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10 answers

Yes, because they satisfy the pythagorean theorem:
a^2=b^2+c^2, a=13, b=12, c=5, a being the hypotenuse (the side below the (right) angle). In fact, they can only represent a right triangle (the theorem works both ways).

2007-01-10 08:24:22 · answer #1 · answered by supersonic332003 7 · 2 0

If the lenght of two of the three factors have been any of the three numders you may make authentic triangles. you will have a real triangle with a base of 12 and altitude of 8 however the 0.33 section could be approximately 14.5 or so. this could be a consequence of a regulation reguarding authentic attitude triangles which any grade pupil would desire to understand.

2016-10-30 13:41:48 · answer #2 · answered by ? 4 · 0 0

Yes, because 5² + 12² = 169 = 13².

2007-01-10 08:51:24 · answer #3 · answered by steiner1745 7 · 0 0

Yes, the three integers form a Pythagorean triple. 5^2 + 12^2 = 13^2.

2007-01-10 08:23:28 · answer #4 · answered by NietzcheanCowboy 3 · 1 0

Really, krystal r!! You already asked two of the same type of question and were given clearly explained answers. Are you trying to learn anything from the answers, or just getting homework answers written down?

2007-01-10 08:26:10 · answer #5 · answered by Hy 7 · 0 0

Any chance you might try to do a couple of these on your own? I think every time you have asked the question someone explains how to get the answer. Why not and try to use what they have told you!!!!! LAZY

2007-01-10 08:26:53 · answer #6 · answered by Ray 5 · 1 0

why don't you work out the math problem, justify your answer and then tell all of us the answer? Nice try, but stop trying to get people to do your homework

2007-01-10 08:25:12 · answer #7 · answered by karma 7 · 1 0

Could krystalr do her own homework? Justify your answer.

2007-01-10 08:25:16 · answer #8 · answered by bequalming 5 · 1 0

Surely you must know how to do these problems on your own by now.

2007-01-10 08:24:43 · answer #9 · answered by Rebecca P 2 · 0 0

they do

2007-01-10 08:23:31 · answer #10 · answered by Bob H 1 · 1 0

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