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nexus9112 [7]
3 years ago
7

Help me plsss i am soo bad at this

Mathematics
1 answer:
lara [203]3 years ago
8 0

Step-by-step explanation:

GIVEN: O is the mid point of AC & BD.

Therefore, AO = CO & BO = DO

\therefore \: AO   \: \cong \: CO.. (given) \\  \\  \:  \:  \:  \:  \angle \: AOB  \: \cong \:\angle \:COD\\..(vertically\:opposite\:angles) \:  \\  \\  \& \: \:  \:   BO  \: \cong \: DO.. (given)  \\  \\  \therefore \triangle \: AOB  \:  \cong \:  \triangle \:COD  \\ ..(by \: SAS \: test  \: of  \: congruence ) \\  \\   \therefore AB =  CD.. \\ (Corresponding  \: sides  \: of  \: congruent  \: triangles)

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Select the correct answer.
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Answer:

<em>The translation corresponds to a translation of 13 units to the right</em>

Step-by-step explanation:

<u>Translation of a Graph</u>

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and the given function is

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The translation corresponds to a translation of 13 units to the right

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Prove that the segments joining the midpoint of consecutive sides of an isosceles trapezoid form a rhombus.
sergiy2304 [10]

Answer:

See explanation

Step-by-step explanation:

a) To prove that DEFG is a rhombus, it is sufficient to prove that:

  1. All the sides of the rhombus are congruent:  |DG|\cong |GF| \cong |EF| \cong |DE|
  2. The diagonals are perpendicular

Using the distance formula; d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

|DG|=\sqrt{(0-(-a-b))^2+(0-c)^2}

\implies |DG|=\sqrt{a^2+b^2+c^2+2ab}

|GF|=\sqrt{((a+b)-0)^2+(c-0)^2}

\implies |GF|=\sqrt{a^2+b^2+c^2+2ab}

|EF|=\sqrt{((a+b)-0)^2+(c-2c)^2}

\implies |EF|=\sqrt{a^2+b^2+c^2+2ab}

|DE|=\sqrt{(0-(-a-b))^2+(2c-c)^2}

\implies |DE|=\sqrt{a^2+b^2+c^2+2ab}

Using the slope formula; m=\frac{y_2-y_1}{x_2-x_1}

The slope of EG is m_{EG}=\frac{2c-0}{0-0}

\implies m_{EG}=\frac{2c}{0}

The slope of EG is undefined hence it is a vertical line.

The slope of  DF is m_{DF}=\frac{c-c}{a+b-(-a-b)}

\implies m_{DF}=\frac{0}{2a+2b)}=0

The slope of DF is zero, hence it is a horizontal line.

A horizontal line meets a vertical line at 90 degrees.

Conclusion:

Since |DG|\cong |GF| \cong |EF| \cong |DE| and DF \perp FG , DEFG is a rhombus

b) Using the slope formula:

The slope of DE is m_{DE}=\frac{2c-c}{0-(-a-b)}

m_{DE}=\frac{c}{a+b)}

The slope of FG is m_{FG}=\frac{c-0}{a+b-0}

\implies m_{FG}=\frac{c}{a+b}

5 0
3 years ago
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