Question

I really need a help with one question.

Answer 1

So basically there is initial volume of 125 ml of ice halving its quantity every 5 minutes

Answer 2

how can I express this in the function form ?

Answer 3

Hint: You need to set up an exponential equation

Answer 4

Final quantity after=125ml^0.5

Answer 5

What a shame.

Answer 6

I am stuck with this simple problem jaja

Answer 7

Any ideas?

Answer 8

f(x) = 125 (0.5)^x

where x is the number of 5 minute intervals

Answer 9

so at x = 0
fX) = 125
at x = 1

f(x) = 62.5

Answer 10

Not sure if thats the best answer though......

Answer 11

125(0.5)^(x/5)

Answer 12

you're looking at a half life

https://en.wikipedia.org/wiki/Half-life#Formulas_for_half-life_in_exponential_decay

we can use "proper" exponents, if you like !!!

Answer 13

Yes - of course!!

Answer 14

okay so the volume initially \(V\)=\(125ml\)

after say time\(T\) the volume will be \(V'\)
well the volume will go like this->
after 5mins->\(\large\frac{125}{2^1}\)
after 5x2mins->\(\large\frac{125}{2^2}\)
after 5x3mins->\(\large\frac{125}{2^3}\)

we convert it to a general form for any time \(T\) we can say that-
\(\large{V'=\frac{125}{2^{[\frac{T}{5}]}}}\)

here \([.]\) is the greatest integer function
an explanation to greatest integer function-
if T=18
then T/5 is 3.6
and when u put that in greatest integer function u get this->
\([3.6]=3\)

Answer 15

That's amazing

Answer 16

start with \(V = V_o e^{- \lambda t}\)

or \(\dfrac{V}{ V_o} = e^{- \lambda t}\)

and you know that
\(\dfrac{1}{ 2} = e^{- \lambda \times 5}\)

so \(\lambda = {1 \over 5} \ln 2\)

Answer 17

It's amazing how math can express what language cannot

Answer 18

:)

Answer 19

Thanks so much guys

Answer 20

np (: