The doubling period of a baterial population is 10 minutes. At time t=120 minutes, the baterial population was 80000.
What was the initial population at time t=0
Find the size of the baterial population after 3 hours.

i think that you should ask google for the answer to that problem because if we don't know it google does

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)

33/100 is what i got

a) 40 more; b) old factory:   10%, new factory:   16%; c) week 4

explanation:

a) at week 0, we see in the graph that the new factory produces 190 specialty items.  

using the function, substituting 0 in for w,

this is a difference of 230-190 = 40.

b) the function for the old factory is written in the form f(x) = a(1+r)ˣ, where a is the original amount, x is the amount of time and r is the growth rate.   the equation for the old factory is ; 230 is in place of a and is the original amount; w is the amount of time; which means 1+r = 1.1.   this means that r must be 0.1 or 10%.

for the new factory, we find the percent of increase to find the growth rate; the formula for this is

between the first two points, the amount of change is 220-190 = 30; this gives us

30/190 = 0.158 ≈ 0.16 = 16%

c) at week 0, the old factory has 230 while the new factory has 190.

at week 1, the old factory has 230(1.1) = 253 while the new factory has 220.

at week 2, the old factory has 230(1.1²) = 278.3 while the new factory has 252.

at week 3, the old factory has 230(1.1³) = 306.13 while the new factory has 290.

at week 4, the old factory has 230(1.1⁴) = 336.743 while the new factory has 337.   this is the first week that the new factory surpasses the old.