+0  
 
0
305
2
avatar+280 

\(The function h(x) is defined as: h(x) = \left\{ \begin{array}{cl} \lfloor 4x \rfloor & \text{if } x \le \pi, \\ 3-x & \text{if }\pi < x \le 5.2, \\ x^2& \text{if }5.2< x. \end{array}\right. Find h(h(\sqrt{2})).\)

 

I got 32 when I tried to solve this but it was incorrect.. ):

 May 6, 2019
 #1
avatar+25556 
+3

I got 32 when I tried to solve this but it was incorrect.. ):

 

\(\text{The function $h(x)$ is defined as: $h(x) = \left\{ \begin{array}{cl} \lfloor 4x \rfloor & \text{if } x \le \pi, \\ 3-x & \text{if }\pi < x \le 5.2, \\ x^2& \text{if }5.2< x. \end{array}\right. $}\\ \text{Find $h\Big(h\left(\sqrt{2} \right) \Big)$.} \)

 

\(\begin{array}{|rcll|} \hline h\left(\sqrt{2} \right) &=& \lfloor 4\cdot \sqrt{2} \rfloor \\ &=& \lfloor 4\cdot 1.41421356237 \rfloor \\ &=& \lfloor 5.65685424949 \rfloor \\ &=& 5 \\\\ h(5) &=& 3-5 \\ &=& -2 \\\\ \mathbf{h\Big(h\left(\sqrt{2} \right) \Big)} & =& \mathbf{ -2 } \\ \hline \end{array} \)

 

laugh

 May 6, 2019
 #2
avatar+280 
+2

Thanks!

 May 6, 2019

15 Online Users

avatar
avatar