Name of the geometric solid suggested by a number cube
1 answer:
You might be interested in
Answer:
2x - y - 3z = 0
Step-by-step explanation:
Since the set
{i, j} = {(1,0), (0,1)}
is a base in
and F is linear, then
<em>{F(1,0), F(0,1)} </em>
would be a base of the plane generated by F.
F(1,0) = a+b = (2i-2j+k)+(i+2j+k) = 3i+2k
F(0,1) = a+c = (2i-2j+k)+(2i+j+2k) = 4i-j+3k
Now, we just have to find the equation of the plane that contains the vectors 3i+2k and 4i-j+3k
We need a normal vector which is the cross product of 3i+2k and 4i-j+3k
(3i+2k)X(4i-j+3k) = 2i-j-3k
The equation of the plane whose normal vector is 2i-j-3k and contains the point (3,0,2) (the end of the vector F(1,0)) is given by
2(x-3) -1(y-0) -3(z-2) = 0
or what is the same
2x - y - 3z = 0
Given the polynomial expression:
(y + 5)²
(y - 5)(y + 5)
Let's simplify each of the given expression:
a.) (y + 5)²
The given equation is a factor of a perfect square trinomial. For this type of expression, the following is the formula for expanding it.
We get,
b.) (y - 5)(y + 5)
To be able to simplify the following expression. We will be using the formula for the difference of two squares.
We get,