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:
6/3 I think so try c and if it’s not I’m not sure
Step-by-step explanation:
:)
Answer:
i think its 70
Step-by-step explanation:
35+35=70
because line ## is same as ## and ##
1 peso = 0.053 dollars.
x pesos = 400 US dollars.
1/x = 0.053 / 400 Cross multiply
400 * 1 = 0.053 x Divide by 0.053
400/0.053 = 0.053*x / 0.053
7547.17 pesos which rounded is
Answer: 7547 pesos
The law of cosines for this particular situation is b^2 = a^2 + c^2 - 2ac cosB.
Filling in what you know, you have b^2 = 25 + 49 - [2(5)(7)-.1908], which simplifies to b^2 = 74 - 69.8092 which gives you a b^2 value of 4.1908, but you have to take the square root of that so you get a side value for b of 2.05.
