Graph -5, 0, 2,and 4 on a numberline
1 answer:
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Answer:
Step-by-step explanation:
Let f be a function from N to N.
N_set of all natural numbers
i) one to one but not onto
consider the function
When two numbers have same square we find that the numbers should be the same because they are positive.
So one to one but not onto because consider 3 it does not have square root in N.
ii) Onto but not one to one
Consider
this is onto because every number has a preimage in N.
But not onto because consider 6 and 3, f(6) = 3 and f(3) =3
So not one to one
iii) both onto and one-to-one
f(x) =
=x+1, x even
This is both one to one and onto since we consider only integers
iv) Neither one to one nor onto
Consider the function
f(x) = 2
This is not onto because 3 cannot have a preimage in N, not one to one because f(1) = f(2) where 1 not equals 2
An expression for this timet-half will be given as ln(2) / λ ≈ 0.693 / λ
<h3>What is an expression?</h3>
Expression in maths is defined as the collection of the numbers variables and functions by using signs like addition, subtraction, multiplication and division.
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.
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 / 2
So, 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
Therefore expression for this timet-half will be given as ln(2) / λ ≈ 0.693 / λ
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