Answer:
158.4 pounds
Step-by-step explanation:
first find the volume of the tank
![V = 30 \times 30 \times 80](https://tex.z-dn.net/?f=V%20%3D%2030%20%5Ctimes%2030%20%5Ctimes%2080)
![V = 72000 cm^3](https://tex.z-dn.net/?f=V%20%3D%2072000%20cm%5E3)
to convert this to m^3 we know that 1m = 100cm
![V = 72000 cm^3 \times \left(\dfrac{1\,\text{m}}{100\,\text{cm}}\right)^3](https://tex.z-dn.net/?f=V%20%3D%2072000%20cm%5E3%20%5Ctimes%20%5Cleft%28%5Cdfrac%7B1%5C%2C%5Ctext%7Bm%7D%7D%7B100%5C%2C%5Ctext%7Bcm%7D%7D%5Cright%29%5E3)
![V = 72000 cm^3 \times \left(\dfrac{1\,\text{m}^3}{100^3\,\text{cm}^3}\right)](https://tex.z-dn.net/?f=V%20%3D%2072000%20cm%5E3%20%5Ctimes%20%5Cleft%28%5Cdfrac%7B1%5C%2C%5Ctext%7Bm%7D%5E3%7D%7B100%5E3%5C%2C%5Ctext%7Bcm%7D%5E3%7D%5Cright%29)
the cm^3 cancel out.
![V = \dfrac{72000}{100^3} m^3](https://tex.z-dn.net/?f=V%20%3D%20%5Cdfrac%7B72000%7D%7B100%5E3%7D%20m%5E3)
![V = 0.072\,m^3](https://tex.z-dn.net/?f=V%20%3D%200.072%5C%2Cm%5E3)
by simply converting the side-lengths to meters beforehand can give you this answer directly:
![V = 0.3 \times 0.3 \times 0.8 = 0.072 m^3](https://tex.z-dn.net/?f=V%20%3D%200.3%20%5Ctimes%200.3%20%5Ctimes%200.8%20%3D%200.072%20m%5E3)
it is given that 1 meter cube of water weighs about 2200 pounds.
![1\,m^3 = 2200 \,\text{lb}](https://tex.z-dn.net/?f=1%5C%2Cm%5E3%20%3D%202200%20%5C%2C%5Ctext%7Blb%7D)
multiply both sides with 0.072 to find our answer:
![1\,m^3 \times 0.072 = 2200 \times 0.072 \,\text{lb}](https://tex.z-dn.net/?f=1%5C%2Cm%5E3%20%5Ctimes%200.072%20%3D%202200%20%5Ctimes%200.072%20%5C%2C%5Ctext%7Blb%7D)
![0.072\,m^3 = 158.4 \,\text{lb}](https://tex.z-dn.net/?f=0.072%5C%2Cm%5E3%20%3D%20158.4%20%5C%2C%5Ctext%7Blb%7D)
hence the weight of the water in the tank would be 158.4 pounds!
Well first you do 7+5 to equal 12, and then you multiply 12 by 6 and get 72, then you add the 3 to get 75. 6(7+5)+3=75. Hope this helps!!
if the diameter is 20, the its radius must be half that or 10.
![\textit{area of a sector of a circle}\\\\ A=\cfrac{\theta \pi r^2}{360}~~ \begin{cases} r=radius\\ \theta =\stackrel{degrees}{angle}\\[-0.5em] \hrulefill\\ A=5\pi \\ r=10 \end{cases}\implies \begin{array}{llll} 5\pi =\cfrac{\theta \pi (10)^2}{360}\implies 5\pi =\cfrac{5\pi \theta }{18} \\\\\\ \cfrac{5\pi }{5\pi }=\cfrac{\theta }{18}\implies 1=\cfrac{\theta }{18}\implies 18=\theta \end{array}](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20sector%20of%20a%20circle%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7B%5Ctheta%20%5Cpi%20r%5E2%7D%7B360%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%20%5Ctheta%20%3D%5Cstackrel%7Bdegrees%7D%7Bangle%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20A%3D5%5Cpi%20%5C%5C%20r%3D10%20%5Cend%7Bcases%7D%5Cimplies%20%5Cbegin%7Barray%7D%7Bllll%7D%205%5Cpi%20%3D%5Ccfrac%7B%5Ctheta%20%5Cpi%20%2810%29%5E2%7D%7B360%7D%5Cimplies%205%5Cpi%20%3D%5Ccfrac%7B5%5Cpi%20%5Ctheta%20%7D%7B18%7D%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B5%5Cpi%20%7D%7B5%5Cpi%20%7D%3D%5Ccfrac%7B%5Ctheta%20%7D%7B18%7D%5Cimplies%201%3D%5Ccfrac%7B%5Ctheta%20%7D%7B18%7D%5Cimplies%2018%3D%5Ctheta%20%5Cend%7Barray%7D)
The answer to your question is -1 < y < 3
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False. Intrinsic motivation is a reflection that needs to do something.