we know that
The volume of the box is equal to
V=area of the base*height
solve for the height
height=Volume/(area of the base)
![V=2x^{3}+3x^{2}- 11x- 6\ ft^{3} \\ Area\ of\ the\ base=x^{2}+x-6\ ft^{2}](https://tex.z-dn.net/?f=V%3D2x%5E%7B3%7D%2B3x%5E%7B2%7D-%2011x-%206%5C%20ft%5E%7B3%7D%20%5C%5C%20Area%5C%20of%5C%20the%5C%20base%3Dx%5E%7B2%7D%2Bx-6%5C%20ft%5E%7B2%7D)
using a graph tool-----> find the roots
see the attached figures
so
![2x^{3}+3x^{2}- 11x- 6=2*(x+3)*(x+0.5)*(x-2)](https://tex.z-dn.net/?f=2x%5E%7B3%7D%2B3x%5E%7B2%7D-%2011x-%206%3D2%2A%28x%2B3%29%2A%28x%2B0.5%29%2A%28x-2%29)
![x^{2}+x-6=(x+3)*(x-2)](https://tex.z-dn.net/?f=x%5E%7B2%7D%2Bx-6%3D%28x%2B3%29%2A%28x-2%29)
substitute
height=Volume/(area of the base)
![height= \frac{2*(x+3)*(x+0.5)*(x-2)}{(x+3)*(x-2)} = 2*(x+0.5)](https://tex.z-dn.net/?f=height%3D%20%5Cfrac%7B2%2A%28x%2B3%29%2A%28x%2B0.5%29%2A%28x-2%29%7D%7B%28x%2B3%29%2A%28x-2%29%7D%20%3D%202%2A%28x%2B0.5%29)
![height=2x+1](https://tex.z-dn.net/?f=height%3D2x%2B1)
therefore
<u>the answer is</u>
![height=2x+1](https://tex.z-dn.net/?f=height%3D2x%2B1)
Step-by-step explanation:
T78 = 5 + (78-1)(3)
= 5 + 77(3)
= 5+231
= 236
Answer:
36 ,36 30,2
Step-by-step explanation:
hope it's okay to you
Can you put some more info in, its hard to answer the question without the full thing.