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frutty [35]
1 year ago
10

Identify the digit with the given place value in number 173.514 hundredths

Mathematics
1 answer:
Llana [10]1 year ago
5 0

Answer: We are given the number 173.514

Each of the digits in this number has the following place value:

\begin{gathered} 1\rightarrow\text{ Hundreds} \\ 7\rightarrow\text{ Tens} \\ 3\rightarrow\text{ Ones} \\ \text{.}\rightarrow\text{  Decimal Point} \\ 5\rightarrow\text{ Tenth} \\ 1\rightarrow\text{  Hundreth} \\ 4\rightarrow\text{  Thousandth} \end{gathered}

Secondly, We have to identify the numbers in digit 185.712

\begin{gathered} 1\rightarrow\text{ Thousand} \\ 8\rightarrow\text{ Hundred} \\ 5\rightarrow\text{ Tens} \\ \text{. Decimal point} \\ 7\text{ }\rightarrow\text{Thenth} \\ 1\rightarrow\text{ Hundreth} \\ 2\text{ }\rightarrow\text{ Thousandth} \\  \end{gathered}

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Make a substitution:

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Then the system becomes

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Simplifying the equations gives

\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}

which is to say,

\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}

\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81

\implies\dfrac uv=\pm27

\implies u=\pm27v

Substituting this into the new system gives

\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1

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Then

\begin{cases}x=\dfrac{u+v}4\\\\y=\dfrac{u-v}2}\end{cases}\implies x=\pm7,y=\pm13

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Step-by-step explanation:

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A wheelchair ramp is 8.0 m long

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