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Brilliant_brown [7]
4 years ago
15

What construction does the image below demonstrate?

Mathematics
2 answers:
MA_775_DIABLO [31]4 years ago
8 0

Answer:(D) A regular hexagon inscribed in a circle

Step-by-step explanation:

Since here the length of  arcs FAD, EAC DAB, and CAG from the points E, point D, point C and  Point B respectively have the same measurement. (Shown in given diagram)

Therefore, All chords of circle A that are FE, ED, DC, CB, BG, and GF have the same length.

Therefore here we get a figure which having 6 edges and inscribed in the circle A.

Which is a  regular hexagon BCDEFG.

Therefore, Option D) is correct.



Dovator [93]4 years ago
5 0

That is a regular hexagon inscribed in a circle.

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5v^2 -- 30v + 40<br> Factoring polynomials
Valentin [98]

First let's factor a 5 from every term.

5v^2-30v+40

=5(v^2-6v+8)

The v^2-6x+8 can be further factored using the x method(as I call it, its not its official name of course), You can get

5(v^2-6v+8)

=5(v-2)(v-4)

Done! any questions just ask in the comment section.

**Edit: You needed the area model so here is how it goes, I cannot draw lines but I will try my best.

we have original equation 5v^2-30v+40

first simplify to 5(v^2-6v+8)

Now factor the second part but using area model.

         v          -2

    _____________

v    |    v^2   |     -2v    |

    --------------------------

-4   |   -4v  |     8      |

      _____________

This is basically the area model the thing in the middle is suppose to be a rectangle split into 4 parts and then the numbers on top and left are suppose to be the factored equations while they multiplies to the number in the box.

<u>Is this efficient?</u>

Definitely not, it took me longer to draw the box than write my whole first explanation.

<u>Is this easy?</u>

No, again it took me about 170 times longer to think of this answer this way.

<u>Should you use this method?</u>

Sure if your teacher wants you to, but just saying later in algebra these methods are way to slow and there are way better methods to factor, so if you arn't solid in the area model just don't worry about it.

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