Area of a trapezoid: [(top +bottom)*height]/2
suppose the length of one base is x, then the other base is 5x-19
[(x+5x-19)*18]/2=477
[6x-19]*18=954
6x-19=53
6x=72
x=12
The longer base is 5*12-19=41
1/2(4x-2)-2/3(6x+9)
(2x-1)-(4x+6)
2x-1-4x+62x-4x-1+6
-2x+5=2.5
So I believe that it is greater than 4
Answer:
A i think
Step-by-step explanation:
The position function of a particle is given by:

The velocity function is the derivative of the position:

The particle will be at rest when the velocity is 0, thus we solve the equation:

The coefficients of this equation are: a = 2, b = -9, c = -18
Solve by using the formula:
![t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D)
Substituting:
![\begin{gathered} t=\frac{9\pm\sqrt[]{81-4(2)(-18)}}{2(2)} \\ t=\frac{9\pm\sqrt[]{81+144}}{4} \\ t=\frac{9\pm\sqrt[]{225}}{4} \\ t=\frac{9\pm15}{4} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20t%3D%5Cfrac%7B9%5Cpm%5Csqrt%5B%5D%7B81-4%282%29%28-18%29%7D%7D%7B2%282%29%7D%20%5C%5C%20t%3D%5Cfrac%7B9%5Cpm%5Csqrt%5B%5D%7B81%2B144%7D%7D%7B4%7D%20%5C%5C%20t%3D%5Cfrac%7B9%5Cpm%5Csqrt%5B%5D%7B225%7D%7D%7B4%7D%20%5C%5C%20t%3D%5Cfrac%7B9%5Cpm15%7D%7B4%7D%20%5Cend%7Bgathered%7D)
We have two possible answers:

We only accept the positive answer because the time cannot be negative.
Now calculate the position for t = 6:
Are you just looking for the green space or more