Answer: No, we don't have a right triangle
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Explanation:
If a triangle with sides a,b,c makes the equation a^2+b^2 = c^2 true, where c is the longest side, then this triangle is a right triangle. This is the converse of the pythagorean theorem.
Here we have a = 2, b = 5 and c = 7.
So...
a^2+b^2 = c^2
2^2+5^2 = 7^2
4+25 = 49
29 = 49
The last equation is false, so the first equation is false for those a,b,c values. Therefore, we do <u>not</u> have a right triangle.
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In contrast, consider the classic 3-4-5 right triangle
a = 3, b = 4 and c = 5 would make a^2+b^2 = c^2 true because 3^2+4^2 = 5^2 is a true equation (both sides lead to 25).
3.14159265358979 is the first ten digits of pi.
Answer:
your answer would be -135
Find the critical points of 
:
All three points lie within 
, and 
takes on values of
Now check for extrema on the boundary of . Convert to polar coordinates:
Find the critical points of 
:
where 
is any integer. There are some redundant critical points, so we'll just consider 
, which gives
which gives values of
So altogether, has an absolute maximum of 65/16 at the points (0, -1/2) and (0, 1/2), and an absolute minimum of 3 at (-1, 0).
