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

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

What does it mean to have diagonals bisect each other?

Degger [83]

Answer:

I think your answer would be A

Step-by-step explanation:

8 0

2 years ago

If f(x) = 2x2 + 2 and g(x) = x2 - 1, find (f - g)(x)I

erastovalidia [21]

9514 1404 393

Answer:

C. x^2 + 3

Step-by-step explanation:

Substitute for f(x) and g(x) and simplify.

(f -g)(x) = f(x) -g(x)

(f -g)(x) = (2x^2 +2) -(x^2 -1) = 2x^2 +2 -x^2 +1

(f -g)(x) = x^2 +3

4 0

3 years ago

HELPPPP ME PLEASE ITS THE LAST QUESTION

Vitek1552 [10]

Answer:

first one 5/8

Step-by-step explanation:

2 1/2 / 4 = 5/8

3 0

2 years ago