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3 years ago

10

(5⅐)⁴=( ) OPTIONS : A.) (5⁴)⅐ B.)5²⁸ C.)(5)¹/²⁸ D.)(5)⁷/⁴

1 answer:

NemiM [27]3 years ago

5 0

Answer:

a

Step-by-step explanation:

4*1/7

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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 $\lim_{t \to \infty} E(t)$

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

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

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Suppose we want to build a rectangular storage container with open top whose volume is $$12 cubic meters. Assume that the cost o

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Step-by-step explanation:

Given

Volume of box $=12 m^3$

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4=B^2\times L[/tex]

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Cost of side walls $ $8/m^2$

Cost of base $ $12/m^2$

Cost of side walls $=(2LH+2BH)\cdot 8$

Cost of base $=12LB$

Total cost $C=16LH+16BH+12LB$

$C=48LB+48B^2+12LB$

$C=\frac{192}{B}+48B^2+\frac{48}{B}$

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