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s344n2d4d5 [400]

3 years ago

14

All the sides of a hexagon become three times the original length. find the ratio of areas of the new and old hexagons.

1 answer:

Anastaziya [24]3 years ago

7 0

To determine the ratio, we need to know the formula of the area of an hexagon in terms of the length of its sides. We cannot directly conclude that the ratio would be 3, the same as that of the ratio of the lengths of the side, since it may be that the relationship of the area and length is not equal. The area of a hexagon is calculated by the expression:

A = (3√3/2) a^2

So, we let a1 be the length of the original hexagon and a2 be the length of the new hexagon.

A2/A1 = (3√3/2) a2^2 / (3√3/2) a1^2
A2/A1 = (a2 / a1)^2 = 3^2 = 9

Therefore, the ratio of the areas of the new and old hexagon would be 9.

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gayaneshka [121]

Answer:

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Step-by-step explanation:

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vodomira [7]

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Vrite your definition of square numbers. Square numbers: <br>

oksian1 [2.3K]

Answer:

Square numbers are numbers that have been multiplied by themselves.

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3 years ago

How to compare 10.5 to 11 and 19

Marizza181 [45]

Answer:

Line up the decimal points. **Answer is Least to greatest: 10.5 , 11, 19**

Step-by-step explanation:

You need to line each number up.

10.5

11.0

19.0

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This problem is also a little common sense because 10.5 is the smallest, 19 is the largest while 11 is in the middle.

4 0

2 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Csum_%7Bn%20%3D%201%7D%5E%7B%20%5Cinfty%20%7D%20%20%7B4%7D%5E%7B%20-%20

r-ruslan [8.4K]

Answer:

$\dfrac{1}{3}$

Step-by-step explanation:

<u>Given</u>:

$\displaystyle \sum^{\infty}_{n=1} 4^{-n}$

The <u>sigma notation</u> means to find the sum of the given <u>geometric series</u> where the first term is when n = 1 and the last term is when n = ∞.

To find the <u>first term</u> in the series, substitute n = 1 into the expression:

$\implies a_1=4^{-1}=\dfrac{1}{4}$

The <u>common ratio</u> of the geometric series can be found by dividing one term by the previous term:

$\implies r=\dfrac{a_2}{a_1}=\dfrac{4^{-2}}{4^{-1}}=\dfrac{1}{4}$

As | r | < 1 the series is <u>convergent.</u>

When a series is convergent, we can find its sum to infinity (the limit of the series).

<u>Sum to infinity formula</u>:

$S_{\infty}=\dfrac{a}{1-r}$

where:

a is the first term in the series
r is the common ratio

Substitute the found values of a and r into the formula:

$S_{\infty}=\dfrac{\frac{1}{4}}{1-\frac{1}{4}}=\dfrac{1}{3}$

Therefore:

$\displaystyle \sum^{\infty}_{n=1} 4^{-n}=\dfrac{1}{3}$

Learn more about Geometric Series here:

brainly.com/question/27737241

brainly.com/question/27948054

6 0

2 years ago