48 = 2(w+4) + 2w
48 = 4w+8
40 = 4w
w= 10 yd
A= 10 * 14
A = 140 yd
Answer:
here i finished!
hope it helps yw!
Step-by-step explanation:
The doubling period of a bacterial population is 15 minutes.
At time t = 90 minutes, the bacterial population was 50000.
Round your answers to at least 1 decimal place.
:
We can use the formula:
A = Ao*2^(t/d); where:
A = amt after t time
Ao = initial amt (t=0)
t = time period in question
d = doubling time of substance
In our problem
d = 15 min
t = 90 min
A = 50000
What was the initial population at time t = 0
Ao * 2^(90/15) = 50000
Ao * 2^6 = 50000
We know 2^6 = 64
64(Ao) = 50000
Ao = 50000/64
Ao = 781.25 is the initial population
:
Find the size of the bacterial population after 4 hours
Change 4 hr to 240 min
A = 781.25 * 2^(240/15
A = 781.25 * 2^16
A= 781.25 * 65536
A = 51,199,218.75 after 4 hrs
Answer:
B: 12.9
Step-by-step explanation:
If A B and C are the alternatives, the answer is B: 12.9
The reason is that the hypotenuse can't be smaller than one of its sides, that eliminates alternative C.
And also, the hypotenuse can't be bigger than segments AC + BC, since the side AC is bigger and AC measures 10.8, the hypotenuse would have to measure < 21.6
Any questions, comment.
H = 4(x + 3y) + 2
H = 4x + 12y + 2
H - 12y - 2 = 4x
(H - 12y - 2) / 4 = x or 1/4H - 3y - 1/2 = x