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Katyanochek1 [597]
3 years ago
11

Please help me with these. These are so hard.​

Mathematics
1 answer:
LuckyWell [14K]3 years ago
4 0

\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.

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The correct answer is:  [B]:  "40 yd² " .
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First, find the area of the triangle:

The formula of the area of a triangle, "A":

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in which:  " A = area (in units 'squared') ;  in our case, " yd² " ; 

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→  A = (1/2) * b * h = (1/2) * (6 yd) * (8 yd) = (1/2) * (6) * (8) * (yd²) ; 
                                                                 
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Now, find the area, "A", of the square:

The formula for the area, "A" of a square:

   A = s² ;

in which:  "A = area (in "units squared") ; in our case, " yd² " ;
                 
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Now, we add the areas of BOTH the triangle AND the square:
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        →  " 24 yd²  +  16 yd² " ; 

to get:  " 40 yd² " ;  which is:  Answer choice:  [B]:  " 40 yd² " .
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