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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
\[x+y\neq\sqrt{x^2+y^2}?\]

 Jul 12, 2015

Best Answer 

 #1
avatar+26364 
+6

$$\\ a) \text{
Compute x+y and} \sqrt{x^2+y^2} \text{ when } x=5 \text{ and } y=12\\\\
b) \text{ When is } \left[x+y=\sqrt{x^2+y^2}~?\right]\\
\text{ When is } \left[x+y\neq\sqrt{x^2+y^2}~?\right]$$

a)

$$\small{\text{$
\begin{array}{rcl}
x+y &=& 5 + 12\\
x+y &=& 17 \\\\
\sqrt{x^2+y^2}&=& \sqrt{5^2+12^2}\\
\sqrt{x^2+y^2}&=& \sqrt{25+144}\\
\sqrt{x^2+y^2}&=& \sqrt{169}\\
\sqrt{x^2+y^2}&=& \sqrt{13^2}\\
\sqrt{x^2+y^2}&=& 13\\
\end{array}
$}}$$

 

b)

$$x+y \stackrel{\textcolor[rgb]{1,0,0}{?}} = \sqrt{x^2+y^2} \qquad \text{ squaring } \\\\
(x+y)^2 \stackrel{\textcolor[rgb]{1,0,0}{?}} = x^2+y^2\\\\
x^2+2xy+y^2 \stackrel{\textcolor[rgb]{1,0,0}{?}}= x^2+y^2 \\\\
\boxed{~~
x^2+\underbrace{\textcolor[rgb]{1,0,0}{2xy}}_{=0}+y^2 = x^2+y^2
~~}\\\\
\boxed{~~
x^2+\underbrace{\textcolor[rgb]{1,0,0}{2xy}}_{ \neq0}+y^2 \neq x^2+y^2
~~}$$

 

.
 Jul 13, 2015
 #1
avatar+26364 
+6
Best Answer

$$\\ a) \text{
Compute x+y and} \sqrt{x^2+y^2} \text{ when } x=5 \text{ and } y=12\\\\
b) \text{ When is } \left[x+y=\sqrt{x^2+y^2}~?\right]\\
\text{ When is } \left[x+y\neq\sqrt{x^2+y^2}~?\right]$$

a)

$$\small{\text{$
\begin{array}{rcl}
x+y &=& 5 + 12\\
x+y &=& 17 \\\\
\sqrt{x^2+y^2}&=& \sqrt{5^2+12^2}\\
\sqrt{x^2+y^2}&=& \sqrt{25+144}\\
\sqrt{x^2+y^2}&=& \sqrt{169}\\
\sqrt{x^2+y^2}&=& \sqrt{13^2}\\
\sqrt{x^2+y^2}&=& 13\\
\end{array}
$}}$$

 

b)

$$x+y \stackrel{\textcolor[rgb]{1,0,0}{?}} = \sqrt{x^2+y^2} \qquad \text{ squaring } \\\\
(x+y)^2 \stackrel{\textcolor[rgb]{1,0,0}{?}} = x^2+y^2\\\\
x^2+2xy+y^2 \stackrel{\textcolor[rgb]{1,0,0}{?}}= x^2+y^2 \\\\
\boxed{~~
x^2+\underbrace{\textcolor[rgb]{1,0,0}{2xy}}_{=0}+y^2 = x^2+y^2
~~}\\\\
\boxed{~~
x^2+\underbrace{\textcolor[rgb]{1,0,0}{2xy}}_{ \neq0}+y^2 \neq x^2+y^2
~~}$$

 

heureka Jul 13, 2015

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