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EastWind [94]
3 years ago
6

The area of the octagon below is 10,391.9 square millimeters. An octagon has an apothem of 56 millimeters and an area of 10,391.

90 square millimeters. What is the approximate perimeter of the octagon? Round to the nearest whole number. mm What is the approximate length of one side of the octagon? Round to the nearest whole number. mm
Mathematics
2 answers:
pentagon [3]3 years ago
8 0

Answer: 371 mm and 46 mm

Step-by-step explanation: I got it right on edge

VladimirAG [237]3 years ago
3 0

Answer:

371 & 46

Step-by-step explanation:

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Solve the inequality x/5-3<0
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x < 15

Step-by-step explanation:

\frac{x}{5} - 3 < 0 ( add 3 to both sides )

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x < 15

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7 0
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Subtract 7x-9 from 2x^2-11<br> A) 2x^2-7x-2<br> B) 2x^2+7x-20<br> C) 2x^2+7x-2
Natali [406]
We keep the 2x^2 because we can only subtract something from that if it also is squared. From there, we subtract the 7x from 2x^2-11. Since there is no previous x's in the 2x^2-11, we just make it -7x. So, without even continuing the problem, we see that A) is the correct answer because it is the only one with -7x. 
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Find out the number of combinations and the number of permutations for 8 objects taken 6 at a time. Express your answer in exact
umka2103 [35]

Solution:

The permutation formula is expressed as

\begin{gathered} P^n_r=\frac{n!}{(n-r)!} \\  \end{gathered}

The combination formula is expressed as

\begin{gathered} C^n_r=\frac{n!}{(n-r)!r!} \\  \\  \end{gathered}

where

\begin{gathered} n\Rightarrow total\text{ number of objects} \\ r\Rightarrow number\text{ of object selected} \end{gathered}

Given that 6 objects are taken at a time from 8, this implies that

\begin{gathered} n=8 \\ r=6 \end{gathered}

Thus,

Number of permuations:

\begin{gathered} P^8_6=\frac{8!}{(8-6)!} \\ =\frac{8!}{2!}=\frac{8\times7\times6\times5\times4\times3\times2!}{2!} \\ 2!\text{ cancel out, thus we have} \\ \begin{equation*} 8\times7\times6\times5\times4\times3 \end{equation*} \\ \Rightarrow P_6^8=20160 \end{gathered}

Number of combinations:

\begin{gathered} C^8_6=\frac{8!}{(8-6)!6!} \\ =\frac{8!}{2!\times6!}=\frac{8\times7\times6!}{6!\times2\times1} \\ 6!\text{ cancel out, thus we have} \\ \frac{8\times7}{2} \\ \Rightarrow C_6^8=28 \end{gathered}

Hence, there are 28 combinations and 20160 permutations.

7 0
11 months ago
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