All categories

2 years ago

Unit 2 Show Your Work, BYU Geometry Part 1

Mathematics

1 answer:

irga5000 [103]2 years ago

3 0

The true statements about ∠MJK and ∠MJL are

The sum of the two angles is 180 degrees
JK and JL are opposite rays
JK and JL form a straight line

<h3>How to determine the true statements?</h3>

The statement is given as:

∠MJK and ∠MJL are a linear pair of angles and the angles are also supplementary

As a general rule, linear pair angles are adjacent angles that add up to 180 degrees. This means that:

They are supplementary angles
They have a sum of 180 degrees
The angles may or may not be congruent
They have opposite rays i.e. straight line

The above means that, the following options are true:

(e), (f) and (h)

Also, the converse of the statement is:

∠MJK and ∠MJL are supplementary angles and the angles are also linear pair of angles

This is false because not all supplementary angles are linear pair angles

Read more about linear pair angles at:

brainly.com/question/1511741

#SPJ1

You might be interested in

Find the perimeter of Triangle ABC with the verticies A(-5,5) B(3,-5) and C(-5,1

finlep [7]

Hello!

Let's calculate a distance between two points using the Pythagorean theorem:<span>

</span> $d ^ 2_ {AB} = (x_ {B} - x_ {A}) ^ 2 + (y_ {B} - y_ {A}) ^ 2$

Data:

$x_B = 3 

x_A = -5 

y_B = -5 

y_A = 5 

d_ {AB} =?$

Solving: distance from A to B

$d^ 2_ {AB} = (x_ {B} - x_ {A}) ^ 2 + (y_ {B} - y_ {A}) ^ 2 

d^ 2_ {AB} = (3 - (-5)) ^ 2 + (-5 - 5) ^ 2 

d^ 2_ {AB} = (8) ^ 2 + (-10) ^ 2 

d^ 2_ {AB} = 64 + 100 

d^ 2_ {AB} = 164 

d_{AB} = \sqrt {164} 

d_{AB} = \sqrt {2 ^ 2 * 41} 

\boxed {d_ {AB} = 2 \sqrt {41}}$

<span>Solving:

distance from A to C</span>

$d ^ 2_ {AC} = (x_ {C} - x_ {A}) ^ 2 + (y_ {C} - y_ {A}) ^ 2$

Data:

$x_C = -5
 
x_A = -5
 
y_C = 1 

y_A = 5 

d_ {AC} =?$

$d^ 2_ {AC} = (x_ {C} - x_ {A}) ^ 2 + (y_ {C} - y_ {A}) ^ 2 

d^ 2_ {AC} = (-5 - (-5)) ^ 2 + (1 - 5) ^ 2 

d^ 2_ {AC} = (0) ^ 2 + (-4) ^ 2 

d^ 2_ {AC} = 0 + 16 

d^ 2_ {AC} = 16 

d_ {AC} = \sqrt {16} 

\boxed {d_ {AC} = 4}
$ <span>
</span>

Solving:

distance from B to C

$d^ 2_ {BC} = (x_ {C} - x_ {B}) ^ 2 + (y_ {C} - y_ {B}) ^ 2$

Data:

$x_C = -5 

x_B = 3 

y_C = 1 

y_B = -5 

d_ {BC} =?$

$d^ 2_ {BC} = (x_ {C} - x_ {B}) ^ 2 + (y_ {C} - y_ {B}) ^ 2
 
d^ 2_ {BC} = (-5 - 3) ^ 2 + (1 - (-5)) ^ 2

d^ 2_ {BC} = (-8) ^ 2 + (6) ^ 2 

d^ 2_ {BC} = 64 + 36

d^ 2_ {BC} = 100 

d_ {BC} = \sqrt {100}

\boxed {d_ {BC} = 10}$

<span>Now let's calculate the perimeter of the triangle ABC, knowing that the perimeter is the sum of the sides, then:
</span>
$p = d_ {AB} + d_ {AC} + d_ {BC} 

$

$p = 2\sqrt {41} + 4 + 10$

$
\boxed {\boxed {p = 14 + 2 \sqrt {41}}} \end {array}} \qquad \quad \checkmark$

8 0

3 years ago

HELP ILL GIVE BRAINLIEST what’s the surface area?

sdas [7]

Answer:

i dont know im so sorry for making you feel like you

Step-by-step explanation:

i dont know im so sorry for making you feel like you

7 0

3 years ago

Did you get the answer to this question

Harlamova29_29 [7]