Answer:
The liters that the tank will contain at 5:11 PM that day are:
Step-by-step explanation:
Firstly, you must identify the outlet flow of the water pumped from the tank, for this, you must subtract the last volume given from the first volume:
- 19,140 L - 8,097 L = 11,043 L
And the minutes that passed from the first volume until the last volume given (18 minutes from 4:47 PM to 5:05 PM), so, you must divide that two values to obtain the outlet flow:
- Outlet flow =
- Outlet flow =
- Outlet flow = 613.5
![\frac{L}{min}](https://tex.z-dn.net/?f=%5Cfrac%7BL%7D%7Bmin%7D)
Now, you must see the next hour given (5:11 PM), if you see, from 5:05 PM to 5:11 PM has passed 6 minutes, taking into account this, you replace the equation of outlet flow to clear the volume:
- Outlet flow =
- Volume = Outlet flow * time
And replace the values to obtain the new volume pumped:
- Volume = 613.5
* 6 min - Volume = 3681 L.
At last, you must subtract these liters from the last volume identified in the tank:
- New Volume in the tank = 8097 L - 3681 L
- New Volume in the tank = 4416 L
The volume in the tank at 5:11 PM is <u>4416 Liters</u>.
Speed milles 120 ÷ 105 min = 1.14 milles
1.14 x 60 = 68.4 milles per hour
Answer:
Step-by-step explanation:
C
Answer:
5
5Step-by-step explanation:
If you use the distance formula you can solve it pretty easily. The distance formula is:
d = ![\sqrt{(x2 - x1)^2 + (y2 - y1)^2} \\](https://tex.z-dn.net/?f=%5Csqrt%7B%28x2%20-%20x1%29%5E2%20%2B%20%28y2%20-%20y1%29%5E2%7D%20%5C%5C)
The two points you want are A and D. A is (-1, -4) and D is (3, -1). Plug that into the formula...
d = ![\sqrt{(3 - (-1))^2 + (-1 - (-4))^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%283%20-%20%28-1%29%29%5E2%20%2B%20%28-1%20-%20%28-4%29%29%5E2%7D)
d = ![\sqrt{4^2 + 3^2}](https://tex.z-dn.net/?f=%5Csqrt%7B4%5E2%20%2B%203%5E2%7D)
d = ![\sqrt{16 + 9}](https://tex.z-dn.net/?f=%5Csqrt%7B16%20%2B%209%7D)
d = ![\sqrt{25}](https://tex.z-dn.net/?f=%5Csqrt%7B25%7D)
5
I can solve it in a different way if you want me to.
Answer:
![15.394\ \text{feet}](https://tex.z-dn.net/?f=15.394%5C%20%5Ctext%7Bfeet%7D)
Step-by-step explanation:
Let
be the length and width of the base of the prism and,
be the height of the prism.
Now height of the prism is 2 feet less than the length and width.
So ![y=x-2](https://tex.z-dn.net/?f=y%3Dx-2)
Volume of the the square prism is ![3174\ \text{ft}^3](https://tex.z-dn.net/?f=3174%5C%20%5Ctext%7Bft%7D%5E3)
![x^2y=3174](https://tex.z-dn.net/?f=x%5E2y%3D3174)
Now we graph both the functions.
The point at which they intersect each other is ![(15.394,13.394)](https://tex.z-dn.net/?f=%2815.394%2C13.394%29)
So, the length of the prism is
.