U. s. internet advertising revenue grew at the rate of r(t) = 0.82t + 1.14 (0 ≤ t ≤

4 answers
Question:

U. s. internet advertising revenue grew at the rate of r(t) = 0.82t + 1.14 (0 ≤ t ≤ 4) billion dollars/year between 2002 (t = 0) and 2006 (t = 4). the advertising revenue in 2002 was $5.9 billion.† (a) find an expression f(t) giving the advertising revenue in year t.

Answers

[tex]f(t) = \int 0.82 t +1.14 dt[/tex]

[tex]f(t) = \frac{0.82}{2}t^2 +1.14 t +C[/tex]

Where C is a constant, now using the initial condition we got:

[tex]f(2)= 5.9 = 0.41 (2)^2 + 1.14*2 +C[/tex]

And solving for C we got:

[tex]C= 5.9 -0.41(4) -1.14*2 = 1.98[/tex]

And the function desired for the advertising revenue would be given by:

[tex]f(t) = 0.41t^2 1.14t +1.98[/tex]

With f the amount in billions and the the years since 2002 to 2006.

Explanation:

For this case we have the following function who represent the revenue grew rate:

[tex]r(t) = 0.82t +1.14 , 0 \leq t \leq 4[/tex]

And we want to calculate the Advertising revenue so we need to integrate the function r(t) and we can use the inidital condition t=0 , f(2)= 5.9 billion.

If we integrate the function we got:

[tex]f(t) = \int 0.82 t +1.14 dt[/tex]

[tex]f(t) = \frac{0.82}{2}t^2 +1.14 t +C[/tex]

Where C is a constant, now using the initial condition we got:

[tex]f(2)= 5.9 = 0.41 (2)^2 + 1.14*2 +C[/tex]

And solving for C we got:

[tex]C= 5.9 -0.41(4) -1.14*2 = 1.98[/tex]

And the function desired for the advertising revenue would be given by:

[tex]f(t) = 0.41t^2 1.14t +1.98[/tex]

With f the amount in billions and the the years since 2002 to 2006.

a) [tex]\bf f(t)=\displaystyle\frac{0.82t^2}{2}+1.14t+5.9[/tex]

b) 21.85 billion $

Step-by-step explanation:

Let f(t) be the advertising revenue in year t.

(a) Find an expression f(t) giving the advertising revenue in year t.

Since R(t) is the rate of growth of f(t), then R(t) = f'(t) and

[tex]\bf f(t)=\int_{0}^{t}R(x)dx=\int_{0}^{t}(0.82x+1.14)dx=\\\\=0.82\int_{0}^{t}xdx+1.14\int_{0}^{t}dx=\displaystyle\frac{0.82t^2}{2}+1.14t+C[/tex]

where 0 ≤ t ≤ 4 and C is a constant.

The advertising revenue in 2002 (t=0) was $5.9 billion, so f(0) = 5.9. Since f(0) = C, then C=5.9 and  

[tex]\bf \boxed{f(t)=\displaystyle\frac{0.82t^2}{2}+1.14t+5.9}[/tex]

(b) If the trend continued, what was the Internet advertising revenue in 2007?

if t=0 represents the year 2002 then t=5 represents the year 2007 and the revenue was f(5)

[tex]\bf f(5)=\displaystyle\frac{0.82(5^2)}{2}+1.14*5+5.9=\boxed{21.85\;billion}[/tex]

there was no passage listed but using the process of elimination i'd say it was d. the high economy would do the opposite. i don't see how low prices would affect it. high unemployment is possible but i feel d is the better answer.

prokaryotic cells do not contain a nucleus or any other membrane-bound organelle. eukaryotic cells contain membrane-bound organelles, including a nucleus. so i belive the answer is a : p hope it

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