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2 years ago

14

If JM = 5x – 8 and LM = 2x – 6, which expression represents JL?

2 answers:

NNADVOKAT [17]2 years ago

7 0

Answer:

A. 3x – 2 ( on edgenuity)

Step-by-step explanation:

serg [7]2 years ago

5 0

Answer: 3x-2

Step-by-step explanation:

My explanation is if you were to write this down, i drew "J", "K", "L", and "M" down from left to write on a line w/ points and then i wrote that " J M " = 5 x - 8, and that " L M " = 2 x - 6, and then subtracted ( 5 x - 8 ) by ( 2 x - 6 ) which would look like this:

( 5 x - 8 ) - ( 2 x - 6 )

The reason i subtracted the total length of the entire number line is because the length of " L M " is part of the total length so you would have to subtract in order to find " J L " because then you wouldn't have to have " L M ". The correct answer is "3 x - 2"

LOVE YAWL THANK YOU SO MUCH BE SAFE!!!!!!!!!!!!!!!! * <3 <3 *

HOPE THIS HELPS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! . <3 <3 <3 <3 .

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3 years ago

The radius of a circle is 14 meters. What is the circle circumference

seropon [69]

Step-by-step explanation:

HERE,

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<u>According</u><u> </u><u>to</u><u> </u><u>the question</u><u>, </u>

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2 years ago

Under a certain transformation T, AA'BC' is the image of AABC. The perimeter of AA'B'C' is twice the perimeter of ABC. What kind

Anika [276]

The dimensions of the image are twice the dimensions of the preimage,

which indicates that the transformation is a dilation.

The transformation <em>T</em> is; <u>a dilation transformation with a scale factor of 2, D₂</u>

Reasons:

The transformation applied to ΔABC = T

The image of ΔABC following the transformation, T = ΔA'B'C'

The perimeter of ΔA'B'C' = Twice the perimeter of ΔABC

Therefore, we have;

A'B' + B'C' + A'C' = 2 × (AB + BC + AC)

Which gives

A'B' + B'C' + A'C' = 2·AB + 2·BC + 2·AC

By similar triangles, we have the ratio of corresponding sides as follows;

$\displaystyle \frac{AB}{A'B'} = \frac{BC}{B'C'} = \mathbf{ \frac{AC}{A'C'}}$

Which gives;

$\displaystyle \frac{AB}{A'B'} = \frac{AB}{2 \cdot AB} = \frac{1}{2}$

A'B' = 2·AB

Therefore;

Triangle ΔA'B'C' is twice the dimension of triangle ΔABC, and <em>T</em> is <u>a dilation transformation with a scale factor of 2, D₂</u>

Learn more about dilation transformation here:

brainly.com/question/2458912

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