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:
The domain is the span of x-values covered by the function.
From the graph, we can see that the graph covers all the x-values from x=-7 to x=4.
However, note that closed and open circles. There is an open circle at x=-7, which means that the domain excludes x=-7. However, the circle at x=4 is closed, meaning it is included in the domain.
Therefore, the domain is, in interval notation:
We use parentheses on the left because we do not include -7. And we use brackets on the right because we <em>do </em>include the 4.
And in set notation, this is: