Please help me latterly i posted this and no one bothered too help me plz i beg of u i really need help u have no idea i am failing

1. An equation for income, I = 3c + 2p, where c is the number of calendars sold and p is the number of posters sold.

2. When 25 calendars and 18 posters are sold, the income is $111.

3. Income for selling 12 calendars and 15 posters is $66.

4. 3 possible combinations; 1) 30 calendars and 5 posters. 2) 20 calendars and 20 posters. 3) 10 calendars and 35 posters.

Step-by-step explanation:

Step 1; It is given that each calendar costs $3 and each poster costs $2. So if c is the number of calendars sold and p is the number of posters sold.

Total income, I = 3c + 2p.

Step 2; Now if we substitute the value of c with 25 and p with 18, we get

I = 3c + 2p = 3 (25) + 2 (18) = 75 + 36 = $111.

Step 3; If the value of p is 12 and the value of c is 15, we have

I = 3c + 2p = 3 (12) + 2 (15) = 36 + 30 = $66.

Step 4; To find combinations that cost $100, we substitute I with 100. So the equation becomes 3c + 2p = 100.

When c = 30, 3 (30) + 2p = 100, 2p = 100 - 90 = 10, 2p = 10, p =5.

When c = 20, 3 (20) + 2p = 100, 2p = 100 - 60 = 40, 2p = 40, p = 20.

When c = 10, 3 (10) + 2p = 100, 2p = 100 - 30 = 70, 2p = 70, p = 35.

$15 each

step-by-step explanation:

$45/3 children= $15 each

a)(x-2)^2+(y-1)^2=9

step-by-step explanation:

(picture explains)