The mass of cesium left afer 120 years if the original mass of cesium is 8640 with an half-life of 30 years, is 540 grams.
<h3>What is half-life?</h3>
This can be defined as the time taken for half the amount of a radioactive substance to decay.
To calculate the amount of cesium left after 120 years, we use the formua below.
Formula:
- R' = R/()............. Equation 1
Where:
- R = Original amount of Cesium
- R' = New amount of cesium
- T = Total time
- t = Half-life
From the question,
Given:
- R = 8640
- T = 120 years
- t = 30 years
Substitute these values into equation 1
- R' = 8640/()
- R' = 8640/
- R' = 8640/16
- R' = 540 grams.
Hence, The mass of cesium left afer 120 years is 540 grams.
Learn more about half-life here: brainly.com/question/25750315
Answer:
q=1, q=2
Step-by-step explanation: Both values are less than 5 and the others are either equal or higher, which is not included in the inequality. I hope this helps. :)
Here's the whole equation;
tldr divide -6 by -2 and flip the inequality from < to >.
Answer:
T(n) = [n(n + 1)] ÷ 2
Step-by-step explanation:
Sequence = 1, 3, 6
First term = 1
Second term = first term + 2 = 3
Third term = second term + 3 = 6
From the sequence :
Each successive term is doubled together with an increment of '1', that is;
Term 'n' : n(n + 1)
Then, the total is divided by the term 'n'.
Therefore, the nth term is expressed as
T(n) = [n(n + 1)] ÷ 2
For instance,
n = 1
T(1) = [1(1 + 1)] ÷ 2
T(1) = 1
n = 3
T(3) = [3(3 + 1)] ÷ 2
T(3) = 12 ÷ 2 = 6
.......
Answer:
w = -10,6
Step-by-step explanation:
w(4+w)=60
4w + w^2 -60 = 0
w^2 +4w -60 = 0
w^2 +10w -6w -60 = 0
w(w+10) -6(w+10) = 0
w = -10, 6