Answer: The required solution is
Step-by-step explanation: We are given to solve the following differential equation :
Let us consider that
be an auxiliary solution of equation (i).
Then, we have
Substituting these values in equation (i), we get
![m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.](https://tex.z-dn.net/?f=m%5E2e%5E%7Bmt%7D%2B10me%5E%7Bmt%7D%2B25e%5E%7Bmt%7D%3D0%5C%5C%5C%5C%5CRightarrow%20%28m%5E2%2B10y%2B25%29e%5E%7Bmt%7D%3D0%5C%5C%5C%5C%5CRightarrow%20m%5E2%2B10m%2B25%3D0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~%5B%5Ctextup%7Bsince%20%7De%5E%7Bmt%7D%5Cneq0%5D%5C%5C%5C%5C%5CRightarrow%20m%5E2%2B2%5Ctimes%20m%5Ctimes5%2B5%5E2%3D0%5C%5C%5C%5C%5CRightarrow%20%28m%2B5%29%5E2%3D0%5C%5C%5C%5C%5CRightarrow%20m%3D-5%2C-5.)
So, the general solution of the given equation is
Differentiating with respect to t, we get
According to the given conditions, we have
and
Thus, the required solution is
Answer:
Therefore, the budget is 656.76$.
Step-by-step explanation:
We know that a lab orders a shipment of 100 rats a week for experiments that the lab conducts. Suppose the mean cost of the rats turned out to be $ 12.63 per week. We calculate budget of a lab for the next year's.
We know that 1 year have 52 weeks. We get
52 · 12.63 = 656.76
Therefore, the budget is 656.76$.
Using the factor label method, you would set up your ratio like so:
Answer:
No
Step-by-step explanation:
No.
i = prt is correct; its result is the simple interest earned.
If you want to solve for time, t, divide both sides by pr:
i/(pr) = t
Fisrst one is just add the functions
(f+g)(x)=f(x)+g(x)=
8x^2+16x+6+x^3-3x^2-9=
(f+g)(x)=x^3+5x^2+16x-3
multiply
ok
f(x)g(x)=(x^3-4x+2)(x^2+2)
just multiply and expand yourself because I don't have my graphing calculator with me right now
