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3 years ago

What is 129.392 rounded to nearest Ten?

Mathematics

2 answers:

vesna_86 [32]3 years ago

7 0

It’s 129.4 have a good day and get an a

Stels [109]3 years ago

6 0

$\huge\bold\color{purple}{Answer:}$

====================================

Explanation:

To round 129.392 to the nearest tenth consider the hundredths’ value of 129.392, which is 9 and equal or more than 5. Therefore, the tenths value of 129.392 increases by 1 to 4.

129.392 rounded to the nearest tenth = 129.4

 $\huge\bold\color{purple}{Hope \: it \: helps!}$ 

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1) $\dfrac{2}{\sqrt{5} } = \dfrac{2 \cdot \sqrt{5} }{5}$

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Step-by-step explanation:

The rationalization of the denominator of the surds are found as follows;

1) $\dfrac{2}{\sqrt{5} }$

$\dfrac{2}{\sqrt{5} } \times \dfrac{\sqrt{5} }{\sqrt{5} } = \dfrac{2 \cdot \sqrt{5} }{5}$

$\dfrac{2}{\sqrt{5} } = \dfrac{2 \cdot \sqrt{5} }{5}$

2) $-\dfrac{5}{\sqrt{3} }$

$-\dfrac{5}{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = -\dfrac{5 \cdot \sqrt{3} }{3}$

$-\dfrac{5}{\sqrt{3} } = -\dfrac{5 \cdot \sqrt{3} }{3}$

3) $\dfrac{\sqrt{2} + \sqrt{5} }{\sqrt{10} }$

$\dfrac{\sqrt{2} + \sqrt{5} }{\sqrt{10} } \times \dfrac{ \sqrt{10} }{\sqrt{10} } = \dfrac{\sqrt{20} + \sqrt{50} }{10 } = \dfrac{2\cdot \sqrt{5} + 5 \cdot \sqrt{2} }{10} = \dfrac{\sqrt{5} }{5} + \dfrac{ \sqrt{2} }{2}$

$\dfrac{\sqrt{2} + \sqrt{5} }{\sqrt{10} } =\dfrac{\sqrt{5} }{5} + \dfrac{ \sqrt{2} }{2}$

4) $\dfrac{3 + \sqrt{2} }{\sqrt{3} }$

$\dfrac{3 + \sqrt{2} }{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = \dfrac{3 \cdot \sqrt{3}+\sqrt{6} }{3 } = \sqrt{3} + \dfrac{\sqrt{6} }{3}$

$\dfrac{3 + \sqrt{2} }{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = \sqrt{3} + \dfrac{\sqrt{6} }{3}$

5) $\dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2} }$

$\dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2} } = \dfrac{\sqrt{5} - \sqrt{2} }{\sqrt{5} - \sqrt{2} } = \dfrac{\sqrt{15} -\sqrt{6} }{5 - 2} = \dfrac{\sqrt{15} - \sqrt{6} }{3}$

$\dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2} }= \dfrac{\sqrt{15} - \sqrt{6} }{3}$

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$\dfrac{\sqrt{7} }{\sqrt{3} - \sqrt{5} } \times \dfrac{\sqrt{3} + \sqrt{5}}{\sqrt{3} + \sqrt{5}} =\dfrac{\sqrt{21} + \sqrt{35}}{8}$

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What is 129.392 rounded to nearest Ten?​

What is 129.392 rounded to nearest Ten?