. Use the process outlined in the lesson to approximate the number 3√10. Use the approximation √10 ≈3162 277 7.
. Find a sequence of five intervals that contain 3√10 whose endpoints get successively closer to 3√10. Write your
iintervals in the form 3^???? < 3^√10 < 3^s for rational numbers ????and s.

Respuesta :

Answer:

sequence of five intervals

(1) 3³  < [tex]3^{\sqrt{10} }[/tex]   < [tex]3^{4}[/tex]

(2) [tex]3^{3.1}[/tex]  < [tex]3^{\sqrt{10} }[/tex]   < [tex]3^{3.2}[/tex]

(3) [tex]3^{3.16}[/tex] < [tex]3^{\sqrt{10} }[/tex]   < [tex]3^{3.17}[/tex]

(4) [tex]3^{3.162}[/tex] < [tex]3^{\sqrt{10} }[/tex]   < [tex]3^{3.163}[/tex]

(5) [tex]3^{3.1622}[/tex]  < [tex]3^{\sqrt{10} }[/tex]   < [tex]3^{3.1623}[/tex]

Step-by-step explanation:

as per question given data      

√10 ≈ 3.162 277 7    

to find out      

sequence of five intervals

solution      

as we have given that √10 value that is here

√10 ≈ 3.162 277 7           ........................1

so  

when we find [tex]3^{\sqrt{10} }[/tex]           ................2

put here √10 value in equation number  2  

we get  [tex]3^{\sqrt{10} }[/tex]   that is  32.27    

so    

sequence of five intervals

(1) 3³  < [tex]3^{\sqrt{10} }[/tex]   < [tex]3^{4}[/tex]

(2) [tex]3^{3.1}[/tex]  < [tex]3^{\sqrt{10} }[/tex]   < [tex]3^{3.2}[/tex]

(3) [tex]3^{3.16}[/tex] < [tex]3^{\sqrt{10} }[/tex]   < [tex]3^{3.17}[/tex]

(4) [tex]3^{3.162}[/tex] < [tex]3^{\sqrt{10} }[/tex]   < [tex]3^{3.163}[/tex]

(5) [tex]3^{3.1622}[/tex]  < [tex]3^{\sqrt{10} }[/tex]   < [tex]3^{3.1623}[/tex]