How would you prove algebraically that the function: f(x)= |x-2|, x<= 2 , is one to one?
This would be a real bear to prove, mainly because it's not true.
irregular shape! :D x x x x x x x x x x x x
You can prove by making two or more identical triangles.Congruence means when two figures have the same size, form, and shape.
A sheet of A4 paper could be classed as a plane 2D shape, as it has length x width. A shoebox is an example of a 3D shape, as it has length x width x depth.
it a shape the sides have names 5 sides = pent
How would you prove algebraically that the function: f(x)= |x-2|, x<= 2 , is one to one?
This would be a real bear to prove, mainly because it's not true.
irregular shape! :D x x x x x x x x x x x x
Let x be in A intersect B. Then x is in A and x is in B. Then x is in A.
How would you prove algebraically that the following function is one to one? f(x)= (x+3)^2 , x>= -3?
One could not. The shape could be a rectangle.
Prove all x px or all x qx then all x px or qx
You can prove by making two or more identical triangles.Congruence means when two figures have the same size, form, and shape.
3D shape = length x width x height 2D shape = length x height
Looking at the graph of the function can give you a good idea. However, to actually prove that it is even or odd may be more complicated. Using the definition of "even" and "odd", for an even function, you have to prove that f(x) = f(-x) for all values of "x"; and for an odd function, you have to prove that f(x) = -f(-x) for all values of "x".
A sheet of A4 paper could be classed as a plane 2D shape, as it has length x width. A shoebox is an example of a 3D shape, as it has length x width x depth.