What is the solution to the following system? 5x+2y+z=4 x+2z=4 2x+y-z=-1 a. x = 0, y = 1, z = 2 b. x = 1, y = –1, z = 2 c. x = 1, y = 0, z = –1 d. x = 2, y = 1, z = 0

correct choice is A

Step-by-step explanation:

From the second equation express x and substitute it into the first and third equations:

Then

From the third equation and substituting it into the second, you get

If then

and

The solution of the system is

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12

i believe it is 0.1m because the measurement of 30.4m limits how precise you can be

you can only be as precise as the least precise measurement