There is no question. But I'd gladly help if you had a question!
True.
Let p1 and p2 be the two parallel planes. Let n1 be the normal vector of plane p1 (which is a vector perpendicular to the plane). If p2 is parallel to p1, then n1 is also a normal vector for p2.
Let p3 be the third secant plane, and n3 be its normal vector.
The direction vector of the intersection line of two planes is given by the cross products of their normal vectors (this is due to the fact that the cross product of two vectors is orthogonal two both of them, and that the direction vector of the intersection must be orthogonal two both normal vectors). So, the direction vectors of the two lines are:
v1 = n1 × n3
v2 = n1 × n3
The are equal. Hence, the lines are parallel.
Given the expression:
-x^2+18x-99
to solve by completing squares we proceed as follows:
-x^2+18x-99=0
this can be written as:
-x^2+18x=99
x^2-18x=-99.......i
but
c=(-b/2)²
Hence:
c=(-(-18)/2)²=81
adding 81 in both sides of i we get:
x^2-18x+81=-99+81
factorizing the quadratic we obtain:
(x-9)(x-9)=-18
thus
(x-9)²+18=0
the above takes the vertex form of :
y=(x-k)²+h
where (k,x) is the vertex:
the vertex of our expression is:
(9,18)
hence the maximum point is at (9,18)
NOTE: The vertex gives the maximum point because, from the expression we see that the coefficient of the term that has the highest degree is a negative, and since our polynomial is a quadratic expression then our graph will face down, and this will make the vertex the maximum point.
I literally just saw a question like this, but it won’t let me upload my photo, says the answer is incorrect
