솔루션피아

풀이에 필요한 정보가 너무 부족하여 고양이의 질량과 구성성분비는 임의로 정했습니다. $$\begin{cases} \text{Cat} &= 5\ut{kg}\\ 1\ut{mol}&=6.02\times10^{23}\\ 1\ut{u}&=1\ut{g/mol} \end{cases}$$ $$\begin{cases} O : 70\ut{\%}\\ C : 20\ut{\%}\\ H : 10\ut{\%}\\ \end{cases}$$ $$\begin{cases} m_O &= 16\ut{u}\\ m_C &= 12\ut{u}\\ m_H &= 1.0\ut{u}\\ \end{cases}$$ $$ \begin{cases} M_O &= 5\ut{kg}\cdot0.7=3.5\ut{kg}\\ M_C &= 5\ut{kg}\cdot0..

$$\begin{aligned} \Ans&=0.257\ut{kg/day}\\ &=0.257\ut{kg/day}\cdot\frac{10^6\ut{mg}}{1\ut{kg}}\cdot\frac{1\ut{day}}{24\ut{h}}\cdot\frac{1\ut{h}}{60\ut{min}} \\ &=\frac{6425}{36}\ut{mg/min}\\ &\approx178.47222222222223\ut{mg/min}\\ &\approx178\ut{mg/min}\\ \end{aligned}$$

$$ \begin{cases} 43\ut{cm}

$$\begin{cases} V&=40\ut{cm}\times40\ut{cm}\times26\ut{cm} \end{cases}$$ $$\ab{a}$$ $$ \begin{aligned} V&=41600\ut{cm^3}\\ &=41.6\ut{L}\\ &=41.6\ut{L}\cdot\frac{16\ut{bottle}}{11.356\ut{L}}\\ &=\frac{166400}{2839}\ut{bottle} \approx58.61218738992603\ut{bottle}\\ &=\(2\cdot20+16+2+\frac{1738}{2839}\)\ut{bottle}\\ &=2\ut{nebuchadnezzar}+1\ut{balthazar}+1\ut{magnum}+\frac{1738}{2839}\ut{bottle} \..

$$\begin{cases} 1\ut{mol}&=6.02\times10^{23}\\ 1\ut{u}&=1\ut{g/mol}\\ \end{cases}$$ $$\begin{aligned} \Ans&={M\over N}\\ &={68\ut{g}\over 6.8\ut{mol}}\\ &=10\ut{g/mol}\\ &=10\ut{u} \end{aligned}$$

$$\begin{cases} \rho&=40\ut{mi/gal}\\ 1\ut{U.K.~gal}&=4.546~090~0\ut{L}\\ 1\ut{U.S.~gal}&=3.785~411~8\ut{L}\\ \end{cases}$$ $$\ab{a}$$ $$ \begin{aligned} \Ans&={1230\ut{mi}\over40\ut{mi/U.S.~gal}}\\ &={123\over4}\ut{U.S.~gal}\\ &={123\over4}\ut{U.S.~gal}\cdot{3.785~411~8\ut{L}\over1\ut{U.S.~gal}}\\ &=\frac{2~328~028~257}{20~000~000}\ut{L}\\ &\approx116.401~412~85\ut{L}\\ &\approx1.2\times10^2\..

$$\begin{cases} 1\ut{ken}&=1.97\ut{m}\\ \end{cases}$$ $$\ab{a}$$ $$\begin{aligned} \frac{1\ut{ken^2}}{1\ut{m^2}}&=\(\frac{1\ut{ken}}{1\ut{m}}\)^2\\ &=\(\frac{1.97\ut{m}}{1\ut{m}}\)^2\\ &=\frac{38809}{10000}\\ &\approx \frac{388}{100}\\&\approx \frac{97}{25}\\ \end{aligned}$$ $$1\ut{ken^2}:1\ut{m^2}\approx 97:25$$ $$\ab{b}$$ $$ \begin{aligned} \frac{1\ut{ken^3}}{1\ut{m^3}}&=\(\frac{1\ut{ken}..

$$\begin{cases} 1\ut{cord}&=8\ut{ft}\times4\ut{ft}\times4\ut{ft}\\ &=128\ut{ft^3} \end{cases}$$ $$ \begin{aligned} \Ans&=2\ut{cord}\\ &=2\ut{cord}\cdot\frac{128\ut{ft^3}}{1\ut{cord}}\cdot\(\frac{12\ut{in}}{1\ut{ft}}\cdot\frac{2.54\ut{cm}}{1\ut{in}}\cdot\frac{1\ut{m}}{100\ut{cm}}\)^3\\ &=\frac{1769802912}{244140625}\ut{m^3}\\ &\approx7.249112727552\ut{m^3}\\ &\approx7.25\ut{m^3}\\ \end{aligned}..

$$\begin{cases} 1\ut{acre}=&40\ut{perch}\times4\ut{perch}=&160\ut{perch^2}\\ 1\ut{rood}=&40\ut{perch}\times1\ut{perch}=&40\ut{perch^2}\\ 1\ut{perch}=&16.5\ut{ft} \end{cases}$$ $$\ab{a}$$ $$ \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=&..