Solve the homogeneous linear system corresponding to the given coefficient matrix. (If there is no solution, enter NO SOLUTION. If the system has an infinite number of solutions, set x4 = t and x2 = s and solve for x1 and x3 in terms of t and s.)

leftbracket3.gif 1 0 0 1
rightbracket3.gif
0 0 1 0
0 0 0 0
(x1, x2, x3, x4) =

rightparen2.gif

The solution of the system is { (-t. s. 0, t) : s,t ∈ R}

Step-by-step explanation:

Lets multiply the first row by (x1,x2,x3,x4) and equalize to 0

(1,0,0,1) * (x1,x2,x3,x4) = x1+x4

which is equal to 0 only when

x1+x4 = 0

x1 = -x4

Now we can replace x1 with -x4.

If we multiply the second row (0,0,1,0) with (x1,x2,x3,x4) we will obtain (0,0,1,0)*(x1,x2,x3,x4) = x3. Since we want that product to be equal to 0, then x3 = 0. If we multiply the third row of zeros with (x1,x2,x3,x4) we will obtain automatically 0.

As a consecuence, we have 2 conditions:

We call t = x4, s = x2, which can take any value. The solutions of the system is

{ (-t. s. 0, t) : s,t ∈ R}

step-by-step explanation:

i could only see the first page. the other pages were blurry. ; -;

x hot dogs for 2$, y drinks for 3$. 29 items and total 70

2x+3y=70

x+y=29

if x=17and y=12

2(17)+3(12)=70

34+36=70

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2