azhinie6lleY
contestada

5. You are making your weekly mean plans and are working with the following constraints: It costs $8 to go out to dinner. It costs $5 to go out to lunch. You want to go out to dinner at least as many times as you go out to lunch. You can spend at most $42.
What is the greatest number of meals you can eat out?
3
4
5
6

Respuesta :

metchelle
The correct answer for this question would be option D.6. Given that per dinner, you can spend $8 and per lunch is $5. Given that you can spend at most $42, you can only eat out 6 times. 3 times for lunch and 3 times for dinner.
Since 8 times 4 would be $24 in total, and 5 x 3 would be $15 in total, so the overall would be $39. Hope this is the answer that you are looking for.
calculista

Let

x-------> the number of dinner

y-------> the number of lunch

we know that

[tex]8x+5y \leq 42[/tex] -------> equation A

[tex]x=y[/tex] ------> equation B

Substitute equation B in equation A

[tex]8[y]+5y \leq 42[/tex]

[tex]13y \leq 42[/tex]

[tex]y \leq 42/13[/tex]

[tex]y \leq 3.23[/tex]

so

the greatest number of lunch is [tex]y=3[/tex]

[tex]x=y[/tex]

Hence

the greatest number of dinner is [tex]x=3[/tex]

therefore

the greatest number of meals is

[tex]x+y=3+3=6[/tex]

the answer is

[tex]6[/tex]

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