Answer:
Step-by-step explanation:
Let the number of hours Nick does the job alone to be x - 3 hours
Hence, in 1 hour, Nick has done 1/x - 3
Together they complete the job in 9 hours
Hence, let's represent the job as 1
9/x + 9/x - 3 = 1
Multiply through by (x)(x-3)
9(x- 3) + 9(x) = 1
9x - 27 + 9x = (x)(x-3)
-27 = x² - 3x
x² - 3x +27 = 0
Answer:
x= 11.25
Step-by-step explanation:
cross multiply 9 by 5 divided by 4 and that would give you x
Answer:
11
Step-by-step explanation:
2.2x 5 = 11
The first 6 is 600,00 and the second is 60,000, so they are in the hundred thousands place and the thousands place
Answer: t-half = ln(2) / λ ≈ 0.693 / λExplanation:The question is incomplete, so I did some research and found the complete question in internet.
The complete question is:
Suppose a radioactive sample initially contains
N0unstable nuclei. These nuclei will decay into stable
nuclei, and as they do, the number of unstable nuclei that remain,
N(t), will decrease with time. Although there is
no way for us to predict exactly when any one nucleus will decay,
we can write down an expression for the total number of unstable
nuclei that remain after a time t:
N(t)=No e−λt,
where λ is known as the decay constant. Note
that at t=0, N(t)=No, the
original number of unstable nuclei. N(t)
decreases exponentially with time, and as t approaches
infinity, the number of unstable nuclei that remain approaches
zero.
Part (A) Since at t=0,
N(t)=No, and at t=∞,
N(t)=0, there must be some time between zero and
infinity at which exactly half of the original number of nuclei
remain. Find an expression for this time, t half.
Express your answer in terms of N0 and/or
λ.
Answer:
1) Equation given:
← I used α instead of λ just for editing facility..
Where No is the initial number of nuclei.
2) Half of the initial number of nuclei:
N (t-half) = No / 2So, replace in the given equation:
3) Solving for α (remember α is λ)
αt ≈ 0.693
⇒ t = ln (2) / α ≈ 0.693 / α ← final answer when you change α for λ