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.

step-by-step explanation:

what is the figure? attach a picture.

income. 10x+1300

expenses 3x+(1800+1500)

expenses 3x+3300

to break even:

income = expense,

10x + 1300 = 3x + 3300,

solve for x.

step-by-step explanation:

mark brainliest