3. If you were to use Gauss-Jordan elimination on the following linear system, what would the augmented matrix be?
3w - X = 2y + 2 - 4
9x - y + z = 10
4w + 3y - 2 = 7
12x + 17 = 2y - Z + 6
O
O
3 -1 -2 -1
9 -1 0 1 10
A.
4 0 3 -1 7
12 -2 1 0 - 11
[3 -1 -2 -1 -4
0 - 1 1 10
B.
4 0 3 -1 7
LO 12-21 -11]
13-12 1-4
o 9 -1 1 10
C.
0 3 -1
0 12 2 -1 6
O
О
3 -1 -2 -1 - 4
0 9-1 1 10
D.
4 0 3 -1 7
0 12 2 1 6

d) 48

step-by-step explanation:

16,32,48,64,80,96,112,128,144,160,176,192,208,224,240,256,272,288,304,320,336

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!

a

step-by-step explanation: