Twenty emails in 5 minutes

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.)

Now, if we go up to ax³+bx²+cx+d=0, a closed form for “x=” is possible to find, although it’s much bulkier than the quadratic version. It’s also possible, yet ugly, to do this for degree 4 polynomials ax⁴+bx³+cx²+dx+f=0.

Remember the quadratic formula? Given ax²+bx+c=0, the solution is x=(-b±√(b^2-4ac))/(2a), which may have felt arduous to memorize in high school, but you have to admit is a conveniently closed-form solution.

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.

all the angles opposite to the angle 150 would be 150 and all the angles that are next to 150 are 30 because they equal 180 it you add it up.

step-by-step explanation:

on monday rodney ate 1/4th of the sandwich = 3/12

on tuesday he ate 1/3rd of the sandwich = 4/12

therefore in total he has eaten = 3/12 + 4/12 = 7/12

thus we can say that he has eaten 7/12 of the sandwich on monday and tuesday alltogether.

happy to

pls mark as brainliest.