A regular heptagon has a radius of approximately 27.87 cm and the length of each side is 24.18 cm. what is the approximate area
1 answer:
The area of the regular heptagon which has a radius of approximately 27.87 cm and the length of each side is 24.18 cm is 2125 cm².
<h3>What is the area of a heptagon?</h3>Heptagon is the closed shape polygon which has 7 sides and 7 interior angles.
The area of the regular heptagon is found out using the following formula.
Here, (<em>a</em>) is the length of the heptagon.
A regular heptagon has a radius of approximately 27.87 cm and the length of each side is 24.18 cm. Put the value of side in the above formula,
Hence, the area of the regular heptagon which has a radius of approximately 27.87 cm and the length of each side is 24.18 cm is 2125 cm².
Learn more about the area of a heptagon here;
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The z-score corresponding to an observed value of x of 2 is 0.215.
Step-by-step explanation:
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Possible observations of X: -3, -1, 0, 1, 1, 2, 4, 4, 5.
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Mean of X, () =
=
= = 1.44
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=
= 2.603
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Answer:
The amount of heat required to raise the temperature of liquid water is 9605 kilo joule .
Step-by-step explanation:
Given as :
The mass of liquid water = 50 g
The initial temperature = = 15°c
The final temperature = = 100°c
The latent heat of vaporization of water = 2260.0 J/g
Let The amount of heat required to raise temperature = Q Joule
Now, From method
Heat = mass × latent heat × change in temperature
Or, Q = m × s × ΔT
or, Q = m × s × ( -
)
So, Q = 50 g × 2260.0 J/g × ( 100°c - 15°c )
Or, Q = 50 g × 2260.0 J/g × 85°c
∴ Q = 9,605,000 joule
Or, Q = 9,605 × 10³ joule
Or, Q = 9605 kilo joule
Hence The amount of heat required to raise the temperature of liquid water is 9605 kilo joule . Answer