SOLUTION: What are all the possible values for x using the pythagorean theorum of a right triangle when the hypotenuse is 27 and one leg is (2x+5)

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Question 1016498: What are all the possible values for x using the pythagorean theorum of a right triangle when the hypotenuse is 27 and one leg is (2x+5)
Found 2 solutions by ikleyn, macston:
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
What are all the possible values for x using the pythagorean theorum of a right triangle when the hypotenuse is 27 and one leg is (2x+5)
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The second leg is sqrt%2827%5E2+-+%282x%2B5%29%5E2%29.


The only restriction for x is this inequality:

2x + 5 < 27,  

which says that the leg is shorter than the hypotenuse.

The last inequality means that x < 11.

Is it what you want to know?


Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
.
0 < 2x+5 < 27
-5 < 2x < 22
-2.5 < x < 11
.