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IgorLugansk [536]
1 year ago
15

Determine if the given expression is written in simplestradical form.-V 150And why?

Mathematics
1 answer:
seropon [69]1 year ago
7 0
\text{Answer : -5}\sqrt[]{6}

Explanation

\begin{gathered} -\text{ }\sqrt[]{150} \\ \text{This can be express further by applying the laws of surd} \\ \sqrt[]{A\text{ x B}}\text{ = }\sqrt[]{A}\text{ x }\sqrt[]{B} \\ Step\text{ 1: Find the factor of 150 that will give one p}\operatorname{erf}ect\text{ square and non - p}\operatorname{erf}ect\text{ square} \\ Factor\text{ of 150 = 25 and 6} \\ \text{Threfore, }\sqrt[]{150}\text{ can be express as }\sqrt[]{25\text{ x 6}} \\ \sqrt[]{150\text{ }}\text{ = }\sqrt[]{25\text{ x 6}} \\ \sqrt[]{150\text{ }}\text{ = }\sqrt[]{25}\text{ x }\sqrt[]{6} \\ \sqrt[]{25}\text{ = 5} \\ -\sqrt[]{150}\text{ = -5}\sqrt[]{6} \\ \text{Therefore, the simplest form of -}\sqrt[]{150}\text{ is -5}\sqrt[]{6} \end{gathered}

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64, –48, 36, –27, ...<br><br> Which formula can be used to describe the sequence?
nordsb [41]

Answer:

\boxed{a_n \:  =  \: 64 \:  \times  \: ( -  \frac{3}{4} ) ^{n \:  -  \: 1} }

Step-by-step explanation:

  • We first compute the ratio of this geometric sequence.

r \:  =  \:  \frac{ - 48}{64}  \\  \\ r   \:  =  \:  \frac{36}{ - 48}  \\  \\  r \:  =  \:  \frac{ - 27}{36}

  • We simplify the fractions:

r \:  =  \:   -  \frac{3 }{4}   \\  \\ r   \:  =  \:   -  \frac{3 }{4}  \\  \\  r \:  =  \:    -  \frac{3 }{4}

  • We deduce that it is the common ratio because it is the same between each pair.

r \:  =  \:  -  \frac{3 }{4}

  • We use the first term and the common ratio to describe the equation:

a_1 \:  =  \: 64; \: r \:  =  \:  -  \frac{3 }{4}

<h3>We apply the data in this formula:</h3>

\boxed{a_n \:  =  \: a_1 \:   \times  \:  {r}^{ n \:  -  \: 1} }

_______________________

<h3>We apply:</h3>

\boxed {\bold{a_n \:  =  \: 64 \:   \times  \:  {( -  \frac{3}{4} )}^{ n \:  -  \: 1} }}

<u>Data</u>: The unknown "n" is the term you want

<h3><em><u>MissSpanish</u></em></h3>
4 0
2 years ago
Could someone help me please ​
Alexxx [7]

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