No, because they're in the same place value.
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
A and E are corresponding angles so A=E
To find t<span>he relative maximum value of the function we need to find where the function has its first derivative equal to 0.
Its first derivative is -7*(2x)/(x^2+5)^2
</span>7*(2x)/(x^2+5)^2 =0 the numerator needs to be eqaul to 0
2x=0
x=0
g(0) = 7/5
The <span>relative maximum value is at the point (0, 7/5).</span>
This question requires creating a few equations and working through them step-by-step. Now, first let's give each of the shapes a variable: let's say that the blue shape is a, the orange shape is b and the green shape is c.
1. We can technically create six formulas for the magic square, with three for sum of the rows and three for the sum of the columns, however the smartest way to approach this is to observe whether there are any obvious answers that we can get.
We can see in row 2 that there are three of the same shape (a) that add to 57. This makes it very simple to calculate the value of the shape.
Since 3a = 57
a = 57/3 = 19
2. Now we need to find a row or column that includes a and one other shape; we could choose either column 2 or 3, so let's go with column 2. Remembering that the blue shape is a and the orange shape is b:
2a + b = 50
Now, given that a = 19:
2(19) + b = 50
38 + b = 50
b = 12
3. We can now take any of the rows or columns that include the third shape (c) since we already know the values of the other two shapes. Let's take column 1:
a + b + c = 38
19 + 12 + c = 38
31 + c = 38
c = 38 - 31
c = 7
Thus, the value of the blue shape is 19, the value of the orange shape is 12 and the value of the green shape is 7.