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Ymorist [56]
3 years ago
10

A company with $60,000 in sales last year had $69,000 in sales this year. Assuming the company's sales continue to grow at the s

ame rate each year, what will the the annual sales be two years from now
Mathematics
2 answers:
aalyn [17]3 years ago
5 0

Subtract 69000 from 60000. Answer becomes 9000. So you add 9000 to 69000 2 times. Your answer becomes 87000.

ella [17]3 years ago
5 0

Answer:

$91252.5

Step-by-step explanation:

We need to determine the rate at which the sales grow. We can do this by taking this years sales of $69000 by the previous years sales $60000:

69000/60000=1.15

The sales increase 1.15 times each year. For two more years:

=69000\cdot{1.15}=79350

=79350\cdot{1.15}=91252.5

The annual sales will be $91252.5 after two years.

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In 1980, the number of high school dropouts in a country was 5217 thousand. By 2008, this number had decreased to 3120 thousand.
Veseljchak [2.6K]

Answer:

74893 students per year

Step-by-step explanation:

The Average rate of change is calculated as

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= 3120 thousand - 5217 thousands/ 2008 - 1980

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Find the length of the missing side
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8 0
3 years ago
One number added to three times another number is 24. Five times the first number added to three times to other number is 36. Fi
Marrrta [24]

The numbers are 3 and 7

Step-by-step explanation:

Let the first number be x and the second number be y

Then according to given statement

One number added to three times another number is 24

x+3y=24     Eqn 1

Five times the first number added to three times to other number is 36

5x+3y=36      Eqn 2

Subtracting Equation 1 from equation 2

5x+3y-(x+3y)=36-24\\5x+3y-x-3y=12\\4x=12

Dividing both sides by 4

\frac{4x}{4}=\frac{12}{4}\\x=3

Putting x=3 in equation 1

3+3y=24\\3y=24-3\\3y=21

Dividing both sides by 3

\frac{3y}{3} = \frac{21}{3}\\y=7

Hence,

The numbers are 3 and 7

Keywords: Linear equations, Variables

Learn more about linear equations at:

  • brainly.com/question/5059091
  • brainly.com/question/5069437

#LearnwithBrainly

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4 years ago
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