I believe it's not a function because the input (x) 4 has two outputs(y), 2 and 8.
The unit normal for the given plane is <5,2,-1>.
The equation of the plane parallel to the given plane passing through (5,5,4) is therefore
5(x-5)+2(y-5)-1(z-4)=0
simplify =>
5x+2y-z=25+10-4=31
Answer: the plane through (5,5,4) parallel to 5x+2y-z=-6 is 5x+2y-z=31
I think that these ordered pairs would be the answer (-2,-1).
Since they are right triangles use Pythagorean Theorem.
a^2 + b^2 = c^2 where c is the hypotenuse.
a. 12^2 + b^2 = 13^2
144 + b^2 = 169
subtract 144 from both sides
b^2 = 25
take the square root of both sides
t = 5 This is also know as a Pythagorean Triple 5-12-13
b. a^2 + 9^2 = 12^2
a^2 + 81 = 144
subtract 81 from both sides
a^2 = 63
take the square root of both sides
a = √63
a = √(9 * 7)
a = 3√7
c. 6^2 + 9^2 = c^2
36 + 81 = c^2
117 = c^2
take the square root of each side
c = √117
c = √(9*13)
x = 3√13
