The swan below is composed several triangles. Use the given information and the figure to find each angle measure. Note: Figure not drawn to scale.
Given: ABC is equilateral; ZBCD = LBDC; DE = CE = EF; LCGF = ZCFG;
AGCF = AGKF = JHM; AKFH = AKLH; KOFO;
ZHKN = HNK; IN JO
B
(4x)
А
02
(7x)
D
C
(47)
(62)
(32)
E
(102)
(2z + 6)
G
(167)
H
1
(8x + 2)°
(47)
(4z - 7)"
K
M
(x + 6)
(8y)
o
N
1. MLABC
2. m2 BCA
3. MZ CAB
4. mZBCD
5. MBDC
6. mZCBD
7. mZEDC
8. MZECD
9. mZCED
10. mZECF
11. mZEFC
12. mZCEF
13. mZCGF
14. mZCFG
15. mZGCF
16. MZKGF
17. m2 KFG
18. MZGKF
19. mZFKH
20. mZFHK
21. mZKFH
22. m2KHL
23. MZHKL
24. m2 KLH
25. mZHJM
26. mZHM)
27. mZJHM
29. mZOKF
28. mZOFK
30. mZKOF
31. mZHKN
33. mZOKN
32. mZHNK
34. mLJNO
35. MZJON
36. mZNJO

step-by-step explanation:

we are given a graph for system of inequalities.

in order to write the system of inequalities for the given graph, we need to find the inequality for each of the line separately.

let us find the slope of dotted line first.

taking two coordinates on the graph.

let us say (0,0) and (3,1).

let us find the slope of the line now.

slope of the dotted line would be =.

y-intercept of dotted line is y=0 only.

plugging slope m= and b=0 in slope-intercept form y=mx+b, we get

.

the line is shaded up of that dotted line.

therefore, for shading up we would take greater than inequality symbol > .

and because it's a dotted line we would take only > symbol in place of = sign in above equation of line.

therefore, inequality for dotted line would be

let us take two points (3,1) and (0,4) on the solid line.

slope of the solid line would be =.

for the solid line we have y-intercept b= 4.

plugging slope m= -1 and b=4 in slope-intercept form y=mx+b, we get

y = -1x + 4.

the line is shaded down of that solid line.

therefore, for shading down we would take less than inequality symbol < .

and because it's a solid line we would take ≤ symbol in place of = sign in above equation of line.

therefore, inequality for solid line would be

answer: the answer is ann

the answer is: d) o, f, and kr.

atomic radius of bromine (br) is 114 pm.

atomic radius of oxygen (o) is 73 pm.

atomic radius of fluorine (f) is 72 pm.

atomic radius of krypron (kr) is 112 pm.

the atomic radius of a chemical element is a measure of the size of its atom.

the atomic radius varies with increasing atomic number, but usually increases because of increasing of number of electrons.

the atomic radius decreases across the periods because an increasing number of protons, because greater attraction between the protons and electrons.