1) LaTeX: m\angle1=\:\:\:\:\:\:\:\:^\circm ∠ 1 = ∘ 2) LaTeX: m\angle2=\:\:\:\:\:^\circm ∠ 2 = ∘

3) LaTeX: x\:=\:\:\:\:m\:\left(exterior\:angl e\right)\:=

1. m<1 = 139°

2. m<2 = 41°

3. x = 29

Step-by-step explanation:

1. m<1 = 86° + 53° (exterior angle theorem of a triangle)

m<1 = 139°

2. 133° = m<2 + 92° (exterior angle theorem of a triangle)

133° - 92° = m<2 (Subtraction property of equality)

41° = m<2

m<2 = 41°

3. 4x + 15 = 93° + (x + 9)° (exterior angle theorem of a triangle)

Add like terms

4x + 15 = 102 + x

4x - x + 15 = 102 (substraction property of equality)

3x + 15 = 102

3x = 102 - 15

3x = 87

x = 87/3 (division property of equality)

x = 29

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.