7s = 4
/7. /7
s= 4/7
thats the equation right?
Do you mean an extraneous solution? I will tell you what I think. To tell if a solution is extraneous you need to go back to the original problem and check if there actually is a solution or if the solution is actually right. If not then it is extraneous.
Answer:Oy=2x+3
Step-by-step explanation:
Step-by-step explanation:
4 and 5 is one 20/25
Answer:
5.6 days
Step-by-step explanation:
We are given;
Initial Mass; N_o = 25 g
Mass at time(t); N_t = 25/2 = 12.5 (I divide by 2 because we are dealing with half life)
k = 0.1229
Formula is given as;
N_t = N_o•e^(-kt)
Plugging in the relevant values;
12.5 = 25 × e^(-0.1229t)
e^(-0.1229t) = 12.5/25
e^(-0.1229t) = 0.5
(-0.1229t) = In 0.5
-0.1229t = -0.6931
t = -0.6931/-0.1229
t = 5.6 days
