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andrezito [222]
3 years ago
10

A regular hexagon is inscribed in a circle as shown determine the measure of FE

Mathematics
2 answers:
iVinArrow [24]3 years ago
8 0

A regular hexagon is inscribed in a circle as shown. Determine the measure of FE.

<u>FE= 6 CM.</u>

<u></u>

Ostrovityanka [42]3 years ago
3 0
The complete question in the attached figure

we know that
1) <span>The triangles that are formed in the hexagon by joining all the vertices with the center of the hexagon are all equilateral and are equal in size
therefore
the radius of the circle is equals to the length side of the regular hexagon
FE=BP--------> FE=6 cm

the answer is
FE=6 cm </span>

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Jaishree studies for 5 hours daily. She devotes 2 hours of her time for science and mathematics. How muþch time does she devote
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Can someone teach me how to do this type of thing?​
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Do the points (−1,9) and (2,4) represent the vertex and a sample point, respectively, of y=−2(x+1)2+9? Explain
Reil [10]

Answer:

Vertex: (-1,9) is true

Sample Points: (2,4) is not true

Step-by-step explanation:

Given

Vertex: (-1,9)

Sample\ Points: (2,4)

Required

Determine if the vertex and sample points exist for y= -2(x+1)\²+9

Solving for the vertex:

y= -2(x+1)\²+9

Writing the given equation in the form:

y = ax^2 + bx + c

Expand the bracket

y= -2(x+1)(x+1)+9

Open Bracket

y= -2(x+1)(x+1)+9

y = -2(x^2 + 2x + 1) + 9

Open bracket

y = -2x^2 - 4x -2 + 9

y = -2x^2 - 4x +7

Solve for x using:

x = \frac{-b}{2a}

Where a = -2; b = -4;

<em>So:</em>

x = \frac{-(-4)}{2 * (-2)}

x = \frac{4}{-4}

x = -1

Substitute -1 for x in y = -2x^2 - 4x +7

y = -2(-1)^2 -4(-1) + 7

Simplify all brackets

y = -2(1) + 4 + 7

y = -2 + 4 + 7

y = 9

Hence;

<em>The vertex (x,y) is (-1,9)</em>

<em>This is true</em>

Checking sample points: (2,4)

In this case; x = 2 and y = 4

Substitute x = 2 and y = 4 in y= -2(x+1)\²+9

4 = -2(2 + 1)^2 + 9

4 = -2(3)^2 + 9

4 = -2 * 9 + 9

4 = -18 + 9

4 \neq -9

<em>Hence;</em>

<em>This sample points </em>(2,4)<em> does not exist</em>

8 0
3 years ago
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