Think of the equation as the 10 times as parts so if Halle read ten times as more than Janet then we do the equation 264/11 because we have to include Janet's part too so Janet read 24, one part of the total while Halle read 240, 10 parts of the total
Answer:
The amount of the chemical flows into the tank during the firs 20 minutes is 4200 liters.
Step-by-step explanation:
Consider the provided information.
A chemical flows into a storage tank at a rate of (180+3t) liters per minute,
Let 
is the amount of chemical in the take at <em>t </em>time.
Now find the rate of change of chemical flow during the first 20 minutes.
![\int\limits^{20}_{0} {c'(t)} \, dt =\left[180t+\dfrac{3}{2}t^2\right]^{20}_0](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B20%7D_%7B0%7D%20%7Bc%27%28t%29%7D%20%5C%2C%20dt%20%3D%5Cleft%5B180t%2B%5Cdfrac%7B3%7D%7B2%7Dt%5E2%5Cright%5D%5E%7B20%7D_0)
So, the amount of the chemical flows into the tank during the firs 20 minutes is 4200 liters.
C) will not effected (the most frequent is still 600$)
Answer:
Step-by-step explanation:
First we have to find the coordinates of the original ordered pairs then we will do the translation.
Now we will add 3 to every x value and add a -2 to every y value.
When you do that you will get the new prime points.
Answer:
We are given an area and three different widths and we need to determine the corresponding length and perimeter.
The first width that is provided is 4 yards and to get an area of 100 we need to multiply it by 25 yards. This would mean that our length is 25 yards and our perimeter would be 2(l + w) which is 2(25 + 4) = 58 yards.
The second width that is given is 5 yards and in order to get an area of 100 yards we need to multiply by 20 yards. This would mean that our length is 20 yards and our perimeter would be 2(l + w) which is 2(20 + 5) = 50 yards.
The final width that is given is 10 yards and in order to get an area of 100 yards we need to multiply by 10. This would mean that our length is 10 yards and our perimeter would be 2(l + w) which is 2(10 + 10) = 40 yards.
Therefore the field that would require the least amount of fencing (the smallest perimeter) is option C, field #3.
<u><em>Hope this helps!</em></u>
