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Fynjy0 [20]
1 year ago
15

How to solve (2.31*10^-6)+(5.87*10^-4)

Mathematics
1 answer:
S_A_V [24]1 year ago
5 0

We want to calculate the following

2.31\cdot10^{-6}+5.87\cdot10^{-4}

Using the properties of exponents, we have that

10^{-6}=10^{-4}\cdot10^{-2}

So we have

2.31\cdot10^{-6}+5.87\cdot10^{-4}=2.31\cdot10^{-2}\cdot10^{-4}+5.87\cdot10^{-4}

So, if factor 10^-4 on the right side, we have

10^{-4}(2.31\cdot10^{-2}+5.87)

Note that

2.31\cdot10^{-2}=0.0231

Then,

5.87+2.31\cdot10^{-2}=5.87+0.0231=5.8931^{}

So we have that

2.31\cdot10^{-6}+5.87\cdot10^{-4}=5.8931\cdot10^{-4}

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\bf \qquad \qquad \textit{double proportional variation}
\\\\
\begin{array}{llll}
\textit{\underline{y} varies directly with \underline{x}}\\
\textit{and inversely with \underline{z}}
\end{array}\implies y=\cfrac{kx}{z}\impliedby 
\begin{array}{llll}
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\qquad  variation
\end{array}\\\\
-------------------------------

\bf \stackrel{\textit{\underline{y} varies directly with the square of \underline{x} and inversely with \underline{z}}}{y=\cfrac{kx^2}{z}}
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