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Zinaida [17]
3 years ago
7

A business has $15,000 to spend on airline tickets to travel to a conference. It wants 27 of its employees to attend. The busine

ss wants to buy as many business-class seats as possible. The business-class seats cost $700. The economy-class seats cost $375. Create a system of equations that models how many of each type of ticket the business should purchase.
A. 700x + 375y = 27
x + y = 15,000

B. x + 375y = 15,000
700x + y = 27

C. 700x + y = 15,000
x + 375y = 27

D. 700x + 375y = 15,000
x + y = 27
Mathematics
1 answer:
marusya05 [52]3 years ago
3 0

Answer:

<u>The correct answer is :</u>

<u>D. 700x + 375y = 15,000 </u>

<u>x + y = 27</u>

Step-by-step explanation:

1. Let's review the information given to answer the question correctly:

Budget for airline tickets = US$ 15,000

Number of employees attending the conference = 27

Price of business-class ticket = US$ 700

Price of economy-class ticket = US$ 375

2. Let's create a system of equations that models how many of each type of ticket the business should purchase.

x = Number of employees flying business-class

y = Number of employees flying economy-class

<u>x + y = 27 (Number of employees attending the conference)</u>

<u>700x + 375y = 15,000 (Cost of employees flying business plus cost of employees flying economy equal to budget for buying the airline tickets) </u>

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F(x)=x-1/x^2-x-6 which is the graph of
hoa [83]

Answer:

The graph is attached below.

Step-by-step explanation:

<em>As you have not added the graph, so I will be solving the function for a graph.</em>

Given the function

f\left(x\right)=x-\frac{1}{x^2}-x-6

x-\mathrm{axis\:interception\:points\:of\:}-\frac{1}{x^2}-6:

\mathrm{x-intercept\:is\:a\:point\:on\:the\:graph\:where\:}y=0

-\frac{1}{x^2}-6=0

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\mathrm{No\:Solution\:for}\:x\in \mathbb{R}

\mathrm{No\:x-axis\:interception\:points}

y-\mathrm{axis\:interception\:point\:of\:}-\frac{1}{x^2}-6:

y\mathrm{-intercept\:is\:the\:point\:on\:the\:graph\:where\:}x=0

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\mathrm{Domain\:of\:}\:-\frac{1}{x^2}-6\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x0\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:0\right)\cup \left(0,\:\infty \:\right)\end{bmatrix}

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The graph is attached below.

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