I believe that the answer is 1845
the surface area of the cube is 1350
the surface area of the pyramid is 495
the total surface area is :
![1350 + 495 = 1845](https://tex.z-dn.net/?f=1350%20%2B%20495%20%3D%201845%20)
good luck
Answer:
Approximately 67.348 litres can be put in six hemispherical bowl with a diameter of 35 centimetres.
Step-by-step explanation:
The volume of a hemisphere (
), measured in cubic centimetres, is obtained from this formula:
![V = \frac{2\pi}{3}\cdot R^{3}](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B2%5Cpi%7D%7B3%7D%5Ccdot%20R%5E%7B3%7D)
Where
is the radius of the hemisphere, measured in centimetres.
We know that radius is the half of the diameter (
), measured in centimetres, then:
![R = 0.5\cdot D](https://tex.z-dn.net/?f=R%20%3D%200.5%5Ccdot%20D)
(
)
![R = 0.5\cdot (35\,cm)](https://tex.z-dn.net/?f=R%20%3D%200.5%5Ccdot%20%2835%5C%2Ccm%29)
![R = 17.5\,cm](https://tex.z-dn.net/?f=R%20%3D%2017.5%5C%2Ccm)
Now, we get the volume of each hemispherical bowl:
![V = \frac{2\pi}{3} \cdot (17.5\,cm)^{3}](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B2%5Cpi%7D%7B3%7D%20%5Ccdot%20%2817.5%5C%2Ccm%29%5E%7B3%7D)
![V \approx 11224.649\,cm^{3}](https://tex.z-dn.net/?f=V%20%5Capprox%2011224.649%5C%2Ccm%5E%7B3%7D)
The total volume of six hemispherical bowl is:
![V_{T} = 6\cdot V](https://tex.z-dn.net/?f=V_%7BT%7D%20%3D%206%5Ccdot%20V)
![V_{T}= 6\cdot (11224.649\,cm^{3})](https://tex.z-dn.net/?f=V_%7BT%7D%3D%206%5Ccdot%20%2811224.649%5C%2Ccm%5E%7B3%7D%29)
![V_{T} = 67347.894\,cm^{3}](https://tex.z-dn.net/?f=V_%7BT%7D%20%3D%2067347.894%5C%2Ccm%5E%7B3%7D)
From Physics we know that 1 litre equals 1000 cubic centimetres. Then:
![V_{T} = 67.348\,L](https://tex.z-dn.net/?f=V_%7BT%7D%20%3D%2067.348%5C%2CL)
Approximately 67.348 litres can be put in six hemispherical bowl with a diameter of 35 centimetres.
Answer:
![2xy^2](https://tex.z-dn.net/?f=2xy%5E2)
Step-by-step explanation:
The question is:
![\frac{6x^{2}y^3}{3xy}](https://tex.z-dn.net/?f=%5Cfrac%7B6x%5E%7B2%7Dy%5E3%7D%7B3xy%7D)
<em>"6" and "3" cancels out as well as the "x"s and "y"s. Shown below</em>:
![\frac{6x^{2}y^3}{3xy}\\=2xy^2](https://tex.z-dn.net/?f=%5Cfrac%7B6x%5E%7B2%7Dy%5E3%7D%7B3xy%7D%5C%5C%3D2xy%5E2)
Base in your question the coordinate is P is 2, PQ has a coordinate of 8 while PR has a coordinate of 12. To find the mid point of each segment you must first find the coordinate of QR which is from 8-12. So the answer will be
PQ = 5, PR = 6, and QR= 10
The answer is y=3x-6. (B). First you subtract 6x from both sides. Then you get -2y=12-6x. Now, you divide both sides by -2. This gets you the final answer; y=3x-6 or y=-6+3x.