Quadrilateral WXYZ has vertices W(-4,-3), X(0,-1). Y(6,-2), Z(2,-4). Using properties of diagonals prove that WXYZ is a parallelogram, but NOT a rectangle.

The diagonals are bisectors of each other and meet at  (1, -5/2) but are not congruent, so WXYZ is a parallelogram and not a rectangle

Step-by-step explanation:

for parallelograms.

the diagonals intersect at the midpoints

W                             X

Z                             Y

Diagonals are  WY  and  XZ

WY :   (-4, -3) to  (6, -2)

midpoint is   [(-4 + 6)/2  ,  (-3 - 2)/2 ] = ( 1, -5/2 )

XZ :   (0, -1)  to  (2, -4)

midpoint is  [ (0 + 2)/2 ,  (-1 + -4)/2 ]  = (1, -5/2)

The diagonals are bisectors of eachother.

Next show that these diagonals are not congruent.

length of  WY  =  root (  (-4 - 6)^2  +  (-3 - (-2))^2 )

WY = root ( 100  +  1)

WY = root(101)

length of XZ  =  root (  (0 - 2)^2  +  ( -1 - -4)^2 )

XZ =  root (4  +  9) = root (13)

we see that  WY  is not equal to XZ

...

so  ..  quadrilateral WXYZ is a parallelogram but not a rectangle.

answer:

sometimes

*i think

there are many solutions.

step-by-step explanation:

step 1: simplify both sides of the equation.

4x+6=4x^2+10x−4

step 2: subtract 4x^2+10x-4 from both sides.

4x+6−(4x^2+10x−4)=4x^2+10x−4−(4x^2+10x−4)

−4x^2−6x+10=0

step 3: factor left side of equation.

−2(2x+5)(x−1)=0

step 4: set factors equal to 0.

x=-5/2 or x=1