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3 years ago

What is the length of segment FG?

Mathematics

2 answers:

Alenkinab [10]3 years ago

7 0

Answer:

D. 10.49

Step-by-step explanation:

(see attached for reference)

Here we have two right triangles formed by dividing a larger right triangle by the Altitude (height) of the larger right triangle.

In this special case, all 3 triangles (2 small and 1 large) are SIMILAR and hence we can use the ratios of similar triangles to solve this.

In our case, ΔGFE is similar to ΔGHF

hence FG/GE = GH/FG

We are given that GE = 10 and GH = 11, hence

FG/10 = 11/FG

FG² = (10)(11)

FG² = 110

FG = ±√110

FG= 10.49

stich3 [128]3 years ago

3 0

 $the \: answer \: is \: 10.49 \\ please \: see \: the \: attached \: picture \\ for \: full \: solution \\ hope \: it \: helps$ 

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Slope intercept form is y=2x+b

Step-by-step explanation:

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What value of x makes this proportion true? O A. 20 OB. 5 O C. 6 O 0.9

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Find the value of z in the similar figures (round to the nearest tenth)

MrRa [10]

As they are similar figures,we can use the ratio of the side to find out the value of z.

The only side that both have the value indicated is 4 and 6,therefore we would use it to find the ratio.

The eqaution would be:
6/4 = z/8
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3 years ago

In the accompanying diagram, parallelogram ABCD has

Taya2010 [7]

Answer:

The area of the paralellogram is 24 square units.

Step-by-step explanation:

Geometrically speaking, the area of the parallelogram has an equation equivalent to the area formula for a rectangle, that is:

$A = b\cdot h$ (1)

Where:

$A$ - Area.

$b$ - Base.

$h$ - Height.

The base and height of the parallelogram are, respectively:

Base

$b = 8-2$

$b = 6$

Height

$h = 5-1$

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Then, the area of the parallelogram is:

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3 years ago

Please help it will help a lot i really need this!!

Lorico [155]

Answer:

A' = (0,0)

B'=(8,0)

C'=(8,2)

D'=(0,2)

It is not a rigid motion.

Step-by-step explanation:

To use the mapping rule, substitute the original x and y values in it.

The coordinates of A are (0,0).

Using the mapping rule, the x coordinate of A' = 2x0 = 0

Using the mapping rule, the y coordinate of A' = (1/2)x0=0

So A' will not change locations. The image will be at (0,0).

The coordinates of B are (4,0)

Using the mapping rule, the x-coordinate of B' = 2x4 = 8

Using the mapping rule, the y-coordinate of B' = (1/2)x0=0

Therefore the image of B' will be located at coorindate (8,0)

The coorindates of C are (4,4).

Using the mapping rule, the x-coordinate of C' = 2x4=8

Using the mapping rule, the y-coordinate of C' = (1/2)x4=2

Therefore the image of C' will be located at coordinate (8,2)

The coordinates of D are (0,4)

Using the mapping rule, the x-coordinate of D' = 2x0 = 0

Using the mapping rule, the y-coordinate of D' = (1/2)x4=2

Therefore the image of D' will be located at coordinate (0,2)

Is the transformation a rigid motion?

NO, this transformation is not a rigid motion because the relative distance between the points does not stay the same after they have been transformed. The transformation is not a translation , rotation, reflection, nor glide reflection.

6 0

3 years ago