Answer:
The quantity of water drain after x min is 50 [tex](0.9)^{x}[/tex]
Step-by-step explanation:
Given as :
Total capacity of rain barrel = 50 gallon
The rate of drain = 10 gallon per minutes
Let The quantity of water drain after x min = y
Now, according to question
The quantity of water drain after x min = Initial quantity of water × [tex](1-\dfrac{\textrm rate}{100})^{\textrm time}[/tex]
I.e The quantity of water drain after x min = 50 gallon × [tex](1-\dfrac{\textrm 10}{100})^{\textrm x}[/tex]
or, The quantity of water drain after x min = 50 gallon × [tex](0.9)^{x}[/tex]
Hence the quantity of water drain after x min is 50 [tex](0.9)^{x}[/tex] Answer
Answer:
[tex]y=50-10x[/tex]
Step-by-step explanation:
Let x represent number of minutes of draining.
We have been given that a barrel drains at a rate of 10 gallons per minute. So gallons drained after x minutes of draining would be [tex]10x[/tex].
Since the numbers of gallons is decreasing, so slope will be negative as:
[tex]-10x[/tex]
We are also told that the 50-gallon rain barrel is filled to capacity. This means that initial value or y-intercept is 50.
We can represent number of gallons remaining (y) after x minutes as:
[tex]y=50-10x[/tex]
Therefore, the equation [tex]y=50-10x[/tex] shows the amount of water in the barrel after x minutes of draining.
Upon graphing our equation, we will get required graph as shown in the attachment.
