The figure below shows a square ABCD and an equilateral triangle DPC: D
Ted makes the chart shown below to prove that triangle APD is congruent to triangle BPC:
Statements
In triangles APD and BPC; DP = PC
In triangles APD and BPC; AD = BC
Justifications
Sides of equilateral triangle DPC are equal
Sides of square ABCD are equal
Angle ADC = angle BCD = 90° so angle ADP = angle BCP = 30°
Triangles APD and BPC are congruent SAS postulate
Which of the following completes Ted's proof? (1 point)
O In square ABCD; angle ADC= angle BCD
O In square ABCD; angle ADP= angle BCP
O In triangles APD and BPC; angle ADC
angle BCD
O In triangles APD and BPC; angle ADP = angle BCP

what   tables?

step-by-step explanation:

14.29%

step-by-step explanation:

answer: your correct answer is b. d.

step-by-step explanation:

simplifying

5(x + -6) = 2(x + 3)

reorder the terms:

5(-6 + x) = 2(x + 3)

(-6 * 5 + x * 5) = 2(x + 3)

(-30 + 5x) = 2(x + 3)

reorder the terms:

-30 + 5x = 2(3 + x)

-30 + 5x = (3 * 2 + x * 2)

-30 + 5x = (6 + 2x)

solving

-30 + 5x = 6 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-30 + 5x + -2x = 6 + 2x + -2x

combine like terms: 5x + -2x = 3x

-30 + 3x = 6 + 2x + -2x

combine like terms: 2x + -2x = 0

-30 + 3x = 6 + 0

-30 + 3x = 6

add '30' to each side of the equation.

-30 + 30 + 3x = 6 + 30

combine like terms: -30 + 30 = 0

0 + 3x = 6 + 30

3x = 6 + 30

combine like terms: 6 + 30 = 36

3x = 36

divide each side by '3'.

x = 12

simplifying

x = 12

this should .