Answer: in real numbers the equation \displaystyle \-x+sqrt{1-x^{2}}\-=sqrt{ 2}\left(2x^{2}-1\right)\-x+sqrt1−x2\-=sqrt2(2x2− 1).
Step-by-step explanation:
I think the answer would most likely be C.
Answer:
Part A) The graph in the attached figure
Part B) see the explanation
Step-by-step explanation:
Part A) Graph the function
we have the quadratic function

This is a vertical parabola open upward
The vertex is a minimum
using a graphing tool
The graph in the attached figure
Part B) What are the values of a, b and c?
we know that
The values of a and b represent the x-intercepts of the quadratic equation
The x-intercepts are
(-2,0) and (6,0)
so

Find the value of c
we know that
The x-coordinate of the vertex in a vertical parabola is equal to the midpoint of the roots
so
The value of c is equal to

substitute the given values

see the attached figure
Answer:
The correct answer is A..
Step-by-step explanation:
From the Invertible Matrix Theorem (IMT) we have a set of equivalent conditions to determine if a square matrix is invertible or not. In particular, it says that a square matrix of dimension tex]n\times n[/tex] is invertible if and only if, its columns span the vector space tex]R^n[/tex].
In the particular case of this exercise we have a matrix of dimension tex]5\times 5[/tex]. So, by the Invertible Matrix Theorem its columns must span the vector space tex]R^5[/tex]. Now, according to the statement of the exercise this condition does not hold. Hence, the given matrix cannot be invertible.