Respuesta :
Answer:
so looks its an easy one
110 - 39 = 71
71 x 5.75 = 408.25
39 x 3.25 = 126.75
408.25 +126.75 = 535
71 burritos and 39 tacos = 535 dollars
Answer: 71
Step-by-step explanation:
\underline{\text{Define Variables:}}
Define Variables:
May choose any letters.
\text{Let }t=
Let t=
\,\,\text{the number of tacos sold}
the number of tacos sold
\text{Let }b=
Let b=
\,\,\text{the number of burritos sold}
the number of burritos sold
\text{\textquotedblleft can make 110 tacos/burritos"}\rightarrow \text{110 or fewer tacos/burritos}
“can make 110 tacos/burritos"→110 or fewer tacos/burritos
Use a \le≤ symbol
Therefore the total number of tacos and burritos sold, t+bt+b, must be less than or equal to 110:110:
t+b\le 110
t+b≤110
\text{\textquotedblleft a minimum of \$520"}\rightarrow \text{\$520 or more}
“a minimum of $520"→$520 or more
Use a \ge≥ symbol
Each taco sells for $3.25, so tt tacos will bring in 3.25t3.25t dollars. Each burrito sells for $5.75, so bb burritos will bring in 5.75b5.75b dollars. Therefore, the total amount of revenue 3.25t+5.75b3.25t+5.75b must be greater than or equal to \$520:$520:
3.25t+5.75b\ge 520
3.25t+5.75b≥520
\text{Plug in }\color{green}{39}\text{ for }t\text{ and solve each inequality:}
Plug in 39 for t and solve each inequality:
Justin sold 39 tacos
\begin{aligned}t+b\le 110\hspace{10px}\text{and}\hspace{10px}&3.25t+5.75b\ge 520 \\ \color{green}{39}+b\le 110\hspace{10px}\text{and}\hspace{10px}&3.25\left(\color{green}{39}\right)+5.75b\ge 520 \\ b\le 71\hspace{10px}\text{and}\hspace{10px}&126.75+5.75b\ge 520 \\ \hspace{10px}&5.75b\ge 393.25 \\ \hspace{10px}&b\ge 68.39 \\ \end{aligned}
t+b≤110and
39+b≤110and
b≤71and
3.25t+5.75b≥520
3.25(39)+5.75b≥520
126.75+5.75b≥520
5.75b≥393.25
b≥68.39
\text{The values of }b\text{ that make BOTH inequalities true are:}
The values of b that make BOTH inequalities true are:
\{69,\ 70,\ 71\}
{69, 70, 71}
Therefore the maximum number of burritos that Justin must sell is 71.
