(05.03 mc) the figure below shows a square abcd and an equilateral triangle dpc: abcd is a square. p is a point inside the square. straight lines join points a and p, b and p, d and p, and c and p. triangle d ted makes the chart shown below to prove that triangle apd is congruent to triangle bpc: statements justifications in triangles apd and bpc; dp = pc sides of equilateral triangle dpc are equal in triangles apd and bpc; ad = bc 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) in square abcd; angle adc = angle bcd in square abcd; angle adp = angle bcp in triangles apd and bpc; angle adp = angle bcp

Step-by-step explanation:

Here, ABCD is a square, and P is a point inside the square.

Where, Δ DPC is a equilateral triangle.

Therefore, DP=PC

Thus In triangles  APD and BPC,

DP= CP

( because both are of 30°)

AD=CB ( sides of square)

Therefore by SAS postulate,

Δ APD≅Δ BPC

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0

what are ur choices?