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azhinie6lleY
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  1. 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

Verified Answer :

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

\(8x+5y \leq 42\) -------> equation A

\(x=y\) ------> equation B

Substitute equation B in equation A

\(8[y]+5y \leq 42\)

\(13y \leq 42\)

\(y \leq 42/13\)

\(y \leq 3.23\)

so

the greatest number of lunch is \(y=3\)

\(x=y\)

Hence

the greatest number of dinner is \(x=3\)

therefore

the greatest number of meals is

\(x+y=3+3=6\)

the answer is

\(6\)

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