Answer:
The orange one is the answer
Step-by-step explanation:
The easiest way to find the solution is to plug in one of the coordinates that is located in the shaded area into each equation. If the equations make sense, as they do in the orange option, you have your answer.
Hope this helps you!
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:
After simplifying, the exponent of x is 33 and the exponent of y is 0
Step-by-step explanation:
For us to simplify, we bring the x base together and the y base together
Recall from the law of indices;
x^a/x^b = x^(a-b)
and x^a * x^b = x^(a + b)
We shall be applying these here
Let us bring the x terms together
we have these as;
x^8/x^14 * 1/x^-39 = x^( 8 - 14 -(-39))
= x^(8 + 39-14) = x^33
for y, we have it that;
y^-26/y^-5 * 1/y^-21
= y^(-26 -(-5) - (-21)
= y^-26 + 5 + 21 = y^0
