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katovenus [111]
2 years ago
7

Match each justification to its corresponding step in the following solution. (Table shown above)

Mathematics
1 answer:
AveGali [126]2 years ago
5 0

Answer:

D

Step-by-step explanation:

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In the diagram below segment DEF is shown (since all three letters are shown, they are collinear). If
monitta

Answer:

  • x = 4

Step-by-step explanation:

<u>Given</u>

  • DE = 3x + 2, EF = x + 5, DF = 23

<u>As per segment addition postulate</u>

  • DE + EF = DF

<u>Substituting values:</u>

  • 3x + 2 + x + 5 = 23
  • 4x + 7 = 23
  • 4x = 16
  • x = 4
3 0
3 years ago
If u do this, i will love u plss!
Andreas93 [3]

Answer:

1/2

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
What is the area of a rectangle with a width of 4 2/3 meters and a length of 2 1/6 meters?
inessss [21]

Answer:

Area = 10 1/9 meters squared

Step-by-step explanation:

mixed number to improper fraction:

4 2/3 meters = 14/3 meters

2 1/6 meters = 13/6 meters

Area = length times width:

Area = (14/3) x (13/6)

        = 182/18

Simplify fraction:

182/18 = 91/9

improper fraction to mixed number:

91/9 = 10 1/9 meters

AREA = 10 1/9 meters

5 0
3 years ago
Is 3 greater than 6/4
White raven [17]
Yes. 6/4 is equivalent to 1 1/2 so 3 > 6/4
3 0
3 years ago
Read 2 more answers
How to find x using trigonometry
I am Lyosha [343]
Soh Cah Toa

Sine of the angle = opposite side over the hypotenuse

sin(42°) = 6.5/h
rearrange, solve for h
h = 6.5/sin(42°)
h = 9.7 cm

For the other triangle, the angle is unknown. I'd split it into two right triangles; down angle x since it is an isosceles it will be bisected into two equivalent triangles.

The hypotenuse we just found is now split in half. 9.7/2 = 4.9 base of the new smaller right triangle.

The new hypotenuse of the smaller triangle is 7.4 cm

Then you have..

sin(x/2) = 4.9/7.4
sin(x/2) = 0.67
use inverse sine function
arcSin(0.67) = x/2
41 = x/2
82° = x

There are many ways to solve this problem. This is just what I thought of first using trig.
3 0
3 years ago
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