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3 years ago

5

A regular dodecagon (12 sided polygon) is rotated n degrees about it's center, carrying the dodecagon onto itself. What is the m

inimum number of degrees it must be rotated to carry the dodecagon onto itself? *
A. 30 degrees
B. 15 degrees
C. 60 degrees
D. 90 degrees
Which one?

1 answer:

mojhsa [17]3 years ago

7 0

Answer:

30 degrees

Step-by-step explanation:

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Step-by-step explanation:

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2 years ago

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3 years ago

Which of the following is a like radical to RootIndex 3 StartRoot 7x EndRoot?

lana [24]

Answer:

A. 4 (RootIndex 3 StartRoot 7 x EndRoot) or $4(\sqrt[3]{7x})$

Step-by-step explanation:

Given:

A radical whose value is, $r_1=\sqrt[3]{7x}$

Now, we need to find the like radical for $r_1$ .

Let the like radical be $r_2$ .

As per the definition of like radicals, like radicals are those that can be expressed as multiples of each other.

So, if two radicals $r_1\ and\ r_2$ are like radicals, then

$r_1 = n \times r_2
$

Where, 'n' is a real number.

Here, $r_1=\sqrt[3]{7x}$

Now, let us check all the options .

Option A:

4 (RootIndex 3 StartRoot 7 x EndRoot) or $r_2=4\sqrt[3]{7x}$

Now, we observe that $r_2$ is a multiple of $r_1$ because

$r_2=4\times \sqrt[3]{7x}\\\\ r_2=4\times r_1..............(r_1=\sqrt[3]{7x})$

Therefore, option A is correct.

Option B:

StartRoot 7 x EndRoot or $r_2=\sqrt{7x}$

As the above radical is square root and not a cubic root, this option is incorrect.

Option C:

x (RootIndex 3 StartRoot 7 EndRoot) or $r_2=x\sqrt[3]{7}$

As the term inside the cubic root is not same as that of $r_1$ , this option is also incorrect.

Option D:

7 StartRoot x EndRoot or $r_2=7\sqrt{x}$

As the above radical is square root and not a cubic root, this option is incorrect.

Therefore, the like radical is option (A) only.

4 0

2 years ago

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