Consider a radioactive substance with original weight 20grams given by W(x)= 20 x (0.95)^x, where W(x) is with weight after x years. In about how many years is the radioactive substance half of its original weight? Type answer to one decimal place.

13.5 years

Step-by-step explanation:

We are given the expression:

W(x)= 20 x (0.95)^x

Where W(x) = 20grams/2

= 10 grams

Hence:

10 = 20 x (0.95)^x

Divide both sides by 20

10/20 = 20 x (0.95)^x/20

0.5 = (0.95)^x

Take the log of both sides

log 0.5 = log (0.95)^x

log 0.5 = x log (0.95)

Divide both sides by log 0.95

x = log 0.5 / log 0.95

x = 13.513407334 years

Approximately to 1 decimal place =✓13.5 years

Therefore, it will take 13.5 years for the radioactive substance to be half of its original weight.

hello from mrbilldoesmath!

answer:   x (x+6)^2, choice b

discussion:

factor x out of the equation:

x^3 + 12x^2 + 36x = x (x^2 + 12x + 36).

note that 6+6 = 12 and 6*6 = 36 so the above factors into:

x (x+6)^2

which is choice b

you,

mrb