An unknown radioactive material is measured to have a half life of 3 months. When the material was first found, there was 2000mg. a) Write an equation that models the mass of the material, t months. b) Use your equation to determine the mass of material in 4 year c) Calculate around how many months it will take to have 750 mg left

OK!!.

N=N(½)ⁿ

n= Time/half life

N=Remaining Mass

N°=Initial Mass or Mass before decay.

t= time taken to decay(Its in Months in this case)

t½= Half Life of the Material. This is the time taken to decay to half its initial value.

N°= 2000mg

a).Equation that Models this is

since n=t/t½

N=N°(½)ⁿ =

N=N°(½)^t/t¹'². This should be your answer.

b). We're asked to find the remaining mass of substance in 4years.

t= 4years

Our Half life is in Months... So we gotta convert or time t from year to Months too.

4yrs === 4x12 = 48Months.

N° was given as 2000mg

N=N°(½)^t/t½

N= 2000(½)^48/3

N=2000(½)^16

Using your calc to evaluate (½)^16... Then multiply by 2000

N=0.0305mg will remain after 4years.

Or After 16Half Lives since 1 half life is 3months

c). We're looking for t this time

N=N°(½)^t/t½

Since it asked for 750mg to remain ... 750 is now our N --- Remaining Mass

750 = 2000(½)t/3

To Isolate "t" and make it the subject

750/2000 = (½)^t/3

0.375 = (½)^t/3

Taking ln(natural log) of both sides

Ln(0.375) = Ln(0.5)^t/3

From the rule of logarithm...

You can bring the power (I.e t/3) to the front

You'll have

Ln(0.375) = t/3Ln(0.5)

Dividing both sides by Ln(0.5) to isolate t

Ln(0.375)/Ln(0.5) = t/3

t/3 = 1.415

t= 3x1.415

t=4.25months.

Have a great day.

Hope this helps... I'm open to questions if you have any too.

answer: i realy don't get it that much can you me figer it out pleas you

step-by-step explanation:

la parábola es el lugar

geométrico en el cual los puntos tienen una distancia idéntica, equidistantes,

de un punto denominado foco, f, y una recta denominada directriz

el radio vector de una

parábola es el segmento que existe entre un punto cualquiera y el foco.

por la ecuación se observa

que la parábola abre hacia la derecha, por lo cual podemos usar las siguientes fórmulas

la ecuación ordinaria de la

parábola es y^2 = 4px, donde p es la longitud desde el vértice de la parábola al

foco.

la fórmula de la directriz

es x + p = 0

resolviendo:

y^2-9x = 0

y^2 = 9x

4p = 9, entonces p = 9/4

entonces el foco tendrá la

coordenada f(9/4,0)

la directriz vendrá

expresada por

x + p = 0

x = -p

x = -9/4

el problema indica que la

ordenada es igual a 6, entonces

(6)^2=9x

x = 36/9

x = 4

entonces el punto al cual

hay que obtener la distancia será p(4,6)

calculando se tendrá: pf = 

pf = 6,25 u

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step-by-step explanation: