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rewona [7]
3 years ago
15

Please answer ASAP

Mathematics
2 answers:
Kay [80]3 years ago
6 0

Problem 1.

My thinking is that the furthest you can get is have two points at each opposite corner, so the distance between them is sqrt(2). If we have two other points with this property, then all four corners are filled up. It is possible to pick two points where the distance is 1 unit.

Then a fifth point can be placed at the center such that the distance from it to any of the corners is sqrt(2)/2. We placed the fifth point at the center to try to get as far away as possible from the other four points.

Basically we're trying to find the worst case scenario (leading to the largest distance possible) and seeing how we can fill up the square. This establishes the upper bound. Any other kind of scenario will have a distance less than the upper bound.

===================================================

Problem 2.

For this one, I'm not sure what to make of it. The terminology is a bit strange so I'm not going to be fairly helpful here. Sorry about that.

If I had to guess, I'd assume it has something to do with the fact that a plane is uniquely defined by 3 points. That fourth point is not coplanar with the other three, which helps define the semi-spherical portion. The fifth point is just extra. The points can't be all collinear or else a plane won't form. Though to be honest, I'm still not sure about problem 2. I'd get a second opinion.

Rudik [331]3 years ago
3 0
The answer above is correct
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Answer:

\frac{x-3}{2x-3}. hole or removable discontinuity at x=2

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Well generally if you want the simplest form, you factor each the denominator and numerator and then see if you can cancel any of the factors out (because they're in the denominator and numerator)

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Rewrite equation:

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\frac{(x-2)(x-3)}{(2x-3)(x-2)}.

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\frac{x-3}{2x-3}

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90

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when you add all these you get 375

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375 ÷ 60 = 6.15

if he starts at 9 am and he cleans for 6.15 hours he will be finished by 3:15pm.

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2 years ago
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7 0
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