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3 years ago

5

A town has a population of 14000 and grows at 4.5% every year. To the nearest tenth of a year, how long will it be until the pop

ulation will reach 41500?

1 answer:

BARSIC [14]3 years ago

3 0

Answer:

around 6.6 years

Step-by-step explanation:

41,500=14,000*.45*x

41,500=6,300*x

41,000/6,300=x

6.5873=x

6.6=x

hope this helps!

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If f(x)=3x-7 and g(x)=6x+4, what is f(x) + g(x)?

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If g(x) = 3x + 1, evaluate g(-2).

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G(x) = 3x + 1, evaluate g(-2).

Simply plug in (- 3) in g(x)

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3 years ago

The number of views on an interesting video after it's uploaded is represented by the following table:

defon

Answer:

The model for V(t) ,the number of views, t ,months after it's uploaded, best fits the data is $,V(t)=50(6.25)^t$

Step-by-step explanation:

Time (months) Views

0 50

2 313

4 1950

6 12,210

8 76,300

10 476,800

$(x_1,y_1)=(0,50)\\(x_2,y_2)=(2,313)\\(x_3,y_3)=(4,1950)$

$Slope = \frac{y_2-y_1}{x_2-x_1}$

$Slope = \frac{313-50}{2-0}$

Slope =131.5

$Slope = \frac{y_3-y_2}{x_3-x_2}$

$Slope = \frac{1950-313}{4-2}$

Slope =818.5

Since the slope is not same . So, it is not a linear relationship

$\frac{y_2}{y_1}=\frac{313}{50}=6.25\\\frac{y_3}{y_2}=\frac{1950}{313}=6.25$

Since the ratio between the consecutive outputs is same so, it is exponential relationship.

Formula : $y=ab^t$

Where a is the initial value = 50

b = rate of change = 6.25

The model for V(t) ,the number of views, t ,months after it's uploaded, best fits the data

So $,V(t)=50(6.25)^t$

5 0

3 years ago