Answer:
mM = 113 and mN = 61
Step-by-step explanation:
In a Cyclic quadrilateral, the rule states that:
The sum opposite interior angles is equal to 180°
In the above diagram, we have cyclic quadrilateral KLMN
According to the rule stated above:
Angle K is Opposite to Angle M
So, Angle K + Angle M = 180°
Angle K = 67°
67° + Angle M = 180°
Angle M = 180° - 67°
Angle M = 113°
Angle L is Opposite to Angle N
so Angle L + Angle N = 180°
Angle L is given as = 119°
119° + Angle N = 180°
Angle N = 180° - 119°
Angle N = 61°
Therefore, Angle M = 113° and Angle N = 61°
I think the answer is c, 88
Hope this helps :)
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