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3 years ago

What's the anwser to all of these

Mathematics

2 answers:

zavuch27 [327]3 years ago

6 0

9) 0.3
10) 4.1
11) 0.5
12) 4.8
13) 3.2
14) 1.6
15) 3.9
16) 1.5

Hope this helped, please mark brainliest

Ratling [72]3 years ago

5 0

9).9/3= .3

10). 8.2/2= 4.1

11). 3.5/7= .5

12.)9.6/2=4.8

13.) 6.4/2= 3.2

14.) 4.8/3=1.6

15.) 7.8/2= 3.9

16.) 7.5/5=1.5

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The doubling period of a bacterial population is 15 minutes. At time t= 80 minutes, the bacterial population was 90000

Papessa [141]

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

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3 years ago

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Answer:

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Step-by-step explanation:

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Ulleksa [173]

Answer:

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Step-by-step explanation:

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Now let's insert x back in the expressions for angles.

m∠B = (11x − 10)° = (11(5) − 10)° = 45°

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m∠A + m∠B + m∠C = 180°

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