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AlekseyPX
3 years ago
15

One estimate of the population of the world on January 1, 2005, is

Mathematics
1 answer:
inn [45]3 years ago
4 0

Answer:

8,566,379,470 people

Step-by-step explanation:

Let's start simple.  In order to find the population increase on January 1, 2006, we need to multiply 6,486,915,022 by 1.4% and add it to 6,486,915,022.

  1. 6,486,915,022*1.4% = 90,816,810.308
  2. 90,816,810.308+6,486,915,022 = 6,577,731,832.31 people on January 2006.

Note that the above two steps gives the same answer as 6,486,915,022*1.014.

So we need to do this for each year.  20 years pass between 1/1/2005 and 1/1/2025.  

We need to do 6,486,915,022*1.014*1.014*1.014... 20 times.

This is equivalent to 6,486,915,022*1.014^{20}.

Multiplying it out gives us 8566379470.2 = 8,566,379,470 people.

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matt is standing on top of acliff 305 feet above a lake. the measurement of the angle of depression to a boat on the lake is 42.
Alex17521 [72]

Answer:

Matt is 338.7 feet.

Step-by-step explanation:

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3 years ago
There are 32 male performers in a dance recital. The ratio of men to women is 4:5. How many women are in the recital
Yanka [14]

Answer:

40

Step-by-step explanation:

if the ratio from men to woman is 4:5 and men are 32 that means w need to find a way to get to 32 using 4 and 8 x 4 is 32. so if we use 8 to get to 32 we need to use 8 on 5 aswell for it to be an equal ratio.

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3 years ago
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son4ous [18]

Answer:

C

Step-by-step explanation:

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Therefore, the answer is C. Hope this helps!

5 0
3 years ago
Consider the function f ( x ) = − 2 x 3 + 27 x 2 − 108 x + 9 . For this function there are three important open intervals: ( − [
Darya [45]

Answer:

<em>A=3 and B=6</em>

Step-by-step explanation:

<u>Increasing and Decreasing Intervals of Functions</u>

Given f(x) as a real function and f'(x) its first derivative.

If f'(a)>0 the function is increasing in x=a

If f'(a)<0 the function is decreasing in x=a

If f'(a)=0 the function has a critical point in x=a

As we can see, the critical points may define open intervals where the function has different behaviors.

We have

f ( x ) = - 2 x^3 + 27 x^2 - 108 x + 9

Computing the first derivative:

f' ( x ) = - 6 x^3 + 54 x - 108

We find the critical points equating f'(x) to zero

- 6 x^3 + 54 x - 108=0

Simplifying by -6

x^2 -9 x +18=0

We get the critical points

x=3,\ x=6

They define the following intervals

(-\infty,3),\ (3,6),\ (6,+\infty)

Thus A=3 and B=6

3 0
3 years ago
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