According to the given information, segment ab is parallel to segment dc and segment bc is parallel to segment ad.. using a straightedge, extend segment ab and place point p above point b. by the same reasoning, extend segment ad and place point t to the left of point a. angles are congruent by the alternate interior angles theorem. angles are congruent by the corresponding angles theorem. by the transitive property of equality, angles bcd and bad are congruent. angles abc and bat are congruent by the alternate interior theorem. angles bat and cda are congruent by the corresponding angles theorem. by the transitive property of equality, ∠abc is congruent to ∠cda. consequently, opposite angles of parallelogram abcd are congruent. what angles accurately complete the proof?

Step-by-step explanation:

The trick is to identify the angle that can be shown to be congruent to both BCD and BAD. There are two: one with D as a vertex, and one with B as a vertex.

The placement of point P suggests that that point will be involved in angle naming, so the angle of interest (CBP) is the one that uses point P in its name and has B as its vertex.

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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hello from mrbilldoesmath!

answer:   x^2 + y^2 = 1024

discussion:

the equation of a circle with center (h,k) and radius "r" is

(x-h)^2 + (y-k)^2 = r^2.

in our case

(x-0)^2 + (y-0) ^2 = 32^2 = 1024

you,

mrb