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3 years ago

13

Rx + sx/t = 1 solve for x

2 answers:

MAXImum [283]3 years ago

4 0

(rx + sx) / t = 1....multiply both sides by t
rx + sx = t...factor the left side
x(r + s) = t ...divide both sides by (r + s)
x = t / (r + s) <===

abx - cdx = 15m...factor left side
x(ab - cd) = 15m...divide both sides by (ab - cd)
x = (15m) / (ab - cd) <====

Lubov Fominskaja [6]3 years ago

3 0

Thanks for your question!

First thing we have to do here is take t to the other side by multiplication:

rx + sx = t

Now, group together, x as the GCF:

x(r+s) = t

Now, divide

x = t/(r+s)

Hope this helps!

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Alisiya [41]

Answer:

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Step-by-step explanation:

In this equation, a=5, b=-1, c=-4.

Plug these into the quadratic formula:

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Now, simplify the equations:

x = $\frac{1+\sqrt{81} }{10}$ = $\frac{1+9}{10}$ = $\frac{10}{10}$ = 1

and

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5 0

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

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2 years ago

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Step-by-step explanation:

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2 years ago

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3 years ago

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garri49 [273]

Answer:

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3 0

3 years ago