We have the next function
[tex]r(t)=900-35t[/tex]
We have a linear function
d) for
[tex]0\le t\le45[/tex]
When does the reservoir have the most water
when t=0
e) for
[tex]0\le t\le45[/tex]
Let's calculate t when r(t)=0 because it is the value when the reservoir has the least water
[tex]0=900-35t[/tex]
we isolate the t
[tex]t=\frac{900}{35}=25.71[/tex]
t=25.71
f)
the inequality is given by the next numbers
r(0)=900-35(0)=900
r(45)=900-35(45)=-675
[tex]-675\le\: r\le900[/tex]
