Alright so I'm coming up with this on the fly; you have the first six letters (a,b,c,d,e,f) and 0-9 and your ten numbers. calculate the amount of possible combinations for the letters by simply writing them down.
ab, ac, ad, ae, a f- five
bc, bd , be, bf- four
cd, ce, cf- three
de, df- two
ef,- one.
adding these all together gets a total of 15 for the letters. now the numbers
01, 02, 03, 04, 05, 06, 07, 08, 09- nine
12, 13, 14, 15, 16, 17, 18, 19- eight
23, 24, 25, 26, 27, 28, 29- seven
34, 35, 36, 37, 38, 39- six
45, 46, 47, 48, 49- five
56, 57, 58, 59- four
67, 68, 69- three
78, 79- two
89- one
added together with a total of 45 combinations.
alright so, 45 different number combinations and 15 letter combinations. multiplying 15 by 45 should tell you the total possible combinations for a two letter and two number serial-number
Answer:
The initial population was 2810
The bacterial population after 5 hours will be 92335548
Step-by-step explanation:
The bacterial population growth formula is:
where P is the population after time t, 
is the starting population, i.e. when t = 0, r is the rate of growth in % and t is time in hours
Data: The doubling period of a bacterial population is 20 minutes (1/3 hour). Replacing this information in the formula we get:
Data: At time t = 100 minutes (5/3 hours), the bacterial population was 90000. Replacing this information in the formula we get:
Data: the initial population got above and t = 5 hours. Replacing this information in the formula we get:
Answer:
Step-by-step explanation:a = m + (p-1)*d
b = m + (q-1)*d
c = m + (r-1)*d
p(b-c) = p*(q-r)*d
q(c-a) = q*(r-p)*d
r(a-b) = r*(p-q)*d
p(b-c)+q(c-a)+r(a-b)
= p*(q-r)*d + q*(r-p)*d +r*(p-q)*d
= (pq-pr+qr-pq+rp-qr)*d
= 0*d = 0
So i prove p(b-c)+q(c-a)+r(a-b)=0 hope this is helpfull
