(Sin x)/ n = six = 6
It has two variables. How are you supposed to solve it?
I already solved it. The problem was originally (Sin x)/n and was meant to be a joke, since cancelling the n on the top and bottom would create six.
(Sin x)/ n = six = 6
It has two variables. How are you supposed to solve it?
I already solved it. The problem was originally (Sin x)/n and was meant to be a joke, since cancelling the n on the top and bottom would create six.
Let cup be a function of beverages.
1cup(coffee) = happiness
2cup(lemonade) = thirst quencher
3cup(pure H20) = no solution
playing monopoly = not my cup(tea)
Your cup(tea) = ?
Playing monopoly?
This one is a joke too. I said one cup of coffee equals happiness, ...., playing monopoly is not my cup of tea. And then I asked, what is your cup of tea?
Pamela Anderson weighs 69 pounds. That is too, too, too much. She wants to weigh 51 pounds. So she goes to Dr. X, who gives her 8 different medications, and then becomes _______________.
its the side side side theorem
has something to do with triangle similarities
i forgot what exactly
Find the area and perimeter. Also, apologies for the fact that it is turned sideways. Consider it an extra layer of difficulty.
Prove that the square root of two cannot be expressed as the ratio between two whole numbers.
Ah, this beauty. I’ll use the “classic way” of doing it.
Lets day that we were wrong all along, and root(2) is rational. Then it must be a fraction of two natural numbers, a/b. Now let’s make sure that they are in simplest form, no need for 8/16 when we can say 1/2. So, we have:
root(2) = a/b, a fraction in simplest form
Algebra time:
2 = a^2/b^2
a^2 = 2(b^2)
So we see a^2 is an even integer. This also means that a itself is an even integer. So, a = 2k for some integer k. Substituting and continuing onward gives us:
(2k)^2=2(b^2)
4(k^2)=2(b^2)
b^2=2(k^2)
Which show b^2 to be an even integer, making b even.
But wait! If a is even and b is even, that means the fraction is not in simplest form, in contradiction to our original assumption. Thus root(2) cannot be a/b or a rational number, therefor it is irrational.
I see. Well, then the number is already in base 12.