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

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

For the problem 9+2(3) do you get the final answer if you had first as you do if you multiply first

mel-nik [20]

Answer:

Step-by-step explanation:

you have to multiply first because of the order of operations, so 2 × 3 = 6, and 6 + 9 = 15

6 0

3 years ago

Read 2 more answers

The percent increase of the airline ticket is ___%

JulsSmile [24]

Answer:

Step-by-step explanation:

%change=100(final-initial)/(initial)

For just the ticket

%change=100(375-300)/300=25%

For the ticket and baggage fee of $50

%change=100(375+50-300)/300=42%

7 0

3 years ago