In a radioactive substance half-life is 40 years about how much of a 30g sample was left after 160 yearsa 1.9b 3.8c 7.5d 15

A 1.9

Step-by-step explanation:

Nt = No e^(-kt)

Where Nt is the ending amount

No is the starting amount

k is the decay constant

and t is the time

We need to determine the half life

Nt/No = e ^ -kt

1/2 of the original amount is left  for 1/2 life.  The time is 40 years.  We will calculate k

1/2 = e ^ -k * 40

Take the natural log of each side

ln(1/2) = ln (e ^ -k * 40)

ln (1/2) = -40k

-ln(2) = -40k

ln 2 = 40k

1/40 ln (2) = k

Now we can calculate now much is left after 160 years of a 30 g sample

Nt = 30 e^(- 1/40 ln(2) * 160)

    = 30 e^(-4ln(2))

    = 15/8

       =1.875

f(x) = y

f(-3) = -5

step-by-step explanation:

-362

step-by-step explanation:

(12)(4)+2(12)-34(12)+2(12)-3

  \/         \/         \/         \/

  48 +   24 -   408 + 24 - 3

      \   /               \   /

          72     -     432 - 3 = -363