What is the circumference of the circle when the r is 19 m (use 3.14 for pi)

givi [52]

Answer:

The definition for the given piecewise-defined function is: $\boxed{\displaystyle\sf\ Option\:D:\:\: f(x) = \begin{cases}\displaystyle\sf\ x + 2 & \sf\:{if\:\:x \leq -1} \\\displaystyle\sf\ 2x + 4 & \sf\:{if\:\:x > -1}\end{cases}}$ .

Step-by-step explanation:

<h3>General Concepts:</h3>

Piecewise-defined functions.
Interval notations.

<h3>What is a piecewise-defined function?</h3>

A piecewise-defined function represents specific rules over different intervals of the domain.

<h3>Symbols used in expressing interval notations:</h3>

Open interval: This means that the endpoint is <em>not</em> included in the interval.

We can use the following symbols to indicate the <u>exclusion</u> of endpoints in the interval:

Left or right parenthesis, "( )" (or both).

Greater than (>) or less than (<) symbols.
Open dot " $\circ$ " is another way of expressing the exclusion of an endpoint in the graph of a piecewise-defined function.

Closed interval: This implies the inclusion of endpoints in the interval.

We can use the following symbols to indicate the <u>inclusion</u> of endpoints in the interval:

Open- or closed brackets (or both), "[ ]."

Greater than or equal to (≥) or less than or equal to (≤) symbols.
Closed circle or dot, "•" is another way of expressing the <em>inclusion</em> of the endpoint in the graph of a piecewise-defined function.

<h2>Determine the appropriate function rule that defines different parts of the domain. </h2>

The best way to determine which piecewise-defined function represents the graph is by observing the <u>endpoints</u> and <u>orientation</u> of both partial lines.

Open circle on (-1, 2): The graph shows that one of the partial lines has an <em>excluded</em> endpoint of (-1, 2) extending towards the <u>right</u>. This implies that its domain values are defined when x > -1.

Closed circle on (-1, 1): The graph shows that one of the partial lines has an <em>included</em> endpoint of (-1, 1) extended towards the <u>left</u>. Hence, its domain values are defined when x ≤ -1.

Based on our observations from the previous step, we can infer that x > -1 or x ≤ -1 apply to piecewise-defined functions A or D. However, only one of those two options represent the graph.

<h2>Solution:</h2><h3>a) Test option A:</h3>

$\boxed{\displaystyle\sf Option\:A)\:\:\:f(x) = \begin{cases}\displaystyle\sf\ 2x + 2 & \sf\:{if\:\:x \leq -1} \\\displaystyle\sf\ x + 4 & \sf\:{if\:\:x > -1}\end{cases}}$

<h3>Piece 1: If x ≤ -1, then it is defined by f(x) = 2x + 2. </h3>

We must choose a domain value that falls within the interval of x ≤ -1 whose output is included is included in the graph of the partial line with a <u>closed dot</u>.

Substitute x = -2 into f(x) = 2x + 2:

f(x) = 2x + 2

f(-2) = 2(-2) + 2
f(-2) = -4 + 2
f(-2) = -2 ⇒ <em>False statement</em>.

⇒ The output value of f(-2) = -2 is <u>not</u> included in the graph of the partial line whose endpoint is at (-1, 1).

<h3>Piece 2: If x > -1, then it is defined by f(x) = x + 4. </h3>

We must choose a domain value that falls within the interval of x > -1 whose output is included in the graph of the partial line with an <u>open dot</u>.

Substitute x = 0 into f(x) = x + 4:

f(x) = x + 4
f(0) = (0) + 4
f(0) = 4 ⇒ <em>True statement</em>.

⇒ The output value of f(0) = 4 <u>is</u> included in the graph of the partial line whose endpoint is at (-1, 2).

Conclusion for Option A:

Option A is not the correct piecewise-defined function because one of the pieces, f(x) = 2x + 2, does not specify the interval (-∞, -1].

<h3>b) Test option D:</h3>

$\boxed{\displaystyle\sf Option\:D)\:\:\:f(x) = \begin{cases}\displaystyle\sf\ x + 2 & \sf\:{if\:\:x \leq -1} \\\displaystyle\sf\ 2x + 4 & \sf\:{if\:\:x > -1}\end{cases}}$

<h3>Piece 1: If x ≤ -1, then it is defined by f(x) = x + 2. </h3>

We must choose a domain value that falls within the interval of x ≤ -1 whose output is included is included in the graph of the partial line with a <u>closed dot</u>.

Substitute x = -2 into f(x) = x + 2:

f(x) = x + 2
f(-2) = (-2) + 2
f(-2) = 0 ⇒ <em>True statement</em>.

⇒ The output value of f(-2) = 0 <u>is</u> included the graph of the partial line whose endpoint is at (-1, 1).

<h3>Piece 2: If x > -1, then it is defined by f(x) = 2x + 4.</h3>

We must choose a domain value that falls within the interval of x > -1 whose output is included is included in the graph of the partial line with an <u>open dot</u>.

Substitute x = 0 into f(x) = 2x + 4:

f(x) = 2x + 4
f(0) = 2(0) + 4
f(0) = 0 + 4 = 0 ⇒ <em>True statement</em>.

⇒ The output value of f(0) = 4 <u>is</u> included in the graph of the partial line whose endpoint is at (-1, 2).

<h2>Final Answer: </h2>

We can infer that the piecewise-defined function that represents the graph is:

$\boxed{\displaystyle\sf\ Option\:D:\:\: f(x) = \begin{cases}\displaystyle\sf\ x + 2 & \sf\:{if\:\:x \leq -1} \\\displaystyle\sf\ 2x + 4 & \sf\:{if\:\:x > -1}\end{cases}}$ .

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Learn more about piecewise-defined functions here:

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