 6890 km is;
T=6890/35
or
T=196.857 million-yrs

2.) Note that 120 Ma means 120 Mega-annum...meaning
120 Million-yrs...that been being the yr of flight, it means the seperation distance S, in that period is S=35*120=4200 km...so the duration P, of the specie's flight is P=4200/60=70 hrs or equivalently it 'll take 2-days 22-hours

you can keep ur Recharge Card

smurfy: If y = ln[tan(1 - x)], show that
dy/dx = -2/[sin(2 - 2x)]

benbuks: let's have tan(1-x)=u
=>lnu=y

-sec^2 (1-x)=du/dx
1/u=dy/du
dy/du *du /dx=dy/dx
=> -sec^2(1-x)/tan(1-x)=dy/dx (replacing the value of u)
since -sec^2 (1-x)=-1/(cos^2 (1 -x) & sin(1-x)/cos(1-x)=tan(1-x)

=> -1/[sin(1-x).cos(1-x)]=dy/dx
now , to get to where we are going let's do some manipulations

we can have -2/2[sin(1-x)cos(1-x) (multiplying up and down by 2/2)
from trig idntities we can write 2sin(1-x).cos(1-x) as sin2(1-x) =sin(2-2x) (since, 'say' 2sinACosA=sin2A)

hence
dy/dx=-2/sin(2-2x)

Q.E.D

Laplacian: ...pls i need help with this questoin...either a solution or an idea is welcome...

Three non-zero integers are in arithmetic progression, the product of any two when when increased by one is a perfect square...find the numbers......

benbuks: ..hint

we can have a , (a+d) ,( a+2d)

remember to have a perfect square b^2 -4ac=0 (from theorem of quadratic equation) , hope it helps ,?

smurfy:

Is it 2, 4, 6?

(2 * 4) + 1 = 9
(4 * 6) + 1 =25

Remember also that if a, b, c are in AP, then b = 1/2(a + c).

benbuks: ..hint

we can have a , (a+d) ,( a+2d)

remember to have a perfect square b^2 -4ac=0 (from theorem of quadratic equation) , hope it helps ,?
we have any series of ood/even numbers
e.g 3,5,7,. . .

Laplacian:
but 2*6+1=13

smurfy:

Are you sure the question isn't talking about any two consecutive terms?

Laplacian:
no

smurfy: In how many different ways can the letters P, Q, R, S, T be arranged if the letter R must be to the right of the letter S?

Laplacian:
case1: if R appears as first letter to d right,
number of ways of arranging d other letters, n=4!=24
case2: if R appears as second letter to d right,
n=4P3*3P1=24*3=72
case3: if R appears as third letter to d right,
n=4P2*3P2=12*6=72
case4: if R appears as fourth letter to d right,
n=4P1*3P3=4*6=24

so there are 192 possible ways of arrangement

Generally, if u are to arrange r objects so that one is always to d right of the other, then
case i: if object appears as i-th object to d right,
n=(r-1)P(r-i)*(r-2)P(i-1)

smurfy: In how many different ways can the letters P, Q, R, S, T be arranged if the letter R must be to the right of the letter S?

**Goes out to buy gift for whoever thrashes the question**

Laplacian: ...pls i need help with this questoin...either a solution or an idea is welcome...

Three non-zero integers are in arithmetic progression, the product of ANY two when when increased by one is a perfect square...find the numbers......

Laplacian: ...pls i need help with this questoin...either a solution or an idea is welcome...

Three non-zero integers are in arithmetic progression, the product of ANY two when when increased by one is a perfect square...find the numbers......

Laplacian:
case1: if R appears as first letter to d right,
number of ways of arranging d other letters, n=4!=24
case2: if R appears as second letter to d right,
n=4P3*3P1=24*3=72
case3: if R appears as third letter to d right,
n=4P2*3P2=12*6=72
case4: if R appears as fourth letter to d right,
n=4P1*3P3=4*6=24

so there are 192 possible ways of arrangement

Generally, if u are to arrange r objects so that one is always to d right of the other, then
case i: if object appears as i-th object to d right,
n=(r-1)P(r-i)*(r-2)P(i-1)

jackpot: Hi Mr Laps, the numbers are 0, 2, 4.

akpos4uall:

Using your logic, I think I have the solution
Case 1: S as first from left
n = 4! = 24

Case 2: S as 2nd from left
We are left with any 3 of P,Q,T for the 1st slot and the remaining 3 slots can be done in 3! ways
n = 3 * 3! = 18

Case 3: S as 3rd from left
We are left with 2 of the 3 P,Q,T for 1st & 2nd slot and the remaining 2 slots will be done in 2! ways
n = (3*2)*2! = 12

Case 4: S as 4th from left
Here P,Q,T can be arranged in any order in 3! ways for the first 3 slots and R in 1 way for the last slot
n = 3!*1 = 6

Total possible = 6 +12 +18 +24 = 60

smurfy: In how many different ways can the letters P, Q, R, S, T be arranged if the letter R must be to the right of the letter S?

**Goes out to buy gift for whoever thrashes the question**

smurfy:

Beautifully correct!

Actually, it's such a simple question. The answer is 5!/2 = 60.

How? Well, let's look at the letters A, B, C. Total possible arrangements are ABC, ACB, BAC, BCA, CAB, CBA. Total possible arrangements is 6. In how many ways, for example, is C to the right of A? Three ways.

You can pick any two letters and you'll always get n!/2 ways of having a particular letter to the right/left of another. This is always so since n! is always even for any positive integral value of n greater than 1.

A big THANK YOU to akpos4uall, Laplacian and aysuccess99 for doing justice to this question.

smurfy: In how many different ways can the letters P, Q, R, S, T
be arranged if the letter R must
be to the right of the letter S?

jaryeh:

If the letter R must be to the right of S,it means S can occupy only any of the first four positions. That can be done in four ways. The remaining four letters could be arranged in 4! ways which is 24,which means everything could be done in 24*4 ways,which is 96. But out of this 96 ways,3+2+1 of them may have S to the right of R which is contrary to the condition stated.

Hence,going by the condition,it can be done in (96-6)ways which is 90 ways.

smurfy:

Which condition are you referring to? If you're referring to the condition in the original question, then I can assure you that the answer provided is as correct as they get. It's 60 and not 90.

And, where have you been all these million years?

jaryeh:

Oga mi,I just go buy land for Mars ni o. You know say heaven and earth go soon pass away.

As per the question,I will take a better look at it when I get home in less than two hours' time.

Laplacian:
what u have reasoned is correct because the arrangement is symmetrical....it follows immediately that we have deceitfully proved that;

{summation i=1 to n-1}
(n-2)P(i-1)*(n-i)!=n!/2