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:
d=0.031746
Step-by-step explanation:
1. Add 4d to both sides: 3.2=2.3d+3+4d
2. Simplify 2.3d +3+ 4d to 6.3d+3: 3.2=6.3d+3
3. Subtract 3 from both sides: 3.2-3=6.3d
4. Simplify 3.2-3 to 0.2: 0.2-6.3d
5. Divide both sides by 6.3: 0.2/6.3=d
6. Simplify 0.2/6.3 to 0.031746: 0.031746=d
7. Switch sides: d=0.031746
Hope this helped!
The horizontal distance between light house and boat is 1588.78 feet approximately.
The figure is given by,
Here, AB = height of the lighthouse bacon light above the water = 139 feet
Now angle ACB = 5 degree
Let the horizontal distance of light house from the boat = BC = x feet
So by trigonometric function we get,
tan 5 = AB/BC
tan 5 = 139/x
x = 139/tan 5 = 1588.78 (approximately)
Hence the horizontal distance between light house and boat is 1588.78 feet approximately.
To know more about Trigonometric Function refer to:
brainly.com/question/1143565
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<span>5.17490 rounded to the nearest thousandth is 5.175</span>
Answer:
DIRECT CONTROL. Definition. HOLDING EXTRANEOUS FACTORS CONSTANT SO THAT THEIR EFFECTS ARE NOT CONFOUNDED WITH THOSE OF THE EXPERIMENTAL CONDITIONS. Term.
