Tanya, you like triangles.
a = sqrt(c^2-b^2);
test_1 a=sqrt(3) is longer than b=1
the good solution would be
a= min(sqrt(c^2 -b^2),b)
Kudos for finding this solution - it did indeed meet all the test cases, but wasn't quite what I had in mind! I will add a new test.
Reverse the vector
middleAsColumn: Return all but first and last element as a column vector
03 - Matrix Variables 2
Test Problem; Create a 5x5 array containing all ones
Calculate Amount of Cake Frosting
Side of a rhombus
Find a Pythagorean triple
Is this triangle right-angled?
Length of the hypotenuse
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