What is the distance AB rounded to the nearest tenth?

2.80

Step-by-step explanation:

Seeing that your scale is 1cm to 1unit to on both axes

If you add a point C then draw lines connecting them it should look like the diagram above

But the distance between A to C and B to C are both 2 cm

Giving you a right angled triangle

So using the Pythagoras theorem

AB²=AC²+BC²

AB²=2²+2²

AB²=4+4

AB²=8

Take square root of both sides

√AB²=√8

AB= 2.828

To the nearest tenth that will be

AB=2.80

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

-2 is the right