Aand b a) if v is an eigenvector for both the matrices with corresponding eigenvalues 2 and 3, then eigenvector for the matrix v is also an (ab) and find b) suppose that a2 5a -6/ = 0. prove that and find its inverse. a is invertible c) suppose that t: v v is a linear transformation. prove that t: vv such that ii(v) = 2t(v) v is also a linear transformation

2.56 is the width of the garden

step-by-step explanation:

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

(x) - (y) = (1 + 4i)

step-by-step explanation:

to determine whether a given function is even, we simply substitute x with -x in the expression. if the result after simplification is identical to the original equation, then the function is said to be even other wise odd.

f(-x) = (-x)^2 - (-x) + 8

      = x^2 + x +8

the resulting function is not identical to the original one;

f(-x) ≠ f(x)

therefore, f(x) is not an even function