$\sf Since \;\sqrt{\boxed{64}}=\boxed{8}\;and\;\sqrt{\boxed{81}}=\boxed{9}\; \textsf{it is known that $\sqrt{75}$ is between}\\\\\sf \boxed{8}\;and\;\boxed{9}\;.$

Step-by-step explanation:

Perfect squares: 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, ...

To find $\sf \sqrt{75}$ , identify the perfect squares immediately before and after 75:

64 and 81

$\begin{aligned}\sf As\;\; 64 < 75 < 81\; & \implies \sf \sqrt{64} < \sqrt{75} < \sqrt{81}\\&\implies \sf \;\;\;\;\;8 < \sqrt{75} < 9 \end{aligned}$

$\sf Since \;\sqrt{\boxed{64}}=\boxed{8}\;and\;\sqrt{\boxed{81}}=\boxed{9}\; \textsf{it is known that $\sqrt{75}$ is between}\\\\\sf \boxed{8}\;and\;\boxed{9}\;.$

See the attachment for the correct placement of $\sf \sqrt{75}$ on the number line.

6 0

1 year ago

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Can I have some help on these two math questions?

BigorU [14]

8. 1,728 i did 12×18÷2=108×32=345÷2= 1,728

8 0

3 years ago