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2 years ago

A regular pentagon has side lengths of 10 cm and an

Mathematics

1 answer:

Fittoniya [83]2 years ago

7 0

Answer: 172 square centimeters.

Step-by-step explanation:

Hi, to answer this question we have to apply the next formula:

Area of a pentagon (A) = 1/2× 5s × a

Where:

s : side of the pentagon

a: apothem

Replacing with the values given and solving for A (area):

A =1/2 x (5 x10) x 6.88

A = 172 square centimeters.

Feel free to ask for more if needed or if you did not understand something.

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Answer:

$\dfrac{m^3}{16p^2q^2}$

Step-by-step explanation:

Given:

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}$

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$2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}$

$(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}$

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$2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}$

$\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}$

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}$

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