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gayaneshka [121]
3 years ago
6

The population of a small town can be modeled by the exponential function p=14,512(1.03)t, where t is the number of years after

2005.
What is the significance of the value of 14,512
Mathematics
2 answers:
Ber [7]3 years ago
8 0
Hahahshxhchxhdhsbsbbsshhs
Rom4ik [11]3 years ago
4 0

Answer:

the significant value of this is the number of people the town started with.

Step-by-step explanation:

the 1.03 is how much the population increased per year after

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Use a benchmark to find an equivalent percent for each fraction. 9/10 and 2/5. Please help! :o will mark brainliest if can!
denpristay [2]

Answer

9/10=<u>90%</u> and 2/5=<u>40%</u>

Explanation

9÷10=0.9   0.9x100=90%

2÷5=0.4   0.4x100=40%

<em>hope this helps!</em>

<em>have a wonderful day :)</em>

3 0
3 years ago
R
Shalnov [3]

Answer:

x = 54

y = 47.5

Step-by-step explanation:

If two lines p and q are parallel and line r is a transversal intersecting these lines at two different points,

(x + 56)° = (2x + 2)° [corresponding angles]

2x - x = 56 - 2

x = 54

Similarly, lines r and s are parallel lines and q is a transversal line intersecting these lines,

(y + 7)° + (3y - 17)°= 180° [Consecutive exterior angles]

4y - 10 = 180

4y = 190

y = 47.5

3 0
3 years ago
1kg-1/4-3/8=<br><br>what is it please help​
olga_2 [115]

Answer: it's either kg-5/8 or kg + -5/8

Step-by-step explanation:

4 0
2 years ago
Consider the following function.
Kryger [21]

Answer:

See below

Step-by-step explanation:

I assume the function is f(x)=1+\frac{5}{x}-\frac{4}{x^2}

A) The vertical asymptotes are located where the denominator is equal to 0. Therefore, x=0 is the only vertical asymptote.

B) Set the first derivative equal to 0 and solve:

f(x)=1+\frac{5}{x}-\frac{4}{x^2}

f'(x)=-\frac{5}{x^2}+\frac{8}{x^3}

0=-\frac{5}{x^2}+\frac{8}{x^3}

0=-5x+8

5x=8

x=\frac{8}{5}

Now we test where the function is increasing and decreasing on each side. I will use 2 and 1 to test this:

f'(2)=-\frac{5}{2^2}+\frac{8}{2^3}=-\frac{5}{4}+\frac{8}{8}=-\frac{5}{4}+1=-\frac{1}{4}

f'(1)=-\frac{5}{1^2}+\frac{8}{1^3}=-\frac{5}{1}+\frac{8}{1}=-5+8=3

Therefore, the function increases on the interval (0,\frac{8}{5}) and decreases on the interval (-\infty,0),(\frac{8}{5},\infty).

C) Since we determined that the slope is 0 when x=\frac{8}{5} from the first derivative, plugging it into the original function tells us where the extrema are. Therefore, f(\frac{8}{5})=1+\frac{5}{\frac{8}{5}}-\frac{4}{\frac{8}{5}^2 }=\frac{41}{16}, meaning there's an extreme at the point (\frac{8}{5},\frac{41}{16}), but is it a maximum or minimum? To answer that, we will plug in x=\frac{8}{5} into the second derivative which is f''(x)=\frac{10}{x^3}-\frac{24}{x^4}. If f''(x)>0, then it's a minimum. If f''(x), then it's a maximum. If f''(x)=0, the test fails. So, f''(\frac{8}{5})=\frac{10}{\frac{8}{5}^3}-\frac{24}{\frac{8}{5}^4}=-\frac{625}{512}, which means (\frac{8}{5},\frac{41}{16}) is a local maximum.

D) Now set the second derivative equal to 0 and solve:

f''(x)=\frac{10}{x^3}-\frac{24}{x^4}

0=\frac{10}{x^3}-\frac{24}{x^4}

0=10x-24

-10x=-24

x=\frac{24}{10}

x=\frac{12}{5}

We then test where f''(x) is negative or positive by plugging in test values. I will use -1 and 3 to test this:

f''(-1)=\frac{10}{(-1)^3}-\frac{24}{(-1)^4}=-34, so the function is concave down on the interval (-\infty,0)\cup(0,\frac{12}{5})

f''(3)=\frac{10}{3^3}-\frac{24}{3^4}=\frac{2}{27}>0, so the function is concave up on the interval (\frac{12}{5},\infty)

The inflection point is where concavity changes, which can be determined by plugging in x=\frac{12}{5} into the original function, which would be f(\frac{12}{5})=1+\frac{5}{\frac{12}{5}}+\frac{4}{\frac{12}{5}^2 }=\frac{43}{18}, or (\frac{12}{5},\frac{43}{18}).

E) See attached graph

5 0
3 years ago
Ok so I need help on this ,asap plz :)
Tju [1.3M]
Hope that makes sense !!!!

3 0
3 years ago
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