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Naddika [18.5K]
3 years ago
10

keith is an elevator operator in a big hotel. A(t)A(t)A, left parenthesis, t, right parenthesis models Keith's altitude (in mete

rs relative to the ground) as a function of time (in minutes).Match each statement with the feature of the graph that most closely corresponds to it.

Mathematics
1 answer:
Lunna [17]3 years ago
3 0

Answer:

Step-by-step explanation:

A relative maximum (or minimum) point is a point that is higher (or lower) than all of the points surrounding it.

This graph has a relative minimum point where A(t)=-12A(t)=−12A, left parenthesis, t, right parenthesis, equals, minus, 12, which means that Keith's lowest altitude was \textit{12}12start text, 12, end text meters below the ground.

A positive (or negative) interval is a domain interval over which the function values are all positive (or negative).

Since A(t)>0A(t)>0A, left parenthesis, t, right parenthesis, is greater than, 0 over the interval [6.5,10][6.5,10]open bracket, 6, point, 5, comma, 10, close bracket, this is a positive interval. This means that between the \textit{6.5}6.5start text, 6, point, 5, end text- and \textit{10}10start text, 10, end text-minute marks, Keith was above the ground.

An increasing (or decreasing) interval is a domain interval over which the function values increase (or decrease) as the input variable increases.

In this graph, the interval, 5, point, 5, comma, 8, close bracket is an increasing interval. This means that between the   end text-minute marks, the elevator was going up.

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A tennis coach took his team out for lunch and bought 8 hamburgers and 5 fries for $24. The players were still hungry so the coa
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The cost of 1 hamburger is $ 2.5 and cost of 1 fries is $ 0.8

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

Let "f" be the cost of 1 fries

Let "h" be the cost of 1 hamburger

<em><u>Given that, tennis coach took his team out for lunch and bought 8 hamburgers and 5 fries for $24</u></em>

8 x cost of 1 hamburger + 5 x cost of 1 fries = 24

8 \times h + 5 \times f = 24

8h + 5f = 24 -------- eqn 1

<em><u>The players were still hungry so the coach bought six more hamburgers and two more fries for $16.60</u></em>

6 x cost of 1 hamburger + 2 x cost of 1 fries = 16.60

6 \times h + 2 \times f = 16.60

6h + 2f = 16.60 ------ eqn 2

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

Multiply eqn 1 by 2

16h + 10f = 48 ------ eqn 3

Multiply eqn 2 by 5

30h + 10f = 83 -------- eqn 4

<em><u>Subtract eqn 3 from eqn 4</u></em>

30h + 10f = 83

16h + 10f = 48

( - ) ----------------------

14h = 35

<h3>h = 2.5</h3>

Substitute h = 2.5 in eqn 1

8(2.5) + 5f = 24

20 + 5f = 24

5f = 4

<h3>f = 0.8</h3>

Thus cost of 1 hamburger is $ 2.5 and cost of 1 fries is $ 0.8

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