A regular decagon is rotated n degrees about its center, carrying the decagon onto itself. The value of n could be:

Mathematics

1 answer:

Delicious77 [7]3 years ago

4 0

Answer:

252°

Step-by-step explanation:

Given

Shape: Decagon

Required

Determine the value of n

A Decagon has 10 sides and a complete rotation takes 360°

First, we need to calculate the angle of each rotation as follows.

Each Rotation = ⅒ * 360°

Each Rotation = 360°

Next, we determine the possible values of n.

The statement is represented as "possible values of n" because n can assume infinite number of values.

However, n must be a multiple of 36

From the list of given options.

10, 150 and 225 are not multiples of 36

Only 252 happens to be a multiple of 36 from the list of given options.

Hence, option D answers the question [base on the given options]

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What are the domain range of f(x)= 1/5 power of x

Snowcat [4.5K]

Answer:

Domain: all real numbers

Range: all real numbers

Step-by-step explanation:

I'm assuming you mean the function $f(x)=x^{\frac{1}{5}}$ . That's usually written

f(x) = x^(1/5) with the ^ meaning "to the power of..." and the fraction exponent in parentheses so as not to be confused with x^1/5 which could mean x to the first power, divided by 5.

Fractional exponents are used to indicate roots. In this case, x is being raised to the 1/5 power, so this is the fifth root of x, written $\sqrt[5]{x}$ . The 5 is called the root index.

For odd roots, like this one, the domain is all real numbers--x can be any number at all. So the domain is all real numbers.

The range is also all real numbers. Attached is a graph of this function. It might not look like it, but the graph rises to the right to any height. The larger x gets, the larger the 5th root gets. A similar thing happens on the left--the smaller x gets, the smaller the 5th root gets.

EDIT: see the comment. For the function $f(x)=(1/5)^x$ , the domain is all real numbers. The range is positive real numbers. I'll attach a graph!