Answer:
Rate of change in elevation = 0.6 in/year
Step-by-step explanation:
Note:
Current elevation (Missing) = 7,602 feet
Given:
Old elevation = 7,602 feet
Number of year = 7,600
Find:
Rate of change in elevation
Computation:
Change in elevation = 7,602 - 7,600
Change in elevation = 2 ft
Change in elevation = 2 x 12 = 24 inches
Rate of change in elevation = 24 / 40
Rate of change in elevation = 0.6 in/year
Your answer is 2.50 (And the rest of the 0s, but they aren't needed...) Since that is greater then 1, but less than 10.
The first thing you must do is to calculate the weight of the gravel (w) that the truck holds (7.5 yards³). So, you have:
If the weight of 1 yard³ is 1.48 tons, then the weight of 7.5 yards³ is:
w=7.5x1.48 tons
w=11.1 tons
The problem says that the gravel will be placed in containers that each hold 3.7 tons of gravel,so, If they need 1 cotainer to place 3.7 tons, how many containers they need to place 11.1 tons?
1 container-------3.7 tons of gravel
x-------11.1 tons of gravel
x=(11.1x1)/3.7
x=11.1/3.7
x=3 containers
H<span>ow many containers of this size are needed to hold all the gravel from one truck?
</span>
The answer is: 3 containers.
Its C! Hope this helps and can you mark me as brainliest :))
Answer:
option 2
in the first equation y is greater than so it would be shaded above the line and in the second equation y is less than so it would be shaded below the line
Step-by-step explanation:
