The line of reflection is x = 1
Not easy to graph but
multily first equaiton by -2 remember to flip sign
-2x+4y>-6
add to other equaiton
2x-2x+y+4y>-6+8
5y>2
divide by 5
y>2/5
subtitute
2x+2/5>8
subtract 2/5 from both sdies
2x>7 3/5
divid eby 2
x>38/10=19/5=3 4/5
x>3 4/5
y>2/5
just shade the area in that zone, not including point (0,0)
I believe it’s a trapezoid but i’m not too sure... sorry :((
Trapezoidal is involving averageing the heights
the 4 intervals are
[0,4] and [4,7.2] and [7.2,8.6] and [8.6,9]
the area of each trapezoid is (v(t1)+v(t2))/2 times width
for the first interval
the average between 0 and 0.4 is 0.2
the width is 4
4(0.2)=0.8
2nd
average between 0.4 and 1 is 0.7
width is 3.2
3.2 times 0.7=2.24
3rd
average betwen 1.0 and 1.5 is 1.25
width is 1.4
1.4 times 1.25=1.75
4th
average betwen 1.5 and 2 is 1.75
width is 0.4
0.4 times 1.74=0.7
add them all up
0.8+2.24+1.75+0.7=5.49
5.49
t=time
v(t)=speed
so the area under the curve is distance
covered 5.49 meters
