In ΔABC shown below, segment DE is parallel to segment AC: Triangles ABC and DBE where DE is parallel to AC The following two-co
lumn proof with missing statements and reasons proves that if a line parallel to one side of a triangle also intersects the other two sides, the line divides the sides proportionally: Statement Reason 1. Line segment DE is parallel to line segment AC 1. Given 2. Line segment AB is a transversal that intersects two parallel lines. 2. Conclusion from Statement 1. 3. ∠BDE ≅ ∠BAC 3. Corresponding Angles Postulate 4. 4. 5. 5. 6. BD over BA equals BE over BC 6. Converse of the Side-Side-Side Similarity Theorem Which statement and reason accurately completes the proof? 4. ΔBDE ~ ΔBAC; Side-Angle-Side (SAS) Similarity Postulate 5. ∠B ≅ ∠B; Reflexive Property of Equality 4. ∠B ≅ ∠B; Reflexive Property of Equality 5. ΔBDE ~ ΔBAC; Angle-Angle (AA) Similarity Postulate 4. ΔBDE ~ ΔBAC; Side-Angle-Side (SAS) Similarity Postulate 5. ∠A ≅ ∠C; Isosceles Triangle Theorem 4. ∠A ≅ ∠C; Isosceles Triangle Theorem 5. ΔBDE ~ ΔBAC; Angle-Angle (AA) Similarity Postulate
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Answer:
Step-by-step explanation:
Not QUITE a parabola . but up until y = 4 it is close
Vertex = (-7-10)
Vertex form of parabola
y = a ( x+7)^2 -10 find a integer-coordinat point to calculate 'a'
I'll use -18,0
0 = a(-18+7)^2 -10
0 = 122 a -10 a = 10/122
y = 10/122 (x+7)^2 - 10 see graph below:
