Answer:
Bet
Step-by-step explanation:
It’s a simple one to write. There are many trios of integers (x,y,z) that satisfy x²+y²=z². These are known as the Pythagorean Triples, like (3,4,5) and (5,12,13). Now, do any trios (x,y,z) satisfy x³+y³=z³? The answer is no, and that’s Fermat’s Last Theorem.
On the surface, it seems easy. Can you think of the integers for x, y, and z so that x³+y³+z³=8? Sure. One answer is x = 1, y = -1, and z = 2. But what about the integers for x, y, and z so that x³+y³+z³=42?
That turned out to be much harder—as in, no one was able to solve for those integers for 65 years until a supercomputer finally came up with the solution to 42. (For the record: x = -80538738812075974, y = 80435758145817515, and z = 12602123297335631. Obviously.)
I can’t see it it too blurry
Answer:
for this equation one factor is (x+6) and the other is (x-4)
Rewriting the left hand side,
csc²t - cost sec t
= (1/sin²t)-(cost)(1/cost)
= 1/sin²t - 1
= 1/sin²t - sin²t/sin²t
= (1-sin²t)/sin²t
= cos²t/sin²t
= cot²t
300÷60 = 5
5 hours of commercials was watched
