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
Answer:
see explanation
Step-by-step explanation:
5x² + 5x + 3 = 0 ← in standard form
with a = 5 , b = 5 , c = 3
Solve using the quadratic formula
x = 
= 
= 
= 
= 
= 
Then
x =
±
i
, that is
x = -
-
i
, -
+
i


for example, let's look at the first set
y+3x =5 or y = -3x+ 5
and y = -3x + 2
y = m + b
the slopes are equal, the y-intercepts differ
that means, they're just parallel lines, no solution
It would be 15 greater than or equal to 14
Step-by-step explanation:
E3, E6, E11, E7: no solution
E1, E12 : 1 solution
E2, E8, E10 : infinite solutions