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Elanso [62]
3 years ago
12

step below. Question 1 Hexagon AGHBGHCGHD GHEGHFGH is a reflected image of hexagon ABCDEF. The midpoints of the sides of hexagon

ABCDEF are also shown. Drag the line of reflection, Għ until the image coincides with the preimage. At this location, if the preimage flips about the line of reflection, it will flip onto itself. In how many different positions can you place GHso the image reflects onto the preimage in this manner? Describe the different positions. Be sure to pass the line of reflection through both vertices and midpoints before answering.​
Mathematics
1 answer:
Anuta_ua [19.1K]3 years ago
5 0

Answer:

Hexagon AGHBGHCGHDGHEGHFGH coincides with hexagon ABCDEF when GH passes through the midpoints of opposite sides; that is, it is a perpendicular bisector of the two sides. HexagonAGHBGHCGHDGHEGHFGH also coincides with hexagon ABCDEF when the line of reflection joins a pair of vertices opposite one another on the hexagon. There are three perpendicular bisectors and three pairs of opposite vertices. In all, there are six lines of reflection that will map the hexagon back onto itself.

Step-by-step explanation:

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3 years ago
John is skiing on a mountain with an altitude of 1200 feet. The angle of depression is 21 About how far does
Agata [3.3K]

Answer:

John ski down the mountain is 1285.37 feet.

Step-by-step explanation:

Given : John is skiing on a mountain with an altitude of 1200 feet. The angle of depression is 21.

To find : About how far does  John ski down the mountain ?

Solution :

We draw a rough image of the question for easier understanding.

Refer the attached figure below.

According to question,

Let AB be the height of mountain i.e. AB=1200 feet

The angle of depression is 21 i.e. \theta=21^\circ

We have to find how far does  John ski down the mountain i.e. AC = ?

Using trigonometric,

\cos\theta = \frac{AB}{AC}

\cos(21)= \frac{1200}{AC}

AC=\frac{1200}{\cos(21)}

AC=1285.37

Therefore, John ski down the mountain is 1285.37 feet.

7 0
3 years ago
PLEASEEEEE HELPPPP MEEEEEEE BRAINLIEST TO THE FIRST ANSWER AND 30 POINTS
icang [17]

Answer:

CD ≠ EF

Step-by-step explanation:

Using the distance formula

d = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2    }

with (x₁, y₁ ) = C(- 2, 5) and (x₂, y₂ ) = D(- 1, 1)

CD = \sqrt{(-1+2)^2+(1-5)^2}

      = \sqrt{1^2+(-4)^2}

      = \sqrt{1+16} = \sqrt{17}

Repeat using (x₁, y₁ ) = E(- 4, - 3) and (x₂, y₂ ) = F(- 1, - 1)

EF = \sqrt{(- 1+4)^2+(-1+3)^2}

     = \sqrt{3^2+2^2}

     = \sqrt{9+4} = \sqrt{13}

Since \sqrt{17} ≈ \sqrt{13} , then CD and EF are not congruent

     

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