CAN I GET HELP ASAP 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?
In square ABCD; angle ADC = angle BCD
In square ABCD; angle ADP = angle BCP
In triangles APD and BPC; angle ADC = angle BCD
In triangles APD and BPC; angle ADP = angle BCP

the correct answer is the second option

8

step-by-step explanation:

just put √(2) * √(32) into a calculator. that will make the answer be 8.

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

cretdit to source

a

step-by-step explanation: