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olga2289 [7]
3 years ago
5

The population of Oak Forest is increasing at a rate of 4% per year. If the population is 74,145 today, what will it be in three

years?
Mathematics
1 answer:
Shtirlitz [24]3 years ago
4 0

Answer:

83,403

Step-by-step explanation: Take 74,145 and multiply it by 4%. Then take that number and add it to the 74,145 and that'll give you year one. For year 2 you'll take your total from year 1 and multiply it by the 4% growth rate then you'll add the 4% to what your ending from year 1 and that'll give you your total growth after 2 years. Then you'll take your ending total from year 2 and multiply it by 4% and then you'll add that 4% to the total end from year 2 and that'll give you your total growth of 4% every year for 3 consecutive years.

Hope this helps!

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Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x4 ln(x) (a) Find the interval on which f is incre
Ainat [17]

Answer: (a) Interval where f is increasing: (0.78,+∞);

Interval where f is decreasing: (0,0.78);

(b) Local minimum: (0.78, - 0.09)

(c) Inflection point: (0.56,-0.06)

Interval concave up: (0.56,+∞)

Interval concave down: (0,0.56)

Step-by-step explanation:

(a) To determine the interval where function f is increasing or decreasing, first derive the function:

f'(x) = \frac{d}{dx}[x^{4}ln(x)]

Using the product rule of derivative, which is: [u(x).v(x)]' = u'(x)v(x) + u(x).v'(x),

you have:

f'(x) = 4x^{3}ln(x) + x_{4}.\frac{1}{x}

f'(x) = 4x^{3}ln(x) + x^{3}

f'(x) = x^{3}[4ln(x) + 1]

Now, find the critical points: f'(x) = 0

x^{3}[4ln(x) + 1] = 0

x^{3} = 0

x = 0

and

4ln(x) + 1 = 0

ln(x) = \frac{-1}{4}

x = e^{\frac{-1}{4} }

x = 0.78

To determine the interval where f(x) is positive (increasing) or negative (decreasing), evaluate the function at each interval:

interval                 x-value                      f'(x)                       result

0<x<0.78                 0.5                 f'(0.5) = -0.22            decreasing

x>0.78                       1                         f'(1) = 1                  increasing

With the table, it can be concluded that in the interval (0,0.78) the function is decreasing while in the interval (0.78, +∞), f is increasing.

Note: As it is a natural logarithm function, there are no negative x-values.

(b) A extremum point (maximum or minimum) is found where f is defined and f' changes signs. In this case:

  • Between 0 and 0.78, the function decreases and at point and it is defined at point 0.78;
  • After 0.78, it increase (has a change of sign) and f is also defined;

Then, x=0.78 is a point of minimum and its y-value is:

f(x) = x^{4}ln(x)

f(0.78) = 0.78^{4}ln(0.78)

f(0.78) = - 0.092

The point of <u>minimum</u> is (0.78, - 0.092)

(c) To determine the inflection point (IP), calculate the second derivative of the function and solve for x:

f"(x) = \frac{d^{2}}{dx^{2}} [x^{3}[4ln(x) + 1]]

f"(x) = 3x^{2}[4ln(x) + 1] + 4x^{2}

f"(x) = x^{2}[12ln(x) + 7]

x^{2}[12ln(x) + 7] = 0

x^{2} = 0\\x = 0

and

12ln(x) + 7 = 0\\ln(x) = \frac{-7}{12} \\x = e^{\frac{-7}{12} }\\x = 0.56

Substituing x in the function:

f(x) = x^{4}ln(x)

f(0.56) = 0.56^{4} ln(0.56)

f(0.56) = - 0.06

The <u>inflection point</u> will be: (0.56, - 0.06)

In a function, the concave is down when f"(x) < 0 and up when f"(x) > 0, adn knowing that the critical points for that derivative are 0 and 0.56:

f"(x) =  x^{2}[12ln(x) + 7]

f"(0.1) = 0.1^{2}[12ln(0.1)+7]

f"(0.1) = - 0.21, i.e. <u>Concave</u> is <u>DOWN.</u>

f"(0.7) = 0.7^{2}[12ln(0.7)+7]

f"(0.7) = + 1.33, i.e. <u>Concave</u> is <u>UP.</u>

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just olya [345]

Answer:

4.32

Step-by-step explanation:

You can multiply 16 by 0.27 to get the answer.

16*0.27=4.32

You can check your work with estimation because 27% is about 1/4 of the number, and just by looking at the number, 4 is 1/4 of the number.

6 0
3 years ago
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In 1934 there was an extreme drought and the Great Plains and the number 1,934 is the value of the nine and the hundredth place
WARRIOR [948]

It should be noted that No. the value of the 9 in the hundred place is not ten times the value of 3 in the tens place.

<h3>How to illustrate the information?</h3>

It should be noted that the value of 3 in 1934 is 30 and the value of 9 is 900.

1934 = 1000 + 900 + 30 + 4

Therefore the value of 9 will be:

= 30 × 3

Therefore, the the value of the 9 in the hundred place is 30 times the value of 3 in the tens place.

Learn more about place values on:

brainly.com/question/2041524

#SPJ1

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1 year ago
To get the echo of a positive integer, we write it twice in a row without a space. For example, the echo of 2022 is 20222022. Is
Lynna [10]

The <em>echo</em> number 20222022202220222022 is the <em>perfect</em> square of 4496890281.

<h3>What echo number is a perfect square</h3>

An <em>echo</em> number has a <em>perfect</em> square if its square root is also a <em>natural</em> number. After some iterations we found that <em>echo</em> number 20222022202220222022 is a <em>perfect</em> square:

\sqrt{20222022202220222022} = 4496890281

The <em>echo</em> number 20222022202220222022 is the <em>perfect</em> square of 4496890281. \blacksquare

To learn more on natural numbers, we kindly invite to check this verified question: brainly.com/question/17429689

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3 years ago
There were 80 people in the parade. A total of 5/8 of the people played a musical instrument.
creativ13 [48]


50 People played an instrument. Every 10 people is represented as a 1 in the fraction. 5=50
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