I believe that B) m<3 would need to double. Since m<6 and m<3 are corresponding angles, they are always equal. So if 6 doubles, 3 needs to as well in order to be equal.
Answer:
An identity matrix, is a matrix that have '1' in the main diagonal. All of the other terms are '0'. When you multiply any matrix by the identity matrix, the result is the same matrix that you multiplied.
Example:
![\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%260%5C%5C0%261%260%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D)
In the set of the real number is the same that the application of identity property.
Every number multiplied by 1 es the same number.
Step-by-step explanation:
Let x = the unknown number to get x = 6
We know that the area of a square = side ²
Therefore, for a known area A, the side length is
side = √ (area)
= √ (175)
= 13.2288 (to 4 places of decimals).
Notice how on both sides of the equation there is one coefficient (number being multiplied by x) and a constant (just a plain number). You want all constants to be on one side and all coefficients on the other. He’s how do to that:
2x+14=-21-5
I am choosing that all constants will be on the right side, so I will do the inverse operation of any constant on the left side to remove it. Here, a constant on the left side is 14. I will subtract 14 from both sides of the equation. Positive 14 minus 14 is zero, so it cancels out and removes it.
2x+14-14 = -21-5x
2x = -35-5x
See how the 14 is now gone? The equation looks much simpler now. Okay next, you can see that there is a coefficient on the “constant side” that I’ve chosen, so I am going to remove that. Negative 5 plus positive 5 equals zero. Do this on both sides of the equation.
2x+5x = -35-5x+5x
7x = -35
Now the equation is just two numbers. All that is left to do is divide by the last coefficient, 7, in this case.
7x/7 = -35/7
x = -5
