Answer:
(b) It is symmetrical about 
Step-by-step explanation:
Given

See attachment for options
Required
True statement about the graph
First, we check the line of symmetric

Expand

Open bracket


A quadratic equation
has the following line of symmetry

By comparison, the equation becomes:



Hence, the line of symmetry is at: 
(b) is true.
Answer:
The cubed-root of v (/v)
Step-by-step explanation:
Like your work on the second question illustrated, the cubed root of v (which is 216) is your expression for the edge length of the box in inches.
Hope this helps!
Given:
Initial point = (–5, 3)
Terminal point = (1, –6)
To find:
The set of parametric equations over the interval 0 ≤ t ≤ 1.
Solution:
The interval is 0 ≤ t ≤ 1, initial point is (–5, 3) and terminal point is (1, –6). It means,


Put t=1 in each parametric equation.
In option A,

In option B,

In option D,

Therefore, options A, B and D are incorrect.
In option C,


Put t=0 in x(t) and y(t).


Since, only in option C
, therefore, the correct option is C.