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a) Compute x+y and \(\sqrt{x^2+y^2}\) when x=5 and  y=12

b) When is

\(x+y=\sqrt{x^2+y^2}?\)
When is

\(x+y\neq\sqrt{x^2+y^2}\)
 

 Mar 15, 2019
 #1
avatar+3994 
+4

(a) If you know the Pythagorean triple (5, 12, 13), the answer is 13. Thus, \(\sqrt{25+144}=\sqrt{169}=\boxed{13}.\)

.
 
 Mar 15, 2019
 #2
avatar+201 
+2

b. the first part is one, the second part is any number aside from one.

 
 Mar 15, 2019
 #3
avatar+98061 
+2

a) x + y  = 5 + 12 = 17

Tertre's answer for  sqrt (x^2 + y^2)  is good

 

b)    x + y =  sqrt (x^2 + y^2)        square both sides

 

x^2 + 2xy + y^2  = x^2 + y^2

 

2xy  = 0

 

xy = 0

 

So....this is true when either x or y   (or both)   = 0

 

c)     this will not be true   when    xy   ≠  0

 

 

cool  cool  cool

 
 Mar 15, 2019

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