The decimals chart in this lesson shows that we line up the decimal points when we add or subtract decimal numbers. Why do we do that?

Rivka Thorpe 2021-06-12 Answered
The decimals chart in this lesson shows that we line up the decimal points when we add or subtract decimal numbers. Why do we do that?

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Expert Answer

Clara Reese
Answered 2021-06-13 Author has 6616 answers
Remember, you can only add numbers together that are in the same place values. Ones are added to ones. Tens are added to tens. Tenths are added to tenths.
RESULT You must line up the decimals because when you add or subtract numbers you must do so according to their place value.
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Relevant Questions

asked 2021-06-19
The decimals chart in this lesson shows that we line up the decimal points when we add or subtract decimal numbers. Why do we do that?
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(a) Write each answer as an equality, and any decimals up to 7 decimal places. Clearly state the value of and \(\displaystyle\delta{\left({\quad\text{or}\quad}{M}{\quad\text{or}\quad}{N}{\quad\text{and}\quad}\delta\right)}\) in each case.
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(iv) How close to 0 do we need to take x so that \(\displaystyle{\left({2}{x}+{9}\right)}{>}{8.999}\)?
(v) How close to 0 do we need to take x so that \(\displaystyle{\left({x}^{{{2}}}+{6}{x}+{9}\right)}{<}{9.001}\) ?
(vi) How close to 0 do we need to take x so that \(\displaystyle{\left({x}^{{{2}}}+{6}{x}+{9}\right)}{<}{9.0001}\) ?
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(viii) How close to -7 do we need to take x so that \(\displaystyle\frac{{1}}{{\left({x}+{7}\right)}^{{{4}}}}{>}{10000}\) ?
(ix) How close to −7 do we need to take x so that \(\displaystyle\frac{{1}}{{\left({x}+{7}\right)}^{{4}}}{>}{100000}\) ?
(x) How close to 0 do we need to take x so that \(\displaystyle{\ln{{x}}}{<}-{10000}\)?
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(i) \(\displaystyle\lim{x}\rightarrow{4}\)
\(\displaystyle{\left(\frac{{x}}{{2}}-{2}\right)}={0}\)
(ii) \(\displaystyle\lim{x}\rightarrow{0}\)
\(\displaystyle{\left({2}{x}+{9}\right)}={9}\)
(iii) \(\displaystyle\lim{x}\rightarrow{0}\)
\(\displaystyle{\left({x}^{{{2}}}+{6}{x}+{9}\right)}={9}\)
(iv) \(\displaystyle\lim{x}\rightarrow-{7}\)
\(\displaystyle\frac{{1}}{{\left({x}+{7}\right)}^{{{4}}}}=\infty\)
(v) \(\displaystyle\lim{x}\rightarrow{0}+\)
\(\displaystyle{\ln{{x}}}=-\infty\)

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...