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vazorg [7]
2 years ago
8

Suppose the population of a town is 15,200 and is growing 2% each year. Write an equation to model the population growth. Predic

t the population after 10 years. y = 15,200 ∙ 2x; about 15,564,800 people y = 2 ∙ 15,200x; about 304,000 people y = 15,200 ∙ 1.02x; about 18,529 people y = 15,200 ∙ 2x; about 18,529 people
Mathematics
1 answer:
larisa [96]2 years ago
5 0
For this case, what you need to know is that to model this problem, you must use equations of the potential type:
 y = A * (b) ^ t
 Where:
 A: initial population.
 b: growth rate.
 t: time.
 Substituting values we have:
 y = 15200 * (1.02) ^ t
 After 10 years we have:
 y = 15200 * (1.02) ^ 10
 y = 18529
 Answer:
 An equation to model of the population growth is:
 y = 15,200 ∙ 1.02 ^ x
 the population after 10 years is:
 about 18,529
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1. First, you must apply the formula for calculate the sum of the interior angles of a regular polygon, which is shown below:


 (n-2) × 180°


 "n" is the number of sides of the polygon (n=5).


 2. Then, the sum of the interior angles of the pentagon, is:


 (5-2)x180°=540°


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 α+α+2α+2α+2α=540°

 8α=540°

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2 years ago
Match each type of function with a characteristic of its graph. Enter the letter of the correct characteristic next to the type
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Answer:

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Quadratic Function: Its graph is a parabola (B).

Inverse Variation Function: Its graph has both a horizontal asymptote and a vertical asymptote (F).

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2 years ago
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For the level 3 course, exam hours cost twice as much as workshop hours, workshop hours cost twice as much as lecture hours. How
natulia [17]
<h2>Answer</h2>

Cost of lectures = $7.33 per hour

<h2>Explanation</h2>

Let e the cost of the exam hours

Let w be the cost of the workshop hours

Let l be the cost of the lecture hours.

We know from our problem that exam hours cost twice as much as workshop, so:

e=2w equation (1)

We also know that workshop hours cost twice as much as lecture hours, so:

w=2l equation (2)

Finally, we also know that 3hr exams 24hr workshops  and 12hr lectures cost $528, so:

3e+24w+12l=528 equation (1)

Now, lets find the value of l:

Step 1.  Solve for l in equation (3)

3e+24w+12l=528

12l=528-3e-24w equation (4)

Step 2. Replace equation (1) in equation (4) and simplify

12l=528-3e-24w

12l=528-3(2w)-24w

12l=528-6w-24w

12l=528-30w equation (5)

Step 3. Replace equation (2) in equation (5) and solve for l

12l=528-30w

12l=528-30(2l)

12l=528-60l

72l=528

l=\frac{528}{72}

l=\frac{22}{3}

l=7.33

Cost of lectures  = $7.33 per hour



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2 years ago
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Lying in the same straight line i think..

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2 years ago
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Aluminium coated dewars are cylindrical flasks used to safely store and transport liquid nitrogen, dry ice etc. A company manufa
Nitella [24]

This question is Incomplete because it lacks the diagram showing the containers as well as the required options. Find attached to this answer the appropriate diagram.

Complete Question

Aluminium coated dewars are cylindrical flasks used to safely store and transport liquid nitrogen, dry ice etc. A company manufactures such dewars of different sizes. Shown here are two dewars, both of the same height but of different diameters. How much aluminium will be required to cover Dewar A compared to that required for Dewar B? (Note: The base and the lid are NOT made up of aluminium.)

A) 1/3 times

B) 3 times

C) 9 times

D) The same amount of aluminum will be required.

Answer:

B) 3 times

Step-by-step explanation:

From the question, we are told that we are to find the amount of aluminum to cover Dewar A and B , we are also told that the base and the lid are not covered with aluminum.

From this information, we can determine that we are to find the curved or Lateral surface area of the cylinder because the base and lid are not included

The curved surface area of a cylinder is calculated as 2πrh

We told that both cylinders have the same height. Hence,

For Dewar A

We have Diameter of 30 cm, radius = Diameter ÷ 2 = 30cm ÷ 2 = 15cm.

Height = h

The curved surface area = 2πrh

= 2 × π × 15 ×h

= 30πhcm²

For Dewar B

We have Diameter of 10 cm, radius = Diameter ÷ 2 = 10cm ÷ 2 = 5cm.

Height = h

The curved surface area = 2πrh

= 2 × π × 5 ×h

= 10πhcm²

When we compare the curved surface Dewar A to the curved surface area of Dewar B

Dewar A : Dewar B

30πhcm² : 10πhcm²

3(10πhcm²) : 10πhcm²

From the above comparison, we can see that 3 times more aluminum is required to cover Dewar A compared to Dewar B.

6 0
1 year ago
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