Mathematics, 29.01.2020 15:47 noeltan12031

Two 600-gram blocks of tofu are enough to make 2.5 chocolate mousse pies and still have 1/3 of a block left over. over the course of a workweek three vegan friends each eat the same amount of tofu, plus they make four chocolate mousse pies which they plan to bring to parties that weekend. if 2750 grams of tofu was consumed (eating or baking) by these people during the workweek, how much tofu did each vegan friend eat over the course of that workweek?

Answers: 2

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Answer from: Ilikepandas2019

$\bf if\ \frac{5}{3}\ of \ 600g\ makes \ \frac{5}{2}\ pies\\\\
\textit{how much tofu will it be in 4 pies?}\qquad well
\\\\

\begin{array}{ccllll}
grams&pies\\
\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\\
600\cdot \frac{5}{3}&\frac{5}{2}\\
x&4
\end{array}\implies \cfrac{600\cdot \frac{5}{3}}{x}=\cfrac{\frac{5}{2}}{4}
\\\\\\
\cfrac{200\cdot 5}{x}=\cfrac{5}{8}\leftarrow \textit{that many grams of tofu in 4 pies}$

now... that's how many grams of tofu in 4pies... so, the three friends used that much, to make the 4 pies,
now
the three friends ate some tofu over the workweek, it was the same amount for each friend.... let's say they ate "a" amount of tofu,
so
three friends, eating "a" each, a + a + a, they ate, 3a of tofu

now, whatever the friends ate, (3a), plus whatever the 4pies is in grams of tofu, sums up to 2750 grams altogether

so
$\bf (a+a+a)+(\textit{4pies worth in grams of tofu})=2750
\\\\
3a+(\textit{4pies worth in grams of tofu})=2750$

solve for "a", to see how much each ate

Answer from: Quest

b is the answer that makes most sense

step-by-step explanation:

Answer from: Quest

(2 x^3 + 7 x^2 + 5 x + 2)/(x (x + 2))

step-by-step explanation:

simplify the following:

x/x^2 + 3 x + 2 - (x (x + 1))/(x + 2)

combine powers. x/x^2 = x^(1 - 2):

x^(1 - 2) + 3 x + 2 - (x (x + 1))/(x + 2)

1 - 2 = -1:

x^(-1) + 3 x + 2 - (x (x + 1))/(x + 2)

put each term in 1/x + 3 x + 2 - (x (x + 1))/(x + 2) over the common denominator x (x + 2): 1/x + 3 x + 2 - (x (x + 1))/(x + 2) = (x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)):

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2))

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)) = ((x + 2) + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2)):

(2 + x + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

3 x^2 (x + 2) = 3 x^3 + 6 x^2:

(2 + x + 3 x^3 + 6 x^2 + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

2 x (x + 2) = 2 x^2 + 4 x:

(2 + x + 6 x^2 + 3 x^3 + 2 x^2 + 4 x - x^2 (x + 1))/(x (x + 2))

-x^2 (x + 1) = -x^2 - x^3:

(-x^2 - x^3 + 3 x^3 + 6 x^2 + 2 x^2 + 4 x + x + 2)/(x (x + 2))

grouping like terms, 3 x^3 - x^3 + 6 x^2 + 2 x^2 - x^2 + 4 x + x + 2 = (3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2:

((3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

3 x^3 - x^3 = 2 x^3:

(2 x^3 + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

6 x^2 + 2 x^2 - x^2 = 7 x^2: