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2 years ago

If each interior angle of a regular polygon is 168 °, then the number of sides of the polygon is ….

Mathematics

2 answers:

Nimfa-mama [501]2 years ago

7 0

the answer to your question is 30 sides

hope this helps

Tems11 [23]2 years ago

6 0

<h3>given:</h3>

each interior angle= 168°

the number of sides the polygon has.

<h3>solution:</h3>

180- 360/n= 168

12= 360/n

n= 360/12

n= 30

therefore, the polygon has 30 sides.

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Find the 11th term of the following geometric sequence. 1,5, 25, 125, ...

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Answer:

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Step-by-step explanation:

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3 years ago

When babies are born, their lengths are recorded. Maddy is 18 1/2 inches. Her twin Alex, is 20 3/4 inches. How much longer is Ma

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Alex is 2 1/4 inches longer than Maddy

Step-by-step explanation:

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3 years ago

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In 2013, voyager 1 traveled 1,468,800 kilometers each day. Whats that in scientific notation

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3 years ago

Could the inverse of a non-function be a function? Explain or give an example.

Kitty [74]

Answer:

The inverse of a non-function mapping is not necessarily a function.

For example, the inverse of the non-function mapping $\lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\!$ is the same as itself (and thus isn't a function, either.)

Step-by-step explanation:

A mapping is a set of pairs of the form $(a,\, b)$ . The first entry of each pair is the value of the input. The second entry of the pair would be the value of the output.

A mapping is a function if and only if for each possible input value $x$ , at most one of the distinct pairs includes $x\!$ as the value of first entry.

For example, the mapping $\lbrace (0,\, 0),\, (1,\, 0) \rbrace$ is a function. However, the mapping $\lbrace (0,\, 0),\, (1,\, 0),\, (1,\, 1) \rbrace$ isn't a function since more than one of the distinct pairs in this mapping include $1$ as the value of the first entry.

The inverse of a mapping is obtained by interchanging the two entries of each of the pairs. For example, the inverse of the mapping $\lbrace (a_{1},\, b_{1}),\, (a_{2},\, b_{2})\rbrace$ is the mapping $\lbrace (b_{1},\, a_{1}),\, (b_{2},\, a_{2})\rbrace$ .

Consider mapping $\lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\!$ . This mapping isn't a function since the input value $0$ is the first entry of more than one of the pairs.

Invert $\lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\!$ as follows:

$(0,\, 0)$ becomes $(0,\, 0)$ .
$(0,\, 1)$ becomes $(1,\, 0)$ .
$(1,\, 0)$ becomes $(0,\, 1)$ .
$(1,\, 1)$ becomes $(1,\, 1)$ .

In other words, the inverse of the mapping $\lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\!$ would be $\lbrace (0,\, 0),\, (1,\, 0),\, (0,\, 1),\, (1,\, 1) \rbrace\!$ , which is the same as the original mapping. (Mappings are sets. There is no order between entries within a mapping.)

Thus, $\lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\!$ is an example of a non-function mapping that is still not a function.

More generally, the inverse of non-trivial ellipses (a class of continuous non-function $\mathbb{R} \to \mathbb{R}$ mappings, including circles) are also non-function mappings.

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2 years ago

find the spectral radius of A. Is this a convergent matrix? Justify your answer. Find the limit x=lim x^(k) of vector iteration

Natasha_Volkova [10]

Answer:

The solution to this question can be defined as follows:

Step-by-step explanation:

Please find the complete question in the attached file.

$A = \left[\begin{array}{ccc} \frac{3}{4}& \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2}& 0\\ -\frac{1}{4}& -\frac{1}{4} & 0\end{array}\right]$

now for given values:

$\left[\begin{array}{ccc} \frac{3}{4} - \lambda & \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2} - \lambda & 0\\ -\frac{1}{4}& -\frac{1}{4} & 0 -\lambda \end{array}\right]=0 \\\\$

$\to (\frac{3}{4} - \lambda ) [-\lambda (\frac{1}{2} - \lambda ) -0] - 0 - \frac{1}{4}[0- \frac{1}{2} (\frac{1}{2} - \lambda )] =0 \\\\\to (\frac{3}{4} - \lambda ) [(\frac{\lambda}{2} + \lambda^2 )] - \frac{1}{4}[\frac{\lambda}{2} - \frac{1}{4}] =0 \\\\\to (\frac{3}{8}\lambda + \frac{3}{4} \lambda^2 - \frac{\lambda^2}{2} - \lambda^3 - \frac{\lambda}{8} + \frac{1}{16}=0 \\\\\to (\lambda - \frac{1}{2}) (\lambda -\frac{1}{4}) (\lambda - \frac{1}{2}) =0\\\\$

$\to \lambda_1=\lambda_2 =\frac{1}{2}\\\\\to \lambda_3 = \frac{1}{4} \\\\\to A = max {|\lambda_1| , |\lambda_2|, |\lambda_3|}\\\\$

$= max{\frac{1}{2}, \frac{1}{2}, \frac{1}{4}}\\\\= \frac{1}{2}\\\\(A) =\frac{1}{2}$

In point b:

Its

spectral radius is less than 1 hence matrix is convergent.

In point c:

$\to c^{(k+1)} = A x^{k}+C \\\\\to x(0) = \left(\begin{array}{c}3&1&2\end{array}\right) , c = \left(\begin{array}{c}2&2&4\end{array}\right)\\\\ \to x^{(k+1)} = \left[\begin{array}{ccc} \frac{3}{4}& \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2}& 0\\ -\frac{1}{4}& -\frac{1}{4} & 0\end{array}\right] x^k + \left[\begin{array}{c}2&2&4\end{array}\right] \\\\$

after solving the value the answer is

$\lim_{k \to \infty} x^k=o = \left[\begin{array}{c}0&0&0\end{array}\right]$

4 0

3 years ago