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nikklg [1K]
2 years ago
9

A summer camp cookout is planned for the campers and their families. There is room for 450 people. Each adult costs $7, and each

camper costs $4. There is a maximum budget of $1,150. Write the system of inequalities to represent this real-world scenario, where x is the number of adults and y is the number of campers. x + y ≤ 1,150 7x + 4y ≤ 450 x + y ≤ 450 7x + 4y ≤ 1,150 x + y ≤ 1,150 4x + 7y ≤ 450 x + y ≤ 450 4x + 7y ≤ 1,150
Mathematics
1 answer:
Tanya [424]2 years ago
3 0

Answer:

\left\{\begin{array}{l}x+y\le 450\\7x+4y\le 1,150\end{array}\right.

Step-by-step explanation:

Let  x be the number of adults and y be the number of campers.

There are rooms for 450 people, so

x+y≤450.

Each adult costs $7, then x adults cost $7x.

Each camper costs $4, then y campers cost $4y.

There is a maximum budget of $1,150, so

7x+4y≤1,150

Hence, you get the system of two inequalities:

\left\{\begin{array}{l}x+y\le 450\\7x+4y\le 1,150\end{array}\right.

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On the first day a total of 40 items were sold for $356. Define the variables and write a system of equations to find the number
Alina [70]
<h3><u><em>Question:</em></u></h3>

On the first day, a total of 40 items were sold for $356. Pies cost $10 and cakes cost $8. Define the variables, write a system of equations to find the number of cakes and pies sold, and state how many pies were sold.

<h3><em><u>Answer:</u></em></h3>

The variables are defined as:

"c" represent the number of cakes sold and "p" represent the number of pies sold

The system of equations used are:

c + p = 40 and 8c + 10p = 356

18 pies and 22 cakes were sold

<h3><em><u>Solution:</u></em></h3>

Let "c" represent the number of cakes sold

Let "p" represent the number of pies sold

Cost of 1 pie = $ 10

Cost of 1 cake = $ 8

Given that total of 40 items were sold

number of cakes + number of pies = 40

c + p = 40 ------ eqn 1

<u><em>Given items were sold for $356</em></u>

number of cakes sold x Cost of 1 cake + number of pies sold x Cost of 1 cake = 356

c \times 8 + p \times 10 = 356

8c + 10p = 356  ----- eqn 2

<u><em>Let us solve eqn 1 and eqn 2</em></u>

From eqn 1,

p = 40 - c    ---- eqn 3

Substitute eqn 3 in eqn 2

8c + 10(40 - c) = 356

8c + 400 - 10c = 356

-2c = - 44

c = 22

<em>Substitute c = 22 in eqn 3</em>

p = 40 - c

p = 40 - 22

p = 18

Thus 18 pies and 22 cakes were sold

3 0
2 years ago
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