8*8 - 4*4*pi*(1/2) = 64 - 8*pi
A because you move down 5 units and over 2
The distance is increasing at a rate that is the speed of the plane multiplied by the cosine of the angle between its flight path and the direct line to the radar station. That cosine is 4/√(3²+4²) = 4/5, so the distance is increasing at
440 mi/h × 4/5 = 352 mi/h
Answer:
If EFGH is an isosceles trapezoid,
if
EH=4x-27+
FG=x%2B9+
EG=3y%2B19
FH=11y-21
then EH=FG and EG=FH
set up equations:
4x-27=x%2B9 .......solve for x
4x-x=27%2B9
3x=36
x=36%2F3
3y%2B19=11y-21.......solve for y
11%2B19=11y-3y
30=8y
y=30%2F8
y=15%2F4
y=3.75
so,x=12 and y=3.75
you can also check the length of equal sides and equal diagonals:
EH=4%2A12-27+=21
FG=12%2B9+=21
EG=3%283.75%29%2B19=30.25
FH=11%283.75%29-21=30.25x=12
Step-by-step explanation:
