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Brilliant_brown [7]
2 years ago
12

Owen is shopping for a coworker's retirement party. Regular sized boxes of forks contain enough for 16 guests and cost $5, while

value-pack boxes contain enough for 25 guests and cost $10. In all, there must be at least enough forks for the 61 guests who plan to attend. Owen only has $60 to spend on cutlery for the party. Write a system of inequalities to describe this scenario and then give two solutions that work for the scenario.
Mathematics
2 answers:
natita [175]2 years ago
6 0

Answer:

3 regular-sized boxes and 2 value-pack boxes

6 regular-sized boxes and 2 value-pack boxes

Step-by-step explanation:

Let x = number of regular-sized boxes of forks

Let y = number of value-pack boxes of forks

<u>Inequality 1</u>

Given:

  • Regular sized boxes contain enough forks for 16 guests
  • Value-pack boxes contain enough forks for 25 guests
  • There must be at least enough forks for the 61 guests

\implies 16x+ 25y \geq 61

<u>Inequality 2</u>

Given:

  • Regular sized boxes cost $5 each
  • Value-pack boxes cost $10 each
  • Maximum money to spend = $60

\implies 5x + 10y \leq 60

Therefore, the system of inequalities to describe the scenario is:

\begin{cases}16x + 25y \geq 61\\5x + 10y \leq 60\end{cases}

To find two solutions that work for the scenario, graph the inequalities.

When <u>graphing inequalities</u>:

< or > : draw a dashed line

≤ or ≥ : draw a solid line

< or ≤ : shade under the line

> or ≥ : shade above the line

Rearrange Inequality 1 to make y the subject:

\implies 25y \geq 61 - 16x

\implies y \geq \dfrac{61}{25} - \dfrac{16}{25}x

Rearrange Inequality 2 to make y the subject:

\implies 5x + 10y \leq 60

\implies 10y \leq 60-5x

\implies y \leq 6-\dfrac{1}{2}x

Graph the two lines by drawing a solid line for each.

Shade <u>above the line of first inequality</u> and <u>below the line of the second inequality</u>.

Solutions to the inequalities are any points in the <u>shaded area</u>.

However, solutions to the inequalities for this scenario are any points in the shaded area where <u>x and y are positive integers</u> (see second attached image).

Therefore, two solutions that work for the scenario are:

  • 3 regular-sized boxes and 2 value-pack boxes
  • 6 regular-sized boxes and 2 value-pack boxes

Learn more about inequalities here:

brainly.com/question/27947009

arlik [135]2 years ago
3 0

Step-by-step explanation:

hope you can understand

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Let's make the first even integer m and the second even integer n. The square of the first, decreased by the second is 130 is:

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Answer:

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Answer:

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