Figure EFGHK as shown below is to be transformed to figure E′F′G′H′K′ using the rule (x, y) → (x 8, y 5): Figure EFGHK is drawn on a 4 quadrant coordinate grid with vertices at E 3, negative 4. F is at 5, 1. G is at 3, 5. H is at negative 4, 3. K is at negative 2, negative 3. Which coordinates will best represent point E′? (1, 11) (4, 8) (8, 4) (11, 1) Figure ABCD is a rhombus where m∠ABC = 84° and m∠ABE = 3x − 6. Solve for x. Rhombus ABCD with diagonals AC and BD and point E as the point of intersection of the diagonals. 16 30 32 62 Question 11(Multiple Choice Worth 4 points) (02.01 MC) Triangle XYZ is shown on the coordinate plane. A triangle on the coordinate plane with vertices X at 0 comma 5, Y at 10 comma 3, and Z at 4 comma negative 1. If triangle XYZ is translated using the rule (x, y) → (x 2, y 3) and then reflected across the x-axis to create triangle X″Y″Z″, what is the location of Z″? (2, −8) (6, −2) (8, −2) (12, −6) Question 12(Multiple Choice Worth 4 points) (02.04 HC) C is the incenter of isosceles triangle ABD with vertex angle ∠ABD. Does the following proof correctly justify that triangles ABC and DBC are congruent? It is given that C is the incenter of triangle ABD, so segment BC is an altitude of angle ABD. Angles ABC and DBC are congruent according to the definition of an angle bisector. Segments AB and DB are congruent by the definition of an isosceles triangle. Triangles ABC and DBC share side BC, so it is congruent to itself by the reflexive property. By the SAS postulate, triangles ABC and DBC are congruent. Triangle ABD with segments BC, DC, and AC drawn from each vertex and meeting at point C inside triangle ABD. There is an error in line 1; segment BC should be an angle bisector. The proof is correct. There is an error in line 3; segments AB and BC are congruent. There is an error in line 5; the ASA Postulate should be used. Question 13(Multiple Choice Worth 4 points) (02.03 MC) Angle N = 40 d