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Bad White [126]

2 years ago

5

Derrick grows vegetables on a circular patch of land. The radius of the patch is 16 meters. What is the approximate area of the

patch of land?

1 answer:

Bezzdna [24]2 years ago

6 0

Answer:

804m²

Step-by-step explanation:

Area = pi × r²

3.14 × 16²

803.84 m²

Approximately, 804m²

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10 points and Brainly to who gets it right first

8_murik_8 [283]

Answer:

A.9

F.27

Step-by-step explanation:

how i got 9 is by counting by 9 and i count 27,and 54

additional 27 times 2= 54

5 0

3 years ago

A bottle is filled with 4oz of perfume. If 20% of the perfume evaporates, how many ounces remain?

marusya05 [52]

0.8 oz.
4x0.20=0.8
I hope this helps

7 0

2 years ago

Read 2 more answers

10,000,000,000×200,000

Snowcat [4.5K]

10,000,000,000×200,000 = 2e+15

5 0

3 years ago

Read 2 more answers

Please solve this ASAP! I WILL GIVE 20 points!

Bond [772]

Answer:

15 parts

Step-by-step explanation:

Sugar: 20/4=5 you are doing five times the recipe because your using 5 times as much sugars. So for the flour 3*5=15

3 0

3 years ago

After the mill in a small town closed down in 1970, the population of that town started decreasing according to the law of expon

wlad13 [49]

Answer:

$P_o = \frac{143000}{e^{-20*0.01303024661}}=110193.69$

And we can round this to the nearest up integer and we got 110194.

Step-by-step explanation:

The natural growth and decay model is given by:

$\frac{dP}{dt}=kP$ (1)

Where P represent the population and t the time in years since 1970.

If we integrate both sides from equation (1) we got:

$\int \frac{dP}{P} =\int kdt$

$ln|P| =kt +c$

And if we apply exponentials on both sides we got:

$P= e^{kt} e^k$

And we can assume $e^k = P_o$

And we have this model:

$P(t) = P_o e^{kt}$

And for this case we want to find $P_o$

By 1990 we have t=20 years since 1970 and we have this equation:

$143000 = P_o e^{20k}$

And we can solve for $P_o$ like this:

$P_o = \frac{143000}{e^{20k}}$ (1)

By 2019 we have 49 years since 1970 the equation is given by:

$98000 = P_o e^{49k}$ (2)

And replacing $P_o$ from equation (1) we got:

$98000=\frac{143000}{e^{20k}} e^{49k} =143000 e^{29k}$

We can divide both sides by 143000 we got:

$\frac{98000}{143000} =0.685 = e^{29k}$

And if we apply ln on both sides we got:

$ln(0.685) = 29k$

And then k =-0.01303024661[/tex]

And replacing into equation (1) we got:

$P_o = \frac{143000}{e^{-20*0.01303024661}}=110193.69$

And we can round this to the nearest up integer and we got 110194.

7 0

2 years ago