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3 years ago

Simplify the expression and find the missing terms.

Mathematics

1 answer:

zysi [14]3 years ago

5 0

What are the terms? I want to help, but I can't without the terms.

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Point A is located at (1, 5), and point M is located at (−1, 6). If point M is the midpoint of segment AB, find the location of

sashaice [31]

Ans: B=(-3, 7)

Step by step explanation:

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3 years ago

How do you solve 75.25-11.92?

slamgirl [31]

Your answer would be 63.33. ((: ❤️

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

8 0

2 years ago

65 points!!

madreJ [45]

I think #1 is -1/2 but im not sure i dont know about the other answers

4 0

3 years ago

Read 2 more answers

for y=x^2-10 do the following . A) Sketch a graph of the equation. B) Identify the vertex. C)Compare the graph of y=f(x) to the

Andreas93 [3]

A ) A graph of the equation is in the attachment.
B ) The vertex ( 0, -10 )
C ) The graph of y = x² is translated for 10 units down.

Download docx

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3 years ago