The calendar obviously has an integral number of years and months in 400 years. If it has an integral number of weeks, then it will repeat itself after that time. The rules of the calendar eliminate a leap year in 3 out of the four century years, so there are 97 leap years in 400 years. The number of excess days of the week in 400 years can be found by ...
(303·365) mod 7 + (97·366) mod 7 = (2·1 + 6·2) mod 7 = 14 mod 7 = 0
Thus, there are also an integral number of weeks in 400 years.
The first day of the week is the same at the start of every 400-year interval, so the calendar repeats every 400 years.
The equation of a hyperbola is:
(x – h)^2 / a^2 - (y – k)^2 / b^2 = 1
So what we have to do is to look for the values of the variables:
<span>For the given hyperbola : center (h, k) = (0, 0)
a = 3(distance from center to vertices)
a^2 = 9</span>
<span>
c = 7 (distance from center to vertices; given from the foci)
c^2 = 49</span>
<span>By the hypotenuse formula:
c^2 = a^2 + b^2
b^2 = c^2 - a^2 </span>
<span>b^2 = 49 – 9</span>
<span>b^2 = 40
</span>
Therefore the equation of the hyperbola is:
<span>(x^2 / 9) – (y^2 / 40) = 1</span>
