Acooling tower for a nuclear reactor is to be constructed in the shape of a hyperboloid of one sheet. the diameter at the base is 300 m and the minimum diameter, 500 m above the base, is 200 m. find an equation describing the shape of the tower in the coordinates where the origin is at the center of the narrowest part of the tower. in particular, use coordinates where the origin is 500 m above the ground.

Okay so the equation you need to use is (x2/a2)+(y2/b2)-(z2/c2)=1

So an equation describing the shape of the tower in the coordinates where the origin is at the center of the narrowest part of the tower would be:

x^2 / 100^2 + y^2 / 100^2 – z^2 / 2 * 100^2 = 1

I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

step-by-step explanation:

12: 00 p.m. to 7: 00 p.m. is a total of 7 hours.

4: 00 p.m. to 7: 00 p.m. is a total of 3 hours.

this means that from 4: 00 p.m to 7: 00 p.m. this represents 3/7 of what was lost from 12: 00 p.m. to 7: 00 p.m.

after knowing this, after doing 11,600 - 6,800 = 4,800

4,800 represents 3/7 of what was lost for 7 hours.

we then divide 4,800 by 3 which gets 1,600.

because we know that 1,600 represents 1/7 of how much water was lost, we just have to multiply by 7, which gets 11,200.   after adding 11,200 to how much water was left at 7: 00 p.m. which was 6,800. 11,200 + 6,800 = 18,000

this means that the pool originally had 18,000 gallons in the pool.

you can also check your work by doing 18,000 - 4(1600), which becomes           1800 - 6400 which gets you 11,600.