You work as a cashier for a grocery store and earn $6 per hour. you also mow lawns and earn $3 per hour. you want to earn at least $30 per week, but would like to work no more than 12 hours per week. how would you solve this in a real world problem

Let  in a week x hours are spent on working as a cashier and y hours are spent on mowing lawns,

Since, the earning as a cashier is $6 per hour and the earning in mowing is $3 per hour.

Thus, the total earning in a week = 6x + 3y

According to the question,

6x + 3y ≥ 30, ------(1)

Also, the total number of hours is not more than 12 hours,

⇒ x + y ≤ 12, -------(2)

Hence, the solution will be the common or feasible region of the inequality (1) and (2),

Graphing inequalities :

The related equation of inequality (1) is,

6x + 3y = 30

Having x-intercept = (5,0)

y-intercept = (0,10),

Also, 6(0) + 3(0) ≥ 30 ( False )

⇒ Inequality (1) will not contain the origin,

Now, The related equation of inequality (2) is,

x + y = 12

Having x-intercept = (12,0)

y-intercept = (0,12),

Also, (0) + (0) ≤ 12 ( True )

⇒ Inequality (2) will contain the origin

Hence, by the above information we can plot the graph. ( shown below )

a.   since the profit made in 2012 is equal to the profit made in 2014, the profit is symmetric about the line y = 120,000.

step-by-step explanation: