솔루션피아

$$ \begin{cases} 1\ut{acre}=&40\ut{perch}\times4\ut{perch}=&160\ut{perch^2}\\ 1\ut{rood}=&40\ut{perch}\times1\ut{perch}=&40\ut{perch^2}\\ 1\ut{perch}=&16.5\ut{ft} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \Ans=&2.80\ut{acre}+13.0\ut{perch}\times3.00\ut{perch}\\ =&2.80\ut{acre}\cdot\frac{160\ut{perch^2}}{1\ut{acre}}+39\ut{perch^2}\\ =&487\ut{perch^2}\\ =&487\ut{perch^2}\cdot\frac{1\ut{rood}}{4..

$$ \text{put}\begin{cases} t_A=&a_A\cdot t_B+b_A\\ t_B=&\text{ref}\\ t_C=&a_C\cdot t_B+b_C\\ \end{cases} $$ $$t_A=a_A\cdot t_B+b_A$$ $$ \begin{cases} 312=&a_A\cdot 125&+b_A\\ 512=&a_A\cdot 290&+b_A\\ \end{cases} $$ $$ \begin{cases} a_A=&\cfrac{40}{33}\\ b_A=&\cfrac{5296}{33}\\ \end{cases} $$ $$t_A=\cfrac{40}{33}\cdot t_B+\cfrac{5296}{33}~\cdots~(1)$$ $$ \begin{cases} 92=&a_C\cdot 25&+b_C\\ 142=&..

$$ \begin{cases} 1\ut{AU}=&92.9\times10^6\ut{mi}\\ 1\ut{ly}=&\cfrac{186000\ut{mi}}{1\ut{sec}}\cdot1\ut{year} \end{cases} $$ $$\ab{a}$$ $$L=r\theta$$ $$ \begin{aligned} 1\ut{AU}=&1\ut{pc}\cdot1\ut{''}\cdot\frac{1\ut{\degree}}{3600\ut{''}}\cdot\frac{\pi\ut{rad}}{180\ut{\degree}}\\ =&\frac{\pi}{648000}\ut{pc}\\ \approx&4.84813681109536\times10^{-6}\ut{pc}\\ \approx&4.85\times10^{-6}\ut{pc}\\ \end{a..

풀이자주:문제의 유효숫자가 애매하여 다양하게 표현했습니다. $$ \begin{cases} 1\ut{Jalapeño}&=4000\ut{SHU}\\ 1\ut{Habanero}&=300000\ut{SHU}\\ \end{cases} $$ $$ \begin{aligned} \Ans=&\frac{312\times 2\ut{Jalapeño}}{1\ut{Habanero}}\\ =&\frac{312\times 2\ut{Jalapeño}\times\cfrac{4000\ut{SHU}}{1\ut{Jalapeño}}}{1\ut{Habanero}\times\cfrac{300000\ut{SHU}}{1\ut{Habanero}}}\\ =&\frac{312\cdot2\cdot4000}{300000}\\ =&\frac{208}{25}\\..

(풀이자주:문제의 유효숫자가 애매하여 다양하게 표현했습니다.) $$\ab{a}$$ $$ \begin{aligned} \Ans=&230\ut{ft^2/gal}\\ =&230\ut{ft^2/gal}\times\(\frac{12\ut{in}}{1\ut{ft}}\cdot\frac{2.54\ut{cm}}{1\ut{in}}\cdot\frac{1\ut{m}}{100\ut{cm}}\)^2\\ &\times \frac{1\ut{gal}}{231\ut{in^3}}\cdot\(\frac{1\ut{in}}{2.54\ut{cm}}\)^3\times\frac{1000\ut{cm^3}}{1\ut{L}}\\ =&\frac{230\cdot12\cdot1000}{100^2\cdot231\cdot2.54}\ut{m^2/L}\\ =&\fr..

$$ \begin{cases} 1\ut{mile}&=1609\ut{m}\\ 1\ut{nautical~mile}&=1852\ut{m}\\ \end{cases} $$ $$ \begin{aligned} \Ans&=45.3\ut{nautical~mile}-45.3\ut{mile}\\ &=45.3(1\ut{nautical~mile}-1\ut{mile})\\ &=45.3(1852\ut{m}-1609\ut{m})\\ &=45.3\times243\ut{m}\\ &=11007.9\ut{m}\\ &\approx11.0\ut{km} \end{aligned} $$

$$ \begin{cases} 1\ut{A.Barrel}&=7056\ut{in^3}\\ 1\ut{C.Barrel}&=5826\ut{in^3}\\ \end{cases} $$ $$ \begin{aligned} \Ans&=73\ut{A.Barrel}-73\ut{C.Barrel}\\ &=73(1\ut{A.Barrel}-1\ut{C.Barrel})\\ &=73(7056\ut{in^3}-5826\ut{in^3})\\ &=73\cdot1230\ut{in^3}\\ &=73\cdot1230\ut{in^3} \cdot \(\frac{2.54\ut{cm}}{\ut{in}}\)^3\cdot\frac{1\ut{L}}{1000\ut{cm^3}}\\ &=\frac{18392430957}{12500000}\ut{L}\\ &=1471..

- wey chaldron bag pottle gill 1 wey= 1 10/9 40/3 640 120240 1chaldron= $\frac{9}{10}=0.900$ 1 $\frac{9}{10}\cdot\frac{40}{3}=12.0$ $\frac{9}{10}\cdot640=576$ $\frac{9}{10}\cdot120240=1.08\times10^5$ 1bag= $\frac{3}{40}=0.0750$ $\frac{3}{40}\cdot\frac{10}{9}=8.33\times10^{-2}$ 1 $\frac{3}{40}\cdot640=48.0$ $\frac{3}{40}\cdot120240=9.02\times10^3$ 1pottle= $\frac{1}{640}=1.56\times10^{-3}$ $\frac..

$$ \begin{cases} 1\ut{ha}&=10^4\ut{m^2}\\ \end{cases} $$ $$ \begin{aligned} V&=57\ut{ha}\times17\ut{m}\\ &=57\ut{ha}\times\frac{10^4\ut{m^2}}{1\ut{ha}}\times17\ut{m}\\ &=969\times10^4\ut{m^3}\\ &=969\times10^4\ut{m^3}\times\(\frac{1\ut{km}}{10^3\ut{m}}\)^3\\ &=9.69\times10^{-3}\ut{km^3}\\ \end{aligned} $$

$$ \begin{cases} 1\ut{u}&=1\ut{g/mol}\\ 1\ut{mol}&=6.022~140~76\times10^{23}\\ m_H&=1.0\ut{u}\\ m_O&=16\ut{u} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} m_{(H_2O)}&=1.0\ut{u}\times2+16\ut{u}\\ &=18\ut{u}\\ &=18\ut{u}\times\frac{1\ut{g}}{1\ut{u}\cdot1\ut{mol}}\times\frac{1\ut{kg}}{1000\ut{g}}\\ &=\frac{18}{6.022~140~76\times10^{23}\cdot1000}\ut{kg}\\ &\approx2.988970320912924\times10^{-26}\ut{k..