$\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

Read 2 more answers

Solve for x -9<4x+3<27 enter your answer as one inequality in the box

ludmilkaskok [199]

If a< c< b then a<c and c<b

Separate the equation into 2 separate ones and solve them:

X-9 < 4x +3

Subtract 3 from both sides:

X-12 < 4x

Subtract x from both sides:

-12< 3x

Divide both sides by 3:

X > -4

4x+3 < 27

Subtract 3 from both sides

4x < 24

Divide both sides by 4

X <6

Combine to get one inequality:

-4<x<6

6 0

3 years ago

(1/p-p)(1/p-q)-q/p^2(1/q+qp^2)​

(1/p-p)(1/p-q)-q/p^2(1/q+qp^2)