Prove that the polynomials are nonnegative a^2-2ab+b^2+c^2

a2 + b2 + c2 - ab - bc - ca

= (1/2)(2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ca)

= (1/2)(a2 - 2ab + b2 + b2 - 2bc + c2 + c2 - 2ca + a2)

= (1/2)[(a-b)2 + (b-c)2 + (c-a)2]

Now we know that, the square of a number is always greater than or equal to zero.

Hence, (1/2)[(a-b)2 + (b-c)2 + (c-a)2] ≥ 0 => it is always non-negative.

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.

pooopoopopoopop

step-by-step explanati∵∵∴∴on: