I'm guessing on the make up of the matrices.
First off let's look at [C][F].
[C]=
[F]=
[C][F]=
where each element of [C][F] comes from multiplying a row of [C] with a column of [F].
Example: First element is product of first row and first column.
.
.
.
Now that we have [C][F], we can subtract it from [B], element by element,
[B]-[C][F]=
[B]-[C][F]=
.
.
.
If this is not how the matrices look,please re-state the problem and be more specific about the make up of the matrices (rows x columns).
Here's an example.
[A] is a 2x2 matrix. A=[1,2,3,4].
The assumption is that [A] looks like this,
[A]=
[B] is a 3x2 matrix. B=[5,6,7,8,9,10]
[B]=
Answer:
third one
Step-by-step explanation:
cuz its a fraction, and fractions arent polynomials
Answer:
{290,315,340, 365, 390, 415, 440, 465} That's the answer. Its hard for me to explain how i got it.
5 times as many. Since dog B has 2 and dog A has 10, A = 2 * X, since you’re solving for the multiplication comparison. 10/2 = 5, so you’ve got 5 times as many.
Solve for x:
x^9 = n x
Subtract n x from both sides:
x^9 - n x = 0
Factor x and constant terms from the left hand side:
-x (n - x^8) = 0
Multiply both sides by -1:
x (n - x^8) = 0
Split into two equations:
x = 0 or n - x^8 = 0
Subtract n from both sides:
x = 0 or -x^8 = -n
Multiply both sides by -1:
x = 0 or x^8 = n
Taking 8^th roots gives n^(1/8) times the 8^th roots of unity:
Answer: x = 0 or x = -n^(1/8) or x = -i n^(1/8) or x = i n^(1/8) or x = n^(1/8) or x = -(-1)^(1/4) n^(1/8) or x = (-1)^(1/4) n^(1/8) or x = -(-1)^(3/4) n^(1/8) or x = (-1)^(3/4) n^(1/8)
