1
is the answer you're looking for.
Answer: The required solution is 
Step-by-step explanation:
We are given to solve the following differential equation :
where k is a constant and the equation satisfies the conditions y(0) = 50, y(5) = 100.
From equation (i), we have
Integrating both sides, we get
![\int\dfrac{dy}{y}=\int kdt\\\\\Rightarrow \log y=kt+c~~~~~~[\textup{c is a constant of integration}]\\\\\Rightarrow y=e^{kt+c}\\\\\Rightarrow y=ae^{kt}~~~~[\textup{where }a=e^c\textup{ is another constant}]](https://tex.z-dn.net/?f=%5Cint%5Cdfrac%7Bdy%7D%7By%7D%3D%5Cint%20kdt%5C%5C%5C%5C%5CRightarrow%20%5Clog%20y%3Dkt%2Bc~~~~~~%5B%5Ctextup%7Bc%20is%20a%20constant%20of%20integration%7D%5D%5C%5C%5C%5C%5CRightarrow%20y%3De%5E%7Bkt%2Bc%7D%5C%5C%5C%5C%5CRightarrow%20y%3Dae%5E%7Bkt%7D~~~~%5B%5Ctextup%7Bwhere%20%7Da%3De%5Ec%5Ctextup%7B%20is%20another%20constant%7D%5D)
Also, the conditions are
and
Thus, the required solution is
>
the answer is “>” because 12.312 is greater than 12.132
Answer:
Transitive property of equality
Step-by-step explanation:
By the definition of transitivity, a relation R is said to be transitive if (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R.
If p = q, q = r then p = r.
Here, we have given that if ZXY = FDE and FDE = CAB, then, ZXY = CAB.
Therefore, it shows the transitive property of equality.
X = 4 ; x = 3 + i ; x = 3 - i
(If you get a zero that is adding or subtracting, you always need to write it twice but change the sign do they cancel out)
f(x) = (x-4)(x-3-i)(x-3+i)
Distributing the last two parenthesis first is always the best way to start off
(x-3-i)(x-3+i) has (x-3) in common so it can be separated to
(x-3)^2 + (-i)(+i)
(x^2 - 6x + 9) ; (-i)(+i) is always +1
(x^2 - 6x + 9) + 1
(x^2 - 6x + 10)
Now multiply this with (x-4)
x^3 - 6x^2 + 10x
- 4x^2 + 24x - 40
x^3 - 10x^2 + 34x - 40 = f(x)
