Cesium-137 is a particularly dangerous by-product of nuclear reactors. It has a half-life of 30 years. It can be readily absorbed into the food chain. Suppose we place 3,000 grams of
cesium-137 in a nuclear waste site.
(a) How much cesium-137 will be present after 30 years, or one half-life?
9
How much cesium-17 will be present after 60 years, or two half-lives?
(b) Find an exponential formula that gives the amount G in grams of cesium-137 remaining in the site after h half-lives.
(c) How many half-lives of cesium-137 is 250 years? Round your answer to two decimal places.
half-lives
(d) Use your answer to part (c) to determine how much cesium-137 will be present after 250 years. Round your answer to the nearest whole number

answer: i would say b

step-by-step explanation:

with the way the question is worded, i believe b is the answer

the cost of the apple depends on the weight

pay attention in your classes and concentrate the best thing you can do to raise your test scores is to pay attention when you're supposed to be learning the material: in class!

take good notes.

do your homework.

use mnemonics and other tricks.

do practice tests.

note: the best i know.

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt