Given: MO bisects ∠PMN and OM bisects ∠PON Prove: ΔPMO ≅ ΔNMO Statements Reasons 1. MO bisects ∠PMN 2. ∠PMO ≅ ∠NMO 3. MO ≅ MO 1. Given 2. Definition of angle bisector 3. Reflexive property 4. OM bisects ∠PON 5. ∠POM ≅ ∠NOM 6. ΔPMO ≅ ΔNMO 4. Given 5. Definition of angle bisector 6. ?

From the given statements and reasons, we can deduce that ΔPMO ≅ ΔNMO because; it fulfils the ASA Congruence postulate.

We are given;

1) MO bisects ∠PMN

2) ∠PMO ≅ ∠NMO which means that ∠PMO is congruent to ∠NMO

3) MO ≅ MO

4) OM bisects ∠PON

5) ∠POM ≅ ∠NOM which means that ∠POM is congruent to ∠NOM

6) ΔPMO ≅ ΔNMO

     Now, from all the statements given, we can see that two corresponding angles of both ΔPMO and ΔNMO are equal.

We also see that they share a common side which is MO.

Thus, we can say that both triangles have two corresponding sides that are equal and a corresponding included angle that is also equal.

     In conclusion, ΔPMO ≅ ΔNMO because it fulfils the ASA Congruence postulate.

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answer: i don't understand. what are the options

step-by-step explanation:

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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