This is false.
The f(3x) means you are multiplying the input 'x' by 3 before plugging into the equation.
3f(x) means you are going to multiply your answer (output) by 3 after you've plugged the input value in and gotten the output.
Answer:
4k^3 - k^2 - 2k
Step-by-step explanation:
-k * -4k^2 = 4k^3
-k * k = -k^2
-k * 2 = -2k
4k^3 - k^2 - 2k
THE CORRECT ANSWER IS N=4
Answer:
We can use the sample about 42 days.
Step-by-step explanation:
Decay Equation:



Integrating both sides


When t=0, N=
= initial amount




.......(1)
.........(2)
Logarithm:
130 days is the half-life of the given radioactive element.
For half life,
,
days.
we plug all values in equation (1)






We need to find the time when the sample remains 80% of its original.







We can use the sample about 42 days.
Answer:
1. 75 minutes
2. 5:50
Step-by-step explanation:
you have to add and subtract