Answer:
, x = 6 ton
Step-by-step explanation:
The truck can be filled with a total of 20 ton.
There are already 14 ton inside the truck, so if we call x the weight of goods that we can still add into the truck, we know that (x+14) must be equal at most at 20. Therefore, the inequality becomes:
Solving it, we find
which means that the most amount of goods that could be added to the truck is 6 tons.
Answer:
2) D: x = [0, 24]
3) R: y = [0, 384]
4) see graph
<u>Step-by-step explanation:</u>
Eric's regular wage is $12 per hour for all hours less than 9 hours.
The minimum number of hours Eric can work each day is 0.
f(x) = 12x for 0 ≤ x < 9
Eric's overtime wage is $18 per hour for 9 hours and greater.
The maximum number of hours Eric can work each day is 24 (because there are only 24 hours in a day).
f(x) = 18(x - 8) + 12(8)
= 18x - 144 + 96
= 18x - 48 for 9 ≤ x ≤ 24
The daily wage where x represents the number of hours worked can be displayed in function format as follows:
2) Domain represents the x-values (number of hours Eric can work).
The minimum hours he can work in one day is 0 and the maximum he can work in one day is 24.
D: 0 ≤ x ≤ 24 → D: x = [0, 24]
3) Range represents the y-values (wage Eric will earn).
Eric's wage depends on the number of hours he works. Use the Domain (given above) to find the wage.
The minimum hours he can work in one day is 0.
f(x) = 12x
f(0) = 12(0)
= 0
The maximum hours he can work in one day is 24 <em>(although unlikely, it is theoretically possible).</em>
f(x) = 18x - 48
f(24) = 18(24) - 48
= 432 - 48
= 384
D: 0 ≤ y ≤ 384 → D: x = [0, 384]
4) see graph.
Notice that there is an open dot at x = 9 for f(x) = 12x
and a closed dot at x = 9 for f(x) = 18x - 48
