(1) let $v = \mathcal{c}(-1,1)$ is a vector space over $\mathbb{r}$, let $f, h \in v$ defined by $f : x \mapsto x^2 , h: x \mapsto x|x|$ , is $\{f, h\}$ linearly independent ? (2) for which values of $t \in \mathbb{r}$ the subset $$ \left\{ \begin{pmatrix}\cos t+i \sin t \\ 1 \\ \end{pmatrix},\begin{pmatrix}1 \\ \cos t- i \sin t\\ \end{pmatrix}\right\}$$of $\mathbb{c}^2$ is linearly dependent ? (3) consider e1 and e2 are projections on m1 and m2 along n1 and n2 respectively. verify the statement: if e1 and e2 commute, then e1 + e2 − e1e2 is a projection (4) let v be a vector space over f and let α; β be projections on v satisfying the condition that im(α) and im(β) are disjoint. is it necessarily true that αβ = βα? (5) let v = c(−1; 1) and let $a > -1/2$ be a real number. is the function $\alpha : v \times v \rightarrow \mathbb{r}$ defined by $\alpha(f, g) =\int_{-1}^1 (1-t^2)^{a-1/2}f(t)g(t) dt $ an inner product on v?

point a is the center of the circle that passes through points e, f, and g and the center of the circle that passes through points x, y, and z.

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!

answer: 5100 just subtract

step-by-step explanation:

do you have an easier question cause this shh is hard like theheck

step-by-step explanation: