Answer:
[tex]\sqrt{1625}[/tex] lies in between 40 and 41 .
Step-by-step explanation:
Given : Dierdra saw a pole that was supported by a wire as shown.
She calculated that the wire was [tex]\sqrt{1625}[/tex] feet long.
We have find between which two numbers is the value of [tex]\sqrt{1625}[/tex] located.
WE know, [tex](40)^2=1600[/tex] and [tex](41)^2=1681[/tex]
Let's find the perfect squares that are closest to 1625.
Since, 1681 lies in between 1600 and 1681 which are perfect squares of 40 and 41 respectively.
Thus, [tex]\sqrt{1625}[/tex] lies in between 40 and 41 .
