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dusya [7]
3 years ago
5

Which shows the numbers 3/7, 5/21, and 0.87 in order from greatest to least?

Mathematics
1 answer:
kipiarov [429]3 years ago
6 0

Answer:

0.87 ,3/7, 5/21

Step-by-step explanation:

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Perform the indicated operations. Write the answer in standard form, a+bi.<br> 5-3i / -2-9i
Vsevolod [243]

\huge \boxed{\mathfrak{Answer} \downarrow}

\large \bf\frac { 5 - 3 i } { - 2 - 9 i } \\

Multiply both numerator and denominator of \sf \frac{5-3i}{-2-9i} \\ by the complex conjugate of the denominator, -2+9i.

\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{\left(-2-9i\right)\left(-2+9i\right)})  \\

Multiplication can be transformed into difference of squares using the rule: \sf\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{\left(-2\right)^{2}-9^{2}i^{2}})  \\

By definition, i² is -1. Calculate the denominator.

\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{85})  \\

Multiply complex numbers 5-3i and -2+9i in the same way as you multiply binomials.

\large \bf \: Re(\frac{5\left(-2\right)+5\times \left(9i\right)-3i\left(-2\right)-3\times 9i^{2}}{85})  \\

Do the multiplications in \sf5\left(-2\right)+5\times \left(9i\right)-3i\left(-2\right)-3\times 9\left(-1\right).

\large \bf \: Re(\frac{-10+45i+6i+27}{85})  \\

Combine the real and imaginary parts in -10+45i+6i+27.

\large \bf \: Re(\frac{-10+27+\left(45+6\right)i}{85})  \\

Do the additions in \sf-10+27+\left(45+6\right)i.

\large \bf Re(\frac{17+51i}{85})  \\

Divide 17+51i by 85 to get \sf\frac{1}{5}+\frac{3}{5}i \\.

\large \bf \: Re(\frac{1}{5}+\frac{3}{5}i)  \\

The real part of \sf \frac{1}{5}+\frac{3}{5}i \\ is \sf \frac{1}{5} \\.

\large  \boxed{\bf\frac{1}{5} = 0.2} \\

3 0
3 years ago
Need to know the population in 2015
Margarita [4]
\bf P=1110.9e^{kt}\quad &#10;\begin{cases}&#10;P=1200 &\textit{in thousands}\\&#10;t=2 &\textit{in 2002}&#10;\end{cases}\implies 1200=1110.9e^{k2}&#10;\\\\\\&#10;\cfrac{1200}{1110.9}=e^{2k}\implies ln\left( \frac{1200}{1110.9} \right)=ln(e^{2k})\implies ln\left( \frac{1200}{1110.9} \right)=2k&#10;\\\\\\&#10;\cfrac{ln\left( \frac{1200}{1110.9} \right)}{2}=k\implies 0.0385755\approx k\implies 0.0386\approx k\\\\&#10;-------------------------------\\\\

\bf P=1110.9e^{0.0386t}\qquad &#10;\begin{cases}&#10;t=14\\&#10;\textit{year 2015}&#10;\end{cases}\implies P=1110.9e^{0.0386\cdot 14}&#10;\\\\\\&#10;P\approx 1907.0747\implies about\ 1,907,075\textit{ once rounded up}
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3 years ago
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