Answer:
It should be option B
Step-by-step explanation:
C' ( 0, 3)
A' ( 2 , 1)
C'A' = CA
C'A' = √(2^2 + 2^2) = √8 = 2√2
C'A' = CA = 2√2
Answer:

Step-by-step explanation:
we are given half-life of PO-210 and the initial mass
we want to figure out the remaining mass <u>after</u><u> </u><u>4</u><u>2</u><u>0</u><u> </u><u>days</u><u> </u>
in order to solve so we can consider the half-life formula given by

where:
- f(t) is the remaining quantity of a substance after time t has elapsed.
- a is the initial quantity of this substance.
- T is the half-life
since it halves every 140 days our T is 140 and t is 420. as the initial mass of the sample is 5 our a is 5
thus substitute:

reduce fraction:

By using calculator we acquire:

hence, the remaining sample after 420 days is 0.625 kg
I don’t know I need points
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
