The ace novelty company wishes to produce two types of souvenirs: type a and type b.
to manufacture:
type a requires: 2 minutes on machine 1 and 1 minute on machine 2
type b requires: 1 minute on machine 1 and 3 minutes on machine 2
there are 180 minutes available on machine 1 and 300 minutes available on machine 2 for processing the order.
let x =# of type a produced
let y = # of type b produced
a. write linear equalities that give appropriate restrictions on x and y
b. if each type a souvenir will result in a profit of $1 and each type b souvenir will result a profit of 1.20 then express the profit, p in terms of x and y.
c. algebraically determine how many souvenirs of each type the ace novelty company should produce so as to maximize profit

Let x be the number of Type A souvenirs, and let y be the number of Type B souvenirs to be made. Then the profit in dollars will be P = x + 1.2y, which is the objective function to be maximized.

Next, we need to set up the constraints. On Machine I, there are at most 180 minutes available, and each Type A souvenir uses 2 minutes and each Type B souvenir uses 1 minutes.
So the total amount of time on Machine I is given by 2x+y and cannot be greater than 180 minutes, so 2x+y ≤ 180 is our first constraint. Similarly for Machine II, the amount of time is x + 3y and cannot exceed 300 minutes, so x + 3y ≤ 300 is our second constraint.
Lastly, we cannot make negative numbers of souvenirs, so x ≥ 0 and y ≥ 0 are also constraints. In total, we need to maximize P = x + 1.2y subject to the constraints
2x + y ≤ 180
x + 3y ≤ 300
x ≥ 0, y ≥ 0 

One shot. if 60% of 10 is 6, then what is 70%? we know it would be seven (since 60% is mathematically 60/100, or 6/10, then 70% is 70/100, or 7/10) but we can check anyway using ratios: 6 : 60% = x : 70%*inverse multiplicationx * 60% = 6 * 70%*divide both sides by 60%x = (6 * 70%) / 60%x = 7and since he has made 6 shots already, he would need just one more to get 7 shots or 70%. if the explanation is too confusing, comment and i will try to