Which graph shows a method for finding the image of point D if the parallelogram is reflected across the dashed line? On a coordinate plane, parallelogram A B C D has points (1, 2), (3, 4), (8, 4), (6, 2). A dashed line goes through points A and C. On a coordinate plane, parallelogram A B C D has points (1, 2), (3, 4), (8, 4), (6, 2). A dashed line goes through points A and C. A red dashed line is drawn from point B to point D. D prime = B. On a coordinate plane, parallelogram A B C D has points (1, 2), (3, 4), (8, 4), (6, 2). A dashed line goes through points A and C. A red dashed line splits the image into 2 equals parts with length of 1.4 centimeters. On a coordinate plane, parallelogram A B C D has points (1, 2), (3, 4), (8, 4), (6, 2). A dashed line goes through points A and C. Point D prime is at (2, 6). On a coordinate plane, parallelogram A B C D has points (1, 2), (3, 4), (8, 4), (6, 2). A dashed line goes through points A and C. A vertical dashed line is drawn through the image.

a complex number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i is a solution of the equation x2 = −1. because no real number satisfies this equation, i is called an imaginary number. has no real solution, since the square of a real number cannot be negative.

huh? im pretty sure it is in 199

step-by-step explanation:

could

step-by-step explanation:

u send me a picture of ur work

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.