Consider the picture.
Let MN be the midsegment of the trapezoid.
That is M is the midpoint of AD, N is the midpoint of BC.
Being the midsegment of the trapezoid, MN is parallel to the bases.
Let O and K be the intersections of the diagonals with the midsegment.
MN//AB, so MO//AB, and since M is the midpoint of DA, O must be the midpoint of DB,
Similarly we prove that K is the midpoint of CA.
Thus O is F and K is E.
O and K lie on the midsegment MN, so F and E lie on the midsegment.
MO is a midsegment of triangle ABD so |MO|=1/2 |AB|=1/2 * 10=5
MK is a midsegment of triangle ADC, so |MK|=1/2 * |DC|=1/2 * 22=11
|OK|=|MK|-|MO|=11-5=6 (units)

It's already a square; we don't need to add anything to complete the square.

That's a vertex at x = -1. The corresponding y is y=0.
Answer: Vertex at (-1,0)
As for the other questions cut off there, the axis is the vertical line x = -1 and the range is y ≥ 0.
Answer:
1/6 is your answer :))
Step-by-step explanation:
There is 1 four out of 6 possible numbers = 1/6
The +5 because that's the y value on the graph
Answer:
(x + 1)^2 + (y - 1)^2 = 74
Step-by-step explanation:
By looking at the graph, we can see the center of the circle is (-1, 1).
Next, we can find the radius. We are given the point (6, -4) is on the circle. The distance from the center to a point on a circle is the radius. We can plug the points (-1, 1) and (6, -4) into the distance formula to get the radius:
r = √([-1-6]^2 + [1-(-4)]^2)
r = √(49 + 25)
r = √74
The equation of a circle is denoted in the form:
(x - h)^2 + (y - k)^2 = r^2
where (h, k) is the center and r is the radius.
We can plug in the values we calculated:
(x + 1)^2 + (y - 1)^2 = 74