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belka [17]
3 years ago
13

. A(n) _____ is a segment that is formed by the intersection of two faces. edge vertex cross section face

Mathematics
2 answers:
Grace [21]3 years ago
5 0
A line segment formed by the intersection of two faces of a polyhedron.
The answer is Edge :D
worty [1.4K]3 years ago
5 0

Answer:

The answer is Edge.

A segment that is formed by the intersection of two faces is called an EDGE. Its located at the boundary of a polygon.

Step-by-step explanation:

An edge is defined as a type of line segment, that joins two faces in a polygon, polyhedron etc. In a  polytope, an edge is a line segment between faces.

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What's the area of the room in this diagram?
GenaCL600 [577]
I believe it should be B.156 square feet because in order to find the area of a square, you gotta multiply 13ft and 12ft. Once that is solved your answer should be 156.
3 0
4 years ago
Is 126 divisible by 3
juin [17]

Answer:

yes

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Rewrite the following equation in slope-intercept form. 5x + 7y = 7 Write your answer using integers, proper fractions, and impr
pogonyaev
Answer:
y = (-5/7) x + 1

Explanation:
The slope-intercept form of the line has the following formula:
y = mx + c
where:
m is the slope
c is the y-intercept

The given is:
5x + 7y = 7

To put this in slope-intercept form, we will need to isolate the y as follows:
5x + 7y = 7
7y = -5x + 7
y = (-5/7) x + (7/7)
y = (-5/7) x + 1
where:
m is the slope = -5/7
c is the y-intercept = 1

Hope this helps :)


6 0
3 years ago
Read 2 more answers
Please please please please just answer quickly i have short time left
Nastasia [14]

9514 1404 393

Answer:

  (x, y) = (10, 4)

Step-by-step explanation:

Opposite sides are the same length, so ...

  ON = LM

  6x -3 = 4x +17

  2x = 20 . . . . . . . . add 3-4x

  x = 10

__

  OL = NM

  x -5 = 3y -7

  10 -5 + 7 = 3y . . . . . substitute x, add 7

  12/3 = y = 4 . . . . . .  simplify, divide by 3

The values of x and y are 10 and 4, respectively.

5 0
3 years ago
P= (x,y) is an arbitrary point on the circle x^2+y^2=36
sesenic [268]

Let (x,y) be an arbitrary point on the circle.

Then d = √[x-3)2 + (y-0)2] = √[x2 - 6x + 9 + y2]

Since x2 + y2 = 1, y2 = 1-x2.

So, d = √[x2 - 6x + 9 + 1 - x2]

     d = √(-6x+10)

Domain:  x is the x-coordinate of a point on the circle centered at (0,0) with radius 1.  So, -1≤x≤1.

             But, for d to be defined, we need -6x+10 ≥ 0.  So, x ≤ 5/3  (True for all x in [-1,1]).

             Domain = [-1,1]

Range:    -1 ≤ x ≤ 1   So, 6 ≥ -6x ≥ -6

                                    16 ≥ -6x+10 ≥ 4

                                     4 ≥ √(-6x+10) ≥ 2   That is, 2 ≤ d ≤4

              Range:  [2,4]

Step-by-step explanation:

3 0
3 years ago
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