Company a charges $0.32 dollars per minute (no fixed monthly charge). company b charges $13.9 per month plus $0.28 per minute. company c charges a fixed rate of $50 per month. let and () denote the monthly charges of company a, b, and c respectively for spending minutes on long distance calls. find a formula for the monthly cost of using company a, a(x): a(x)=  find a formula for the monthly cost of using company b, b( x ): b(x) =  find a formula for the monthly cost of using company c, c(x): c(x) =

For this case, the first thing we must do is define variables.

We have then:

x: number of minutes

A (x), B (x), C (x): total cost of companies A, B and C, respectively.

We must write a function of the form:

Where,

m: is the cost per minute

b: it is the fixed cost.

For the company A we have:

For company B we have:

For company C we have:

The cost of each company monthly is:

For this case, the first thing we must do is define variables.

We have then:

x: number of minutes

A (x), B (x), C (x): total cost of companies A, B and C, respectively.

We must write a function of the form:

Where,

m: is the cost per minute

b: it is the fixed cost.

For the company A we have:

For company B we have:

For company C we have:

The cost of each company monthly 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