Lol
trouble varies directly as distance
lets say t=trouble and d=distance
t=kd
k is constant
given
when t=20, and d=400
find k
20=400k
divide by 400 both sides
1/20=k
t=(1/20)d
given, d=60
find t
t=(1/20)60
t=60/20
t=3
3 troubles
Answer:
perpindicular slope = -2/3
Step-by-step explanation:
Answer:
7.
Step-by-step explanation:
9y+12=75
subtract 12 from 75
divide by 9y
Answer:
ldk Sry
Step-by-step explanation:
Answer:

A maximum of 112 number of 100 - kilograms can be loaded in the container.
Step-by-step explanation:
Given that:
Weight of each crate = 100 kg
The greatest weight that can be loaded in the container = 24000 kg
Weight already loaded in the container = 12800 kg
To find:
The inequality to determine the value
i.e. number of 100 - kilograms that can be loaded in the shipping container?
Solution:
Weight already loaded = 12800 kg
Let the number of 100 - kilograms that can be loaded in the container = 
Weight of
= 100
kg
This combined weight nor be greater than the capacity of the container.
OR we can say, it must be lesser than or equal to greatest weight that can be loaded into the container.


i.e. a maximum of <em>112</em> number of 100 - kilograms can be loaded in the container.