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____ [38]
2 years ago
5

Which postulate or theorem can be used to prove the triangles are congruent?

Mathematics
1 answer:
lianna [129]2 years ago
6 0

Answer:

ASA

Step-by-step explanation:

Hope this helps:)

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The measure of an exterior angle of a triangle is 75 degrees. If one of the non-adjacent interior angles measures 28 degrees, wh
Leokris [45]
The measure of an exterior angle of a triangle is equal to the sum of the measures of the two remote interior angles.

75-28=47
4 0
3 years ago
Can someone take 3 min for there time and help me with this please??
seropon [69]
Since the left side of the equation says f(5), then plug in 5 for x. 

So, it is: 
              
-4|5| + 3
= -20 + 3
= 23 
4 0
3 years ago
21x+28=14 and 9x+12y=6
Westkost [7]
<span>21x+28=14
x= -2/3 or -0.666666666

</span><span>9x+12y=6
x=2/3 or .666666666
y=0</span>
7 0
3 years ago
Give EG = 5x+7 EF =5x and FG=2x-7
Len [333]

Answer:

x = 7

42, 35, 7

Step-by-step explanation:

  • EG = 5x+7
  • EF = 5x  
  • FG = 2x-7

Point F is on line segment of EG, therefore EG = EF + FG

  • 5x + 7 = 5x + 2x - 7
  • 2x = 14
  • x = 7

Then

  • EG = 5*7 + 7 = 42
  • EF = 5*7 = 35
  • FG = 2*7 - 7 = 7
7 0
3 years ago
rank has a circular garden. The area of the garden is 100 ft2. What is the approximate distance from the edge of Frank’s garden
Ray Of Light [21]

It is given that the area of the circular garden = 100 ft^2

Area of circle with radius 'r' = \pi r^2

We have to determine the approximate distance from the edge of Frank’s garden to the center of the garden, that means we have to determine the radius of the circular garden.

Since, area of circular garden = 100

\pi r^2 = 100

\frac{22}{7} \times r^2 = 100

r^2 = \frac{700}{22}

r^2 = 31.8

r = \sqrt{31.8}

So, r = 5.6 ft

r = 6 ft (approximately)

Therefore, the approximate distance from the edge of Frank’s garden to the center of the garden is 6 ft.

So, Option A is the correct answer.

7 0
3 years ago
Read 2 more answers
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