2. What equation models the data in the table if d = number of days and c = cost?

Days
2
3
5
6
Cost
44
66
110
132

A. c = d + 22

B. c = 22d

C. d = 22c

D. c = d + 22d

 B. c = 22d

I'm gonna go with c=22d

Step-by-step explanation:  22d

C = 22d is the correct answer because for every day the cost increases by 22 so two days mean the cost is equal to 22 (2)

The equation that models the data in the table is:

Step-by-step explanation:

The table that represent the days and cost is given by:

      Days Cost

          2          44

          3          66

          5           110

          6          132

We see that the cost is a constant multiple of  the number of days.

i.e. the equation is given by:

            c=22d

i.e. the relationship is linear.

( Since, when d=2

we have: c=22×2=44

when d=3 we have:

c=22×3=66

and so on )

You can see that 44/2 = 22

66/3 = 22

110/5 = 22

132/6 = 22

Then, the relation between C and d is a constant.

C/ d = 22

And you can model C as:

C = 22d.

This model tells that the cost equals 22 times the days.

p12q8u43u2304932456

Step-by-step explanation:

90909090909

B

Step-by-step explanation:

It is B because c would be the total cost and 22d is $22 a day.

Hoped this helped. :)

the answer to the question is a  

it has the same slope and the same y-intercept.

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