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3 years ago

Help me write this two numbers in words 1,328,800,16 15,802

Mathematics

1 answer:

Zigmanuir [339]3 years ago

7 0

one hundred thirty two million, eight hundred eighty thousand and sixteen

(Your commas are in the incorrect spot)

fifteen thousand, eight hundred and two

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Mentally confused, please help.

garik1379 [7]

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2 years ago

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Brad says that if a second number is 125% of the first number, then the first number must be 75% of the second number. Is he cor

olga55 [171]

Answer: He is not correct.

Steps:

Let x1 and x2 be the first and second number, respectively.

$x_2 = \frac{5}{4}x_1\\\frac{4}{5}x_2 =x_1$

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2 years ago

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

nexus9112 [7]

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

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2 years ago

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3 years ago

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How to simplify 4x^-3<br><br><br><br> Is the answer 1/4a^3 ?

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3 years ago