Calculating the Year Using the Gregorian Calendar

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1. So the question for one of the problems(My teacher made up the problem which is divided into parts but I just had a question in one of them) So the problem is let h = 3+(the last digit of the year in which you were born). Use the conventional Western (Gregorian) calendar to determine the year.


3. I tried looking it up and it seemed the year doesn't change for the gregorian calender. In my case my birthday is in 1994. So would it just be h = 3+4 = 7. So h = 7?
 
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hungrymouth said:
1. So the question for one of the problems(My teacher made up the problem which is divided into parts but I just had a question in one of them) So the problem is let h = 3+(the last digit of the year in which you were born). Use the conventional Western (Gregorian) calendar to determine the year.


3. I tried looking it up and it seemed the year doesn't change for the gregorian calender. In my case my birthday is in 1994. So would it just be h = 3+4 = 7. So h = 7?

I can't see anything wrong with that. But why is this a calculus question?
 
Well it's because that value h will be used along with value k and z to calculate the sphere of an equation and using to state whether the spheres intersect each plane. So my h value has to be right. I just wanted to verify if I was doing it correctly. It's vector calculus.
 
hungrymouth said:
Well it's because that value h will be used along with value k and z to calculate the sphere of an equation and using to state whether the spheres intersect each plane. So my h value has to be right. I just wanted to verify if I was doing it correctly. It's vector calculus.

Oh, so it's just a trick to make sure not all of the students get exactly the same question?
 
Yep that's basically it!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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