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Convert to binary........

convert 3.141592 to base 2 with steps please, if possible.

Thank you for any help.

[pre-decimal point position]:

$\begin{array}{|rcrcrr|} \hline && && & \text{Remainder} \\ \hline 3 &:& 2 &=& \color{blue}{1} & \color{red}{1} \\ \color{blue}{1} &:& 2 &=& 0 & \color{red}{1} \\ \hline \end{array}$

decimals:

$\begin{array}{|rcrcrr|} \hline && && & \text{Carry} \\ \hline 0.141592 &*& 2 &=& {\color{red}0}.283184 & \color{red}{0} \\ 0.283184 &*& 2 &=& {\color{red}0}.566368 & \color{red}{0} \\ 0.566368 &*& 2 &=& {\color{red}1}.132736 & \color{red}{1} \\ 0.132736 &*& 2 &=& {\color{red}0}.265472 & \color{red}{0} \\ 0.265472 &*& 2 &=& {\color{red}0}.530944 & \color{red}{0} \\ 0.530944 &*& 2 &=& {\color{red}1}.061888 & \color{red}{1} \\ 0.061888 &*& 2 &=& {\color{red}0}.123776 & \color{red}{0} \\ 0.123776 &*& 2 &=& {\color{red}0}.247552 & \color{red}{0} \\ 0.247552 &*& 2 &=& {\color{red}0}.495104 & \color{red}{0} \\ 0.495104 &*& 2 &=& {\color{red}0}.990208 & \color{red}{0} \\ 0.990208 &*& 2 &=& {\color{red}1}.980416 & \color{red}{1} \\ 0.980416 &*& 2 &=& {\color{red}1}.960832 & \color{red}{1} \\ 0.960832 &*& 2 &=& {\color{red}1}.921664 & \color{red}{1} \\ 0.921664 &*& 2 &=& {\color{red}1}.843328 & \color{red}{1} \\ 0.843328 &*& 2 &=& {\color{red}1}.686656 & \color{red}{1} \\ 0.686656 &*& 2 &=& {\color{red}1}.373312 & \color{red}{1} \\ 0.373312 &*& 2 &=& {\color{red}0}.746624 & \color{red}{0} \\ 0.746624 &*& 2 &=& {\color{red}1}.493248 & \color{red}{1} \\ 0.493248 &*& 2 &=& {\color{red}0}.986496 & \color{red}{0} \\ 0.986496 &*& 2 &=& {\color{red}1}.972992 & \color{red}{1} \\ \cdots \\ \hline \end{array}$

$3.141592_{10} = 11.00100100001111110101\ldots_2$

Convert to binary........

convert 3.141592 to base 2 with steps please, if possible.

Thank you for any help.

[pre-decimal point position]:

$\begin{array}{|rcrcrr|} \hline && && & \text{Remainder} \\ \hline 3 &:& 2 &=& 1 & \color{red}{1} \\ 1 &:& 2 &=& 0 & \color{red}{1} \\ \hline \end{array}$

decimals:

$\begin{array}{|rcrcrr|} \hline && && & \text{Carry} \\ \hline 0.141592 &*& 2 &=& {\color{red}0}.283184 & \color{red}{0} \\ 0.283184 &*& 2 &=& {\color{red}0}.566368 & \color{red}{0} \\ 0.566368 &*& 2 &=& {\color{red}1}.132736 & \color{red}{1} \\ 0.132736 &*& 2 &=& {\color{red}0}.265472 & \color{red}{0} \\ 0.265472 &*& 2 &=& {\color{red}0}.530944 & \color{red}{0} \\ 0.530944 &*& 2 &=& {\color{red}1}.061888 & \color{red}{1} \\ 0.061888 &*& 2 &=& {\color{red}0}.123776 & \color{red}{0} \\ 0.123776 &*& 2 &=& {\color{red}0}.247552 & \color{red}{0} \\ 0.247552 &*& 2 &=& {\color{red}0}.495104 & \color{red}{0} \\ 0.495104 &*& 2 &=& {\color{red}0}.990208 & \color{red}{0} \\ 0.990208 &*& 2 &=& {\color{red}1}.980416 & \color{red}{1} \\ 0.980416 &*& 2 &=& {\color{red}1}.960832 & \color{red}{1} \\ 0.960832 &*& 2 &=& {\color{red}1}.921664 & \color{red}{1} \\ 0.921664 &*& 2 &=& {\color{red}1}.843328 & \color{red}{1} \\ 0.843328 &*& 2 &=& {\color{red}1}.686656 & \color{red}{1} \\ 0.686656 &*& 2 &=& {\color{red}1}.373312 & \color{red}{1} \\ 0.373312 &*& 2 &=& {\color{red}0}.746624 & \color{red}{0} \\ 0.746624 &*& 2 &=& {\color{red}1}.493248 & \color{red}{1} \\ 0.493248 &*& 2 &=& {\color{red}0}.986496 & \color{red}{0} \\ 0.986496 &*& 2 &=& {\color{red}1}.972992 & \color{red}{1} \\ \cdots \\ \hline \end{array}$

$3.141592_{10} = 11.00100100001111110101\ldots_2$

complete _x5=3x10

${\color{red} 6}\times5=3\times10$

what is the area of a circle with a radius of 8 centimeters

$\begin{array}{|rcll|} \hline A &=& \pi r^2 \quad & | \quad r = 8\ cm \\ &=& \pi 8^2 \ cm^2 \\ &=& 64 \pi \ cm^2 \quad & | \quad \pi = 3.14159265359\ldots \\ &=& 201.061929830 \ cm^2 \\ \hline \end{array}$

p(x-3)=15C^3(0.5)3(0.5)15-3

$\begin{array}{|rcll|} \hline && 15C_{3}(0.5)^{3}(0.5)^{15-3} \quad & | \quad 15C_{3} = \frac{15}{3}\cdot \frac{14}{2}\cdot \frac{13}{1} = 5\cdot 7 \cdot 13 = 455 \\ &=& 455 \cdot (0.5)^{3}(0.5)^{15-3} \\ &=& 455 \cdot(0.5)^{3+15-3} \\ &=& 455 \cdot(0.5)^{15} \\ &=& 455 \cdot(\frac12 )^{15} \\ &=& 455 \cdot \frac{1}{2^{15}} \\ &=& \frac{455}{2^{15}} \\ &=& \frac{455}{32768} \\ &=& 0.01388549805 \\ \hline \end{array}$

Find the area between two concentric circles defined by 

Let x_c the center of the circles in x

Let y_c the center of the circles in y

$x2 + y2 -2x + 4y \underbrace{+1}_{=x_c^2+y_c^2-r_1^2} = 0 \\\\ x2 + y2 -2x + 4y \underbrace{-11}_{=x_c^2+y_c^2-r_2^2} = 0$

$\begin{array}{|lrcll|} \hline (1) & 1 &=& x_c^2+y_c^2-r_1^2 \\ (2) & -11 &=& x_c^2+y_c^2-r_2^2 \\ \hline (1)-(2): & 1-(-11) &=& x_c^2+y_c^2-r_1^2-(x_c^2+y_c^2-r_2^2) \\ & 1+11 &=& x_c^2+y_c^2-r_1^2-x_c^2-y_c^2+r_2^2 \\ & 12 &=& -r_1^2 +r_2^2 \\ & \mathbf{r_2^2-r_1^2} & \mathbf{=} & \mathbf{12} \\ \hline \end{array}$

The area between two concentric circles:

$\begin{array}{|rcll|} \hline A &=& \pi r_2^2 - \pi r_1^2 \\ &=& \pi \cdot ( r_2^2 - r_1^2) \quad & | \quad r_2^2-r_1^2 = 12 \\ &=& \pi \cdot 12 \\ &=& 37.6991118431 \\ \hline \end{array}$

The resistance, R (in ohms) of a wire varies directly with the length, L (in cm) of the wire,
and inversely with the cross-sectional area, A (in cm^2).
A 500 cm piece of wire with a radius of 0.2cm has a resistance of 0.025 ohm.
Find an equation that relates these variables in the form of $R = \frac {\square\cdot \pi\cdot L}{\square}$

Let $\rho$ = resistivity
Let L = length
Let A = cross sectional area
Let r = Radius

$\begin{array}{|rcll|} \hline R &=& \dfrac{ \rho \cdot L } {A} \quad & \text{rewrite...}\\ \rho &=& \dfrac{ A \cdot R } {L} \quad & | \quad A = \pi r^2 \\ \rho &=& \dfrac{ \pi r^2 \cdot R } {L} \\ \rho &=& \dfrac{ r^2 \cdot R } {L} \cdot \pi \quad & | \quad L = 500\ cm \quad R= 0.025\ \Omega \quad r = 0.2\ cm \\ \rho &=& \dfrac{ 0.2^2 \cdot 0.025 } {500} \cdot \pi \\ \rho &=& \dfrac{ 0.04 \cdot 0.025 } {500} \cdot \pi \\ \rho &=& \dfrac{ 0.001 } {500} \cdot \pi \\\\ \mathbf{\rho} & \mathbf{=} & \mathbf{0.000002 \cdot \pi } \\ \hline \end{array}$

Find an equation that relates these variables in the form of $R = \frac {\square\cdot \pi\cdot L}{\square}$

$\begin{array}{|rcll|} \hline R &=& \dfrac{ \rho \cdot L } {A} \quad & \quad \rho = 0.000002 \cdot \pi \\\\ \mathbf{R} & \mathbf{=} & \mathbf{ \dfrac{ {\color{red}0.000002} \cdot \pi \cdot L } {{\color{red}A}} } \\ \hline \end{array}$

the radius of the circle is 5 feet. It has a sector with a central angle of 54%.

What is the area of the sector to the nearest tenth

Central angle:

$\begin{array}{|rcll|} \hline && 360^{\circ} \cdot 54\ \% \\ &=& 360^{\circ} \cdot \frac{54}{100} \\ &=& 3.60^{\circ} \cdot 54\\ &=& 194.4^{\circ} \\ \hline \end{array}$

The area of the sector:

$\begin{array}{|rcll|} \hline A_{\text{sector}} &=& \pi r^2 \cdot \frac{\varphi}{360^{\circ}} \quad & | \quad \varphi= 194.4^{\circ} \quad r = 5\ \text{feet} \\ &=& \pi \cdot 5^2 \cdot \frac{194.4^{\circ}}{360^{\circ}} \\ &=& \pi \cdot 25 \cdot 0.54 \\ &=& \pi \cdot 13.5 \\ &=& 42.4115008235 \\ \hline \end{array}$

The area of the sector to the nearest tenth is $42.4\ \text{feet}^2$

Express the equation ${2x^2}+4x-1$

In the form $a{(x+b)^2}+c$

Where $\{a,b,c\}\in\mathbb{Z}$

$\begin{array}{|rcll|} \hline && 2x^2+4x-1 \\ &=& 2\cdot (x^2+2x) - 1 \\ &=& 2\cdot \Big( (x+1)^2 -1 \Big) - 1 \\ &=& 2\cdot (x+1)^2 -2 - 1 \\ &=& 2\cdot (x+1)^2 -3 \\ \hline \end{array}$

what is the sum of entries in row 10 of Pascal trigangle

$\begin{array}{|rcll|} \hline (1+x)^{10} &=& \binom{10}{0} + \binom{10}{1}x \\ &+& \binom{10}{2}x^2 + \binom{10}{3}x^3 \\ &+& \binom{10}{4}x^4 + \binom{10}{5}x^5 \\ &+& \binom{10}{6}x^6 + \binom{10}{7}x^7 \\ &+& \binom{10}{8}x^8 + \binom{10}{9}x^9 \\ &+& \binom{10}{10}x^{10} \\ \hline \end{array}$

Set x = 1:

$\begin{array}{|rcll|} \hline (1+1)^{10} &=& \binom{10}{0} + \binom{10}{1}\cdot 1 \\ &+& \binom{10}{2}\cdot 1^2 + \binom{10}{3}\cdot 1^3 \\ &+& \binom{10}{4}\cdot 1^4 + \binom{10}{5}\cdot 1^5 \\ &+& \binom{10}{6}\cdot 1^6 + \binom{10}{7}\cdot 1^7 \\ &+& \binom{10}{8}\cdot 1^8 + \binom{10}{9}\cdot 1^9 \\ &+& \binom{10}{10}\cdot 1^{10} \\\\ 2^{10} &=& \binom{10}{0} + \binom{10}{1} + \binom{10}{2} + \binom{10}{3} + \binom{10}{4} + \binom{10}{5} \\ &+& \binom{10}{6} + \binom{10}{7} + \binom{10}{8} + \binom{10}{9} + \binom{10}{10} \\ \hline \end{array}$

The sum of entries in row 10 of Pascal trigangle is 2¹⁰