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1 year ago

I am offering another 100 points

Mathematics

2 answers:

Ivahew [28]1 year ago

6 0

Answer:

$\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.

iragen [17]1 year ago

5 0

Answer:8 and 9, since square root of 64 is 8 and square root of 81 is 9, it is known that square root of 75 is between 8 and 9

Step-by-step explanation:

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3 years ago

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Salsk061 [2.6K]

Answer:

a. Mean

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3 years ago

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attashe74 [19]

The sequence is geometric, so

$a_n = r a_{n-1}$

for some constant r. From this rule, it follows that

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Now, by substitution we have

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and so on down to (D)

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(notice how the exponent on r and the subscript on a add up to n)

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2 years ago