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2 years ago

Find the area of a regular octagon with side lengths of 17.2 m and an apothem length of 20.8 m.

Mathematics

1 answer:

jek_recluse [69]2 years ago

5 0

Answer:

Pentagon a = 4.9 in. s = 7.1 in. /l ;/ >cvr'-. ~ j-?.1· '19~~. 5) Octagon a = 20.8 m s= 17.2m. ,4 ~;.>a-rt. ::: i, 17. z • t~f "g. ~ :::::, l'/?1 ,0 nt z, ... What is the length of the apothem to the nearest whole number

Step-by-step explanation:

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Can you help me find the answer to it pls and ty <3 .

grin007 [14]

The expression that represent the area of the cut off part is 81π / 4

The area of the cut off part can be found as follows:

Area of a circle = πr²

where

r = radius

Therefore, the cut off part is like 1 / 4th of a circle.

Hence,

area of the cut off region = πr² / 4

r = 9 inches

area of the cut off region = π9² / 4

area of the cut off region = 81π / 4

learn more on area here: brainly.com/question/15403474

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1 year ago

Which expression forms an equation with the given expression? 24 – 5 • 2 = ___________ A. (5 + 3) • 2 B. 2(6 + 1) C. 5 • 8 – 2 D

zhenek [66]

D. the rest is because of the characters rule

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2 years ago

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PLEAS4EE!!<br>180°- 125° 15'=<br>

boyakko [2]

Answer:

180.25°

Step-by-step explanation:

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2 years ago

An advertising company designs a campaign to introduce a new product to a metropolitan area of population 3 Million people. Let

Advocard [28]

Answer:

$P(t)=3,000,000-3,000,000e^{0.0138t}$

Step-by-step explanation:

Since P(t) increases at a rate proportional to the number of people still unaware of the product, we have

$P'(t)=K(3,000,000-P(t))$

Since no one was aware of the product at the beginning of the campaign and 50% of the people were aware of the product after 50 days of advertising

We have and ordinary differential equation of first order that we can write

$P'(t)+KP(t)= 3,000,000K$

The <em>integrating factor </em>is

$e^{Kt}$

Multiplying both sides of the equation by the integrating factor

$e^{Kt}P'(t)+e^{Kt}KP(t)= e^{Kt}3,000,000*K$

Hence

$(e^{Kt}P(t))'=3,000,000Ke^{Kt}$

Integrating both sides

$e^{Kt}P(t)=3,000,000K \int e^{Kt}dt +C$

$e^{Kt}P(t)=3,000,000K(\frac{e^{Kt}}{K})+C$

$P(t)=3,000,000+Ce^{-Kt}$

But P(0) = 0, so C = -3,000,000

and P(50) = 1,500,000

$e^{-50K}=\frac{1}{2}\Rightarrow K=-\frac{log(0.5)}{50}=0.0138$

And the equation that models the number of people (in millions) who become aware of the product by time t is

$P(t)=3,000,000-3,000,000e^{0.0138t}$

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2 years ago

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VLD [36.1K]

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1 year ago

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