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mrs_skeptik [129]
3 years ago
5

3 minus 2/9b equals 1/3b minus 7

Mathematics
1 answer:
Sphinxa [80]3 years ago
5 0
X=18 get Photomath it helps
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Find x if segment A B is perpendicular to segment C D in the figure below.
ki77a [65]
If segment AB is perpendicular to segment CD, then a right angle (90°) is formed.


7x + 27 = 90

7x = 63

x = 9
3 0
3 years ago
Read 2 more answers
What’s 0.005 recurring as a fraction
PilotLPTM [1.2K]

Answer:

0.005 = 5/1000 = 1/200

Hope it helps

3 0
3 years ago
Elimination method<br> 4x-2y=48<br> 2x+6y=-18
yulyashka [42]
4x-2y=48\\ \\ 2x+6y=-18

First we'll multiply the second equation by -2


-2\cdot (2x+6y)=-2\cdot -18\\ \\ -4x-12y=36

Now let's add the new equation and the first one.

(4x-2y)+(-4x-12y)=48+36\\ \\ 4x-4x-2y-12y=48+36\\ \\ -14y=84\\ \\ y=\frac { 84 }{ -14 } \\ \\ y=-6

We found y's value. We'll plug it in one of the equations to find x's value.

y=-6\\ \\ 4x-2y=48\\ \\ 4x-2\cdot -6=48\\ \\ 4x+12=48\\ \\ 4x=48-12\\ \\ 4x=36\\ \\ x=\frac { 36 }{ 4 } \\ \\ x=9

Solution ;

(9, -6)


4 0
3 years ago
Read 2 more answers
Do the following side lengths form a triangle?<br><br> 6, 18, 14
iren2701 [21]

Answer:

Yes

Step-by-step explanation:

Conditions for the sides to form a triangle is that sum of any two sides should be greater than the third side.

6 + 18 = 24

24 > 14

18 + 14 = 32

32 > 6

6 + 14 = 20

20 > 18

Therefore,

sides 6, 18 and 14 forms a triangle.

5 0
2 years ago
Alexa used her compass to evenly divide the circumference of the circle below.
jok3333 [9.3K]

Answer:

  • triangle
  • square
  • hexagon
  • dodecagon

Step-by-step explanation:

The 12 points marked can be divided into this many equal-size groups:

  3, 4, 6, 12

so the figures that can be constructed are ...

  • triangle . . . . connects every 4th mark
  • square . . . . . connects every 3rd mark
  • hexagon . . . .connects every other mark
  • dodecagon . . connects every mark

_____

12 is also divisible by 1 and 2, but the minimum number of vertices in a regular polygon is 3.

3 0
3 years ago
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