Because the polynomial has degree 2, we can assume that there are 2 solutions (roots), whether real or imaginary.
You can subtract 60 in order to put this in standard form
48x^2+44x-60 = 0
From there, just put a,b, and c into the quadratic formula and you're good to solve for your answers.
(-b+-sqrt(b^2-4ac))/2a
(-44+-sqrt(44^2-4(48)(-60)))/2(48)
Then solve.
There is probably a better way, but this should give you the two roots/solutions.
Anything without a variable in front.
In other words, the constants are -3.7 and 1/3
The cross of the position vectors is
.. [5, 2, 2] × [6, -1, 1] = [4, 7, -17] . . . . . the normal vector of the desired plane
Since the origin is a point in the plane, its equation can be written as
.. 4x +7y -17z = 0
Answer:
The answer to the question is C
Part A. Your model is as good as any. It is hard to tell if the label needs to include the descriptor (length = 12 ft, for example). Certainly your model is sufficient for Part B.
Part B. The total area is the sum of the areas of each of the 6 faces of the box. Opposite faces are the same area, so you have
A = 2(LW +WD +LD) = 2(LW +D(L +W))
Subsituting the given dimensions, you get
A = 2((12 ft)(6 ft) +(3 ft)(12 ft +6 ft))
A = 2(72 ft² +(3 ft)(18 ft))
A = 2(72 ft² +54 ft²) = 252 ft²
The least amount of paper required to cover the box is 252 ft².
