The answer to the question is 3 8/33
Hey there, Lets solve this problem together.
The First step is to line up the numbers.
<span>We calculate </span>

<span>the result of which is </span>

<span>
</span>
<span>We calculate </span>

<span> the result of which is </span>

<span>.
</span>
Since we get a negative number in the next column, we must take 1 from the next column and carry it over to this column. Now the number will be changed to 10.
We calculate

, and the result is

.
<span>We calculate </span>

<span> the result of which is</span>

<span>.
</span>
Therefore,
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
Hello,
y=2xe^x
y'=2(e^x+xe^x)=2(x+1)e^x
y''=2(e^x+(x+1)e^x)=2(x+2)e^x
x |-infinite -2 0 +infinite
e^x | ++++++++++++++++++
x+2 |------------0 +++++++++++
y'' | -----------0 +++++++++++
y''<0 if x<-2
<span>The interval on which the graph is concave down is (-infinite -2[</span>
The slope is -3/2 and the y-intercept is -3. If you need me to explain I can do that :) hope this helps