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irina1246 [14]
3 years ago
8

Six equilateral triangles are connected to create a regular hexagon. The area of the hexagon is 24a^2 – 18 square units. Which i

s an equivalent expression for the area of the hexagon based on the area of a triangle? A. 6(4a^2 – 3) B. 6(8a^2 – 9) C. 6a(12a – 9) D. 6a(18a – 12)
Mathematics
2 answers:
netineya [11]3 years ago
6 0

we know that

The area of the hexagon is equal to the sum of the areas of the six equilateral triangles

Let

x-------> area of one equilateral triangle

so

6x=24a^{2} -18

Divide by 6 both sides

x=4a^{2} -3 -------> area of one equilateral triangle

To find an equivalent expression for the area of the hexagon based on the area of a triangle, multiply the area of one equilateral triangle by 6

6*(4a^{2} -3)

therefore

the answer is

The equivalent expression is equal to 6*(4a^{2} -3)

Leviafan [203]3 years ago
3 0

6(4a2 – 3) is an equivalent expression for the area of the hexagon based on the area of a triangle.

 

<span>A regular </span>hexagon<span> <span>can be cut into six equilateral triangles, and an equilateral triangle can be divided into two 30°- 60°- 90° triangles. So if you're doing a </span></span>hexagon<span> problem, you may want to cut up the figure and use equilateral triangles or 30°- 60°- 90° triangles to help you find the apothem, perimeter, or </span>area.

 

The correct answer between all the choices given is the first choice or letter A. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

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Answer:

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Step-by-step explanation:

The main thing we have to take into account in this question is that we are about to find the probability of a <em>sample mean</em>. The distribution for <em>sample means</em> follows a <em>normal distribution</em> with mean \\ \mu and standard deviation \\ \frac{\sigma}{\sqrt{n}}. Mathematically

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The probability that the mean battery life would be greater than 533.2 minutes

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\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

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\\ z = \frac{3.2}{\frac{58}{8.66025}}

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Most standard normal tables have values for z for only two decimals, so we can round the previous value for z as z = 0.48.

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However, in the question we are asked for \\ P(z>0.48) = P(x>533.2). As well as all normal distributions, the standard normal distribution is symmetrical around the mean, and we have

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\\ P(z>0.48) = 0.3156

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