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:
<h3>200201+1918191=9191919289+</h3>
Step-by-step explanation:
<h3>tnks </h3>
Answer:
4/20
Step-by-step explanation:
8 socks we don't know
1 sock 19 socks
11 white
1 more sock 7 /20 chance to pick a random sock
For a better understanding of the solution provided here, please find the diagram attached.
In the diagram, ABCD is the room.
AC is the diagonal whose length is 18.79 inches.
The length of wall AB is 17 inches.
From the given information, we have to determine the length of the BC, which is depicted a 
, because for the room to be a square, the length of the wall AB must be equal to the length of the wall BC.
In order to determine the length of the wall BC, or , we will have to employ the Pythagoras' Theorem here. Thus:
Thus, 
inches
and hence, the given room is not a square.
