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Readme [11.4K]
3 years ago
11

In July 2005, the internet was linked by a global network of about 353 million host computers. The number of host computers has

been growing approximately exponentially and was about 37.1 million in July 1998. (a) Find a formula for the number, N, of internet host computers (in millions of computers) as an exponential function of t, the number of years since July 1998, using the continuous exponential model N ( t )
Mathematics
1 answer:
aalyn [17]3 years ago
5 0

Answer:

N(t) = 37.1(1.38)^t

Step-by-step explanation:

Exponential function:

An exponential function has the following format:

N(t) = N(0)(1+r)^t

In which N(0) is the initial value and r is the rate of change, as a decimal.

37.1 million in July 1998

This means that N(0) = 37.1

So

N(t) = 37.1(1+r)^t

In July 2005, the internet was linked by a global network of about 353 million host computers.

This is 2005 - 1998 = 7 years, so N(7) = 353. We use this to find r.

N(t) = 37.1(1+r)^t

353 = 37.1(1+r)^7

(1+r)^7 = \frac{353}{37.1}

\sqrt[7]{(1+r)^7} = \sqrt[7]{\frac{353}{37.1}}

1 + r = (\frac{353}{37.1})^{\frac{1}{7}}

1 + r = 1.38

The formula is:

N(t) = 37.1(1.38)^t

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Solve the following systems of equations by substitution method.
Anuta_ua [19.1K]

Answer:

1) x = 3 and y = -2

2) x = 2 and y = 1

3) x = -2 and y = -1

4) x = 3 and y = 2

5) x = 2

6) x = 2 and y = -3

7) x = 4 and y = 3

8) x = -3 and y = -6

Step-by-step explanation:

1. Answer :

y = -2

4x - 3y = 18 ----(1)

Substitute y = -2 in (1).

4x - 3(-2) = 18

4x + 6 = 18

Subtract 6 from both sides.

4x = 12

Divide both sides by 4.

x = 3

So,

x = 3 and y = -2

2. Answer :

y = 6x - 11 ----(1)

2x + 3y = 7 ----(2)

Substitute y = 6x - 11 in (2).

2x + 3(6x - 11) = 7

2x + 18x - 33 = 7

20x - 33 = 7

Add 33 to both sides.

20x = 40

Divide both sides by 20.

x = 2

Substitute x = 2 in (1).

y = 6(2) - 11

= 12 - 11

= 1

So,

x = 2 and y = 1

3. Answer :

x = -3y - 5 ----(1)

4x - 5y = -3 ----(2)

Substitute x = -3y + 5 in (2).

4(-3y - 5) - 5y = -3

-12y - 20 - 5y = -3

-17y - 20 = -3

Add 20 to both sides.

17y = -17

Divide both sides by 17.

y = -1

Substitute y = -1 in (1).

x = -3(-1) - 5

= 3 - 5

= -2

So,

x = -2 and y = -1

4. Answer :

x = 5y - 7 ----(1)

-2x - 3y = -12 ----(2)

Substitute x = 5y - 7 in (2).

-2(5y - 7) - 3y = -12

-10y + 14 - 3y = -12

-13y + 14 = -12

Subtract 14 from both sides.

-13y = -26

Divide both sides by -13.

y = 2

Substitute y = 2 in (1).

x = 5(2) - 7

= 10 - 7

= 3

So,

x = 3 and y = 2

5. Answer :

5x + 3y - 8 = 0 ----(1)

2x - 3y - 6 = 0 ----(2)

Solve for 3y in (1).

5x + 3y - 8 = 0

Subtract 5x from both sides.

3y - 8 = -5x

Add 8 to both sides.

3y = -5x + 8

Substitute 3y = -5x + 8 in (2).

2x - (-5x + 8) - 6 = 0

2x + 5x - 8 - 6 = 0

7x - 14 = 0

Add 14 to both sides.

7x = 14

Divide both sides by 7.

x = 2

6. Answer :

x - 3y - 11 = 0 ----(1)

5x + y - 7 = 0 ----(2)

Solve for x in (1).

x - 3y - 11 = 0

Subtract 3y and 11 to both sides.

x = 3y + 11 ----(3)

Substitute x = 3y + 11 in (2).

5(3y + 11) + y - 7 = 0

15y + 55 + y - 7 = 0

16y + 48 = 0

Subtract 48 from both sides.

16y = -48

Divide both sides by 16.

y = -3

Substitute y = -3 in (3).

x = 3(-3) + 11

= -9 + 11

= 2

So,

x = 2 and y = -3

7. Answer :

2x - 3y = -1 ----(1)

y = x - 1 ----(2)

Substitute y = x - 1 in (1).

2x - 3(x - 1) = -1

2x - 3x + 3 = -1

-x + 3 = -1

Subtract 3 from both sides.

-x = -4

Multiply both sides by -1.

x = 4

Substitute x = x in (3).

y = 4 - 1

= 3

So,

x = 4 and y = 3

8. Answer :

-4x + y = 6 ----(1)

-5x - y = 21 ----(2)

Solve for y in (1).

-4x + y = 6

Add 4x to both sides.

y = 4x + 6 ----(3)

Substitute y = 4x + 6 in (2).

-5x - (4x + 6) = 21

-5x - 4x - 6 = 21

-9x - 6 = 21

Add 6 to both sides.

-9x = 27

Divide both sides by -9.

x = -3

Substitute x = -3 in (3).

y = 4(-3) + 6

= -12 + 6

= -6

So,

x = -3 and y = -6

Learn more:

brainly.com/question/26555025

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brainly.com/question/26539371

Mathematics

brainly.com/question/26537282

Divisibility - Math

brainly.com/question/26525264

Mathematics

Thank You!

Answered by: ms115

6 0
3 years ago
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