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AnnZ [28]
3 years ago
8

PLEASE PLEASE HELP. I’ll give branliest....I’m begging.

Mathematics
1 answer:
morpeh [17]3 years ago
5 0

Answer:

Step-by-step explanation:

Things to remember,

1). Any point on the graph of a function is represented by [x, f(x)].

2). x-coordinates are the input values on x-axis and f(x) is the output value of the function on y-axis.

A). For x = 2, value of f(2) will be,

    f(2) = 4

B). For x = 5, value of f(5) will be,

    f(5) = 7

C). Since, 4 < 7

   Therefore, f(2) is less than f(5).

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The projected rate of increase in enrollment at a new branch of the UT-system is estimated by E ′ (t) = 12000(t + 9)−3/2 where E
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Answer:

The projected enrollment is \lim_{t \to \infty} E(t)=10,000

Step-by-step explanation:

Consider the provided projected rate.

E'(t) = 12000(t + 9)^{\frac{-3}{2}}

Integrate the above function.

E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt

E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c

2000=-\frac{24000}{3}+c

2000=-8000+c

c=10,000

Therefore, E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000

Now we need to find \lim_{t \to \infty} E(t)

\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000

\lim_{t \to \infty} E(t)=10,000

Hence, the projected enrollment is \lim_{t \to \infty} E(t)=10,000

8 0
3 years ago
Write an expression for the sequence of operations described below. divide 10 by the sum of n and m Do not simplify any part of
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Answer:

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Step-by-step explanation:

divide 10 by the sum of n and m

(m+n)/10

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last one

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