Question

Can someone walk me through the steps in solving this question

Answer 1

so, we know that, in 1oz of baked potato, there are 48.3 calories, ok..how many calories in 3 and 1/3 oz then?

now, bear in mind, we first convert the mixed fraction to "improper", and then use that,


`\bf \begin{array}{ccll} \stackrel{\stackrel{baked}{potato}}{oz}&calories\\ \text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\ 1&48.3\\ 3(1)/(3)&p \end{array}\implies \cfrac{1}{3(1)/(3)}=\cfrac{48.3}{p}\implies \cfrac{(1)/(1)}{(3\cdot 3+1)/(3)}=\cfrac{48.3}{p}`


`\bf \cfrac{(1)/(1)}{(10)/(3)}=\cfrac{48.3}{p}\implies \cfrac{1}{1}\cdot \cfrac{3}{10}=\cfrac{48.3}{p}\implies \cfrac{3}{10}=\cfrac{48.3}{p} \\\\\\ 3p=483\implies p=\cfrac{483}{3}\implies \boxed{p=\stackrel{calories}{161}}`

now, we know that in 1oz of chicken gas 24.6 calories, so, how many calories then in 5 and 1/4 oz?


`\bf \begin{array}{ccll} \stackrel{chicken}{oz}&calories\\ \text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\ 1&24.6\\ 5(1)/(4)&c \end{array}\implies \cfrac{1}{5(1)/(4)}=\cfrac{24.6}{c}\implies \cfrac{(1)/(1)}{(5\cdot 4+1)/(4)}=\cfrac{24.6}{c}`


`\bf \cfrac{(1)/(1)}{(21)/(4)}=\cfrac{24.6}{c}\implies \cfrac{1}{1}\cdot \cfrac{4}{21}=\cfrac{24.6}{c}\implies \cfrac{4}{21}=\cfrac{24.6}{c} \\\\\\ 4c=516.6\implies c=\cfrac{516.6}{4}\implies \boxed{c=\stackrel{calories}{129.15}}`

so, how many calories in that meal? well, p + c.