ANSWER
![x= 9 \: in](https://tex.z-dn.net/?f=%20x%3D%209%20%5C%3A%20in)
EXPLANATION
The volume of a rectangular prism is given by the formula,
![V = l \times b \times h](https://tex.z-dn.net/?f=V%20%3D%20l%20%5Ctimes%20b%20%5Ctimes%20h)
We substitute the dimensions to obtain,
![V = 6 \times 3 \times 12 \: {in}^{3}](https://tex.z-dn.net/?f=V%20%3D%206%20%5Ctimes%203%20%5Ctimes%2012%20%5C%3A%20%7Bin%7D%5E%7B3%7D%20)
The volume of the rectangular pyramid is,
![V = \frac{1}{3} \times 6 \times 12 \times x \: {in}^{3}](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B1%7D%7B3%7D%20%5Ctimes%206%20%5Ctimes%2012%20%5Ctimes%20x%20%5C%3A%20%7Bin%7D%5E%7B3%7D%20)
Equating the two volumes gives,
![\frac{1}{3} \times 6 \times 12 \times x = 6 \times 3 \times 12](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7D%20%5Ctimes%206%20%5Ctimes%2012%20%5Ctimes%20x%20%3D%206%20%5Ctimes%203%20%5Ctimes%2012)
We divide through by 6×12 to get,
![\frac{x}{3}=3](https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%7D%7B3%7D%3D3)
We solve for x to obtain,
![x = 3 \times 3](https://tex.z-dn.net/?f=x%20%3D%203%20%5Ctimes%203)
Answer:
x=2, y=21
Step-by-step explanation:
there are two methods: substiution and elimation. I will be doing subtistution in this soultion:
first isolate y in either equation, it would be easier to do it in the second one.
y= 11+5x
Then, subtitute that into the first equation
2(11+5x)+8x=58 (it says 2y+8x+58 but I assume you mean 2y+8x=58?)
22+10x+8x=58
18x = 36
x=2
substitute that back into y = 11+5x, to get y=11+10 = 21
The answer is A.First we substitute 8 into the given equation.
![s = 10 \sqrt{4(8)} +15](https://tex.z-dn.net/?f=s%20%3D%2010%20%5Csqrt%7B4%288%29%7D%20%2B15%20)
Multiply 8 and 5, then find the square root of 40. Finally multiply it by 10.
![s = 10 \sqrt{40} +15](https://tex.z-dn.net/?f=s%20%3D%2010%20%5Csqrt%7B40%7D%20%2B15%20)
![s = 63.245 +15](https://tex.z-dn.net/?f=s%20%3D%2063.245%20%2B15%20)
Then add 15.
And the answer will be 78.245.
Now the question says that s is in
thousands, so the real answer would be about 75,000.
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>.
A triangle is equal to 180. Using that, we can make a equation.
6x-3 + 3x+6 + 5x = 180
Combining the like terms, we get:
14x+3 = 180
Subtract 3 on both sides.
14x = 177
Finally, divide 177 by 14.
x = 177/14 or about 12.6