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3 years ago

50 POINTS LEASE HELPPP!!!!

Mathematics

1 answer:

xxMikexx [17]3 years ago

7 0

A regular octagon can be thought of as being composed of 4 "kite" shaped areas.

The area of a "kite" with diagonals d and w is

AreaKITE=d⋅w2.

(This is fairly easy to prove if it isn't a formula you already know).

Consider the "kite" PQCW in the diagram above.

∠QCW=π2 and |QC|=|WC|=r

s⇒|QW|=√2r (Pythagorean)

Therefore (since |PC|=r)

AreaPQCW=|PC|⋅|QW|2=r⋅√2r2=√2r^22

The octagon is composed of 4 such kites, so

AreaOctagon=2√2r^2

AreaOctogon=2√2(6)^2=101.823376491

AreaOctagon=101.8 units squared

here is the diagram https://d2gne97vdumgn3.cloudfront.net/api/file/80fl6nYTCaa5q8AZOyq0

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Bogdan [553]

Answer:

$x=\frac{-9}{7}$

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Step-by-step explanation:

We have been given the equation:

$3x-2y-1=0$

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Multiply equation (1) by 5 and equation (2) by 3 we get:

$15x-10y=5$ (3)

$15x-3y=-12$ (4)

Now, subtract (4) from (3) we get:

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$y=\frac{-17}{7}$

Now substitute $y=\frac{-17}{7}$ in equation (1) we get:

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$x=\frac[-27}{21}$

$x=\frac{-9}{7}$

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