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3 years ago

8

I have a circle that has a diameter of 10 inches

1 answer:

Stells [14]3 years ago

3 0

The area of the circle is 314.16

The formula is a= pi times the radius squared

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What is the instantaneous rate of change of f(x)=xe^x-(x 2)e^(x-1) at x=0 ?

lozanna [386]

You have to derive for a multiplication in both terms:
=e^x+xe^x-(e^x-1 + (x-2)e^x-1) now apply distributive property in the last term:
=e^x+xe^x+e^x-1-xe^x-1 now replace each x by 0 (x=0)
=1 + 0 + e^-1 + 0 = 1+ e^-1 = 1.3679

8 0

3 years ago

GIVING BRAINLIEST FOR BEST ANSWER!<br> Write a story that could represent this math problem.

xenn [34]

Answer:

Step-by-step explanation:

The third pig is only 2/3 of the way done with his strawberry house, the fourth pig builds a house out of sticks he’s only 2/3 of the way there, the last pig builds a house out of bricks he’s only 2/3 of the way there.

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2 years ago

Alex has collected apples. It has 40 boxes of 8.5 kg each and 6 sacks of 90kg each. Pack the apples in bags of 2.5 kg each. how

zimovet [89]

Answer:

40*8.5=340

6*90=540

total=880/2.5=352bags

3 0

3 years ago

Add the two expressions.

sattari [20]

Answer:

5.29

Step-by-step explanation:

6 0

3 years ago

A common inhabitant of human intestines is the bacterium Escherichia coli. A cell of this bacterium in a nutrient-broth medium d

nikitadnepr [17]

Answer:

a) k=2.08 1/hour

b) The exponential growth model can be written as:

$P(t)=Ce^{kt}$

c) 977,435,644 cells

d) 2.033 billions cells per hour.

e) 2.81 hours.

Step-by-step explanation:

We have a model of exponential growth.

We know that the population duplicates every 20 minutes (t=0.33).

The initial population is P(t=0)=58.

The exponential growth model can be written as:

$P(t)=Ce^{kt}$

For t=0, we have:

$P(0)=Ce^0=C=58$

If we use the duplication time, we have:

$P(t+0.33)=2P(t)\\\\58e^{k(t+0.33)}=2\cdot58e^{kt}\\\\e^{0.33k}=2\\\\0.33k=ln(2)\\\\k=ln(2)/0.33=2.08$

Then, we have the model as:

$P(t)=58e^{2.08t}$

The relative growth rate (RGR) is defined, if P is the population and t the time, as:

$RGR=\dfrac{1}{P}\dfrac{dP}{dt}=k$

In this case, the RGR is k=2.08 1/h.

After 8 hours, we will have:

$P(8)=58e^{2.08\cdot8}=58e^{16.64}=58\cdot 16,852,338= 977,435,644$

The rate of growth can be calculated as dP/dt and is:

$dP/dt=58[2.08\cdot e^{2.08t}]=120.64e^2.08t=2.08P(t)$

For t=8, the rate of growth is:

$dP/dt(8)=2.08P(8)=2.08\cdot 977,435,644 = 2,033,066,140$

(2.033 billions cells per hour).

We can calculate when the population will reach 20,000 cells as:

$P(t)=20,000\\\\58e^{2.08t}=20,000\\\\e^{2.08t}=20,000/58\approx344.827\\\\2.08t=ln(344.827)\approx5.843\\\\t=5.843/2.08\approx2.81$

3 0

3 years ago