Chelsea and Anika are trying to determine whether ΔABC and ΔPRQ can be proven congruent through rigid motions. Chelsea says that ΔABC ≅ ΔPRQ because ΔABC can be rotated 270° clockwise to create ΔPRQ.

Anika says that ΔABC ≅ ΔPRQ because ΔABC can be reflected over the x-axis to create ΔPRQ.

a coordinate grid with triangles ABC and PQR at A negative 7 comma 2, B negative 3 comma 6, C negative 2 comma 1, P negative 7 comma negative 2, Q negative 2 comma negative 1, and R negative 3 comma negative 6

Who is correct?

Chelsea only
Anika only
Both Chelsea and Anika
Neither Chelsea nor Anika

Anika is correct

Step-by-step explanation: A 270* rotation is a 90* rotation in the opposite direction. And when you do  a 90* rotation counterclockwise, the points don’t match up with PRQ, so Chelsea is wrong.

3. parallelogram

step-by-step explanation:

we turn each of our sides into direction vectors

ab=b-a=(3 - 7, 1 - 0) = (-4, 1)

bc=c-b=(1,6)

cd=d-c=(4,-1)

da=a-d=(-1,-6)

we see how the opposite sides have negated direction vectors; this is characteristic of a parallelogram.   we have congruent sides ab=cd and bc=da with unequal lengths because 1^2+6^2 is different from (-4)^2+1^2.   so it's not a square or a rhombus.  

we check the dot product of direction vectors of adjacent sides.   we have

ab dot bc = -4(1) + 1(6) = 2

that's not zero so the sides are not perpendicular.   we've ruled out rectangle.

the direction vectors indicate two pairs of congruent parallel sides, so we choose parallelogram over trapezoid.

i dont know if you cant find a way its best to actually add all of them start from 105 tho

step-by-step explanation: