$\bf \textit{difference and sum of cubes} \\\\ a^3+b^3 = (a+b)(a^2-ab+b^2) ~\hfill a^3-b^3 = (a-b)(a^2+ab+b^2) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \boxed{a^6+b^6}\implies a^{2\cdot 3}+b^{2\cdot 3}\implies (a^2)^3+(b^2)^3 \\[2em] [a^2+b^2] [(a^2)^2-a^2b^2+(b^2)^2]\implies \boxed{(a^2+b^2)(a^4-a^2b^2+b^4)}$

about the second one... well, is a "fait accompli" that using the pythagorean theorem, if x = 8 and y = 5, the hypotenuse must be √(8² + 5²) = √(89), which is neither of those choices.

5, 8, 13 are no dice, namely 5² + 8² ≠ 13

25, 64, 17 is are no dice too, because 25² + 17² ≠ 64²

however, 5,12 and 13 are indeed a pythagorean triple

also is 39, 80, 89.

when looking for a pythagorean triple, recall that c² = a² + b².

so the longest leg is the sum of the square of the small ones.

so what you'd do is, check the small legs, square them, add them up, if they're indeed a pythagorean triple, they "must" add up to the longest leg.

4 0

3 years ago

Y+7=9/10(x+9) in standard form

sammy [17]

Alright,

-9/10x + y = 1 1/10

8 0

3 years ago