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
You can solve this in 2 easy steps: 1, multiply 44.55 by .17 (44.55 * .17 = 7.57) then adding the number you get to 44.55 (44.55 + 7.57 = 52.12). Your answer is D.
Answer:
c = 24.34
Step-by-step explanation:
Here, we can use the cosine rule
Generally, we have this as:
a^2 = b^2 + c^2 - 2bcCos A
12^2 = 14^2 + c^2 - 2(14)Cos 19
144 = 196 + c^2 - 26.5c
c^2 - 26.5c + 196-144 = 0
c^2 - 26.5c + 52 = 0
We can use the quadratic formula here
and that is;
{-(-26.5) ± √(-26.5)^2 -4(1)(52)}/2
(26.5 + 22.23)/2 or (26.5 - 22.23)/2
24.37 or 2.135
By approximation c = 24.34 will be correct
If 1 cm = 3 ft, then 5 cm = 5*3 = 15 ft.<==
