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
The answer for this one is 29 and 13. Hope it help!
Answer:
D
Step-by-step explanation:
Imagine this as a right angle triangle, where the diagonal length is the hypotenuse, the length is one side, and the width is the other.
We can therefore use Pythagoras' Theorem (or Pythagorean Theorem) to solve. The formula for this is a²+b²=c², where c is the hypotenuse, and a and b are the sides.
We can input the values we know to this formula to get the width. This gives 110²+b²=133.14² or 12100+b²=17 726.2596.
From there subtracting 12100 from both sides gives b²=5626.2596.
Square rooting b isolates it, leaving b=75.0083969.
Since the value of the diagonal was approximate, this can be assumed the b is 75m.
**This content involves Pythagoras' Theorem/Pythagorean Theorem, which you may wish to revise. I'm always happy to help!
Answer:
1000 : 1000 : 1500
Step-by-step explanation:
2 + 2 + 3 = 7
3500 ÷ 7 = 500
2 × 500 = 1000
2 × 500 = 1000
3 × 500 = 1500
1000 : 1000 : 1500
