I may be wrong here but I think it is 50.
Answer:
8
Step-by-step explanation:
you solve this by reversing the excercise.
17 + 7 = 24
24/3 = 8
Answer: part a is 5 x 12 = 60 , then 40 divided by ten which is 4 and I’m too stupid for part b lol
Part I
The weight of the truck without cargo is 22150+11300= 33450 pound
The weight of the cargo is given as
![c](https://tex.z-dn.net/?f=c)
and altogether with the weight of the truck, the sum must not exceed 75500
The inequality is given
![33450+c \leq 75500](https://tex.z-dn.net/?f=33450%2Bc%20%5Cleq%2075500)
Part II
![33450+c \leq 75500](https://tex.z-dn.net/?f=33450%2Bc%20%5Cleq%2075500)
![c \leq 75500-33450](https://tex.z-dn.net/?f=c%20%5Cleq%2075500-33450)
![c \leq 42050](https://tex.z-dn.net/?f=c%20%5Cleq%2042050)
Hence, the maximum weight allowed for the cargo is 42050 pounds
Answer:
x = 6, y = 5
Step-by-step explanation:
When x = 6 and y = 5,
3(6) - 5 = 13
18 - 5 = 13
13 = 13
L.H.S = R.H.S