Given:

Two numbers are [tex]\dfrac{9}{7}[/tex]  and [tex]\dfrac{10}{7}[/tex].

To find:

A repeating decimal that is between [tex]\dfrac{9}{7}[/tex]  and [tex]\dfrac{10}{7}[/tex].

Solution:

Using calculator, we get

[tex]\dfrac{9}{7}=1.285714285714…[/tex]

[tex]\dfrac{9}{7}=1.\overline{285714}[/tex]

and,

[tex]\dfrac{10}{7}=1.42857142857
1…[/tex]

[tex]\dfrac{10}{7}=1.\overline{428571}[/tex]

Now, the repeating decimal that is between [tex]\dfrac{9}{7}[/tex]  and [tex]\dfrac{10}{7}[/tex] be x. So,

[tex]1.\overline{285714}<x<1.\overline{428571}[/tex]

On analyzing the numbers to hundredth places, we get 1.28 < 1.33 < 1.42, therefore

[tex]1.\overline{285714}<1.3333…<1.\overline{428571}[/tex]

[tex]1.\overline{285714}<1.\overline{3}<1.\overline{428571}[/tex]

And we know that [tex]1.\overline{3}[/tex] is the decimal form of [tex]\dfrac{4}{3}[/tex].

Therefore, the required number is [tex]1.\overline{3}[/tex] .