The artistic crop isn't helpful; it cuts off some vertex names.
The circumcenter H is the meet of the perpendicular bisectors of the sides, helpfully drawn. We have right triangle ELH, right angle L, so
EH² = HL² + EL²
EL = √(EH² - HL²) = √(5.06²-2.74²) ≈ 4.2539393507665339
Since HL is a perpendicular bisector of EF, we get congruent segments FL=EL.
Answer: 4.25
c(x)=9x+89
I know that you don't have the answer listed, but it is the correct answer.
I think second must b the answee
Given:
2 parallelograms with an area of 9 1/3 yd²
height of each parallelogram is 1 1/3 yd
Area of parallelogram = base * height
We need to divide the combined area into two to get each parallelogram's base.
9 1/3 = ((9*3)+1)/3 = 28/3
28/3 ÷ 2 = 28/3 * 1/2 = 28/6 yd² or 4 4/6 yd² ⇒ 4 2/3 yd²
Area of each parallelogram is 4 2/3 yd²
4 2/3 yd² = base * 1 1/3 yd
14/3 yd² ÷ 4/3 yd = base
14/3 yd² x 3/4 yd = base
14*3 / 3*4 = base
42 / 12 = base
3 6/12 yd = base
or 3 1/2 yd = base
a) the base of each parallelogram is 3 1/2 yards
b) we can assume that the two parallelograms form a rectangle.
area of a rectangle is length times width.
length is 3 1/2 yds * 2 = 7 yds
width is 3 1/2 yds
Area of rectangle = 7 yds * 3 1/2 yds
Area = 7 yd * 7/2 yd
Area = 7*7 / 2 yd²
Area = 49 / 2 yd²
Area = 24 1/2 yd²
*Hint: In order to find the coordinates, you plug in the coordinates into the rule and solve.
(x, y) → (x + 2, y - 8)
(4, -5) → (4 + 2, -5 - 8)
(6, -13)
The coordinates of B' is (6, -13).
