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2 years ago

PLEASE NO SPAM!!!! I REALLY NEED TO GET A GOOD GRADE ON THIS!

Mathematics

1 answer:

kenny6666 [7]2 years ago

3 0

Answer:

I think it is A

Step-by-step explanation:

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A pond is being drained by a pump. After 3 hours, the pond is half empty. A second pump is put into operation and together the t

aleksandr82 [10.1K]

The first pump empties half the pond in 3 hours, so in 1/6 that time (1/2 hour), it empties (1/6)·(1/2) = 1/12 of the pond.

The second pump empties the other 5/12 of the pond in that half hour, so has a pumping rate of (1/2 h)/(5/12 pond) = (6/5 h)/pond.

The second pump could do the entire job alone in 1 hour and 12 minutes.

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3 years ago

Read 2 more answers

HELP I NEED HELP ASAP

Andrei [34K]

Answer:

D its D

Step-by-step explanation:

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3 years ago

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A small military base housing 1,000 troops, each of whom is susceptible to a certain virus infection. Assuming that during the c

slava [35]

Answer:

$I=\frac{1000}{exp^{0,806725*t-0.6906755}+1}$

Step-by-step explanation:

The rate of infection is jointly proportional to the number of infected troopers and the number of non-infected ones. It can be expressed as follows:

$\frac{dI}{dt}=a*I*(1000-I)$

Rearranging and integrating

$\frac{dI}{dt}=a*I*(1000-I)\\\\\frac{dI}{I*(1000-I)}=a*dt\\\\\int\frac{dI}{I*(1000-I)}=\int a*dt\\\\-\frac{ln(1000/I-1)}{1000}+C=a*t$

At the initial breakout (t=0) there was one trooper infected (I=1)

$-\frac{ln(1000/1-1)}{1000}+C=0\\\\-0,006906755+C=0\\\\C=0,006906755$

In two days (t=2) there were 5 troopers infected

$-\frac{ln(1000/5-1)}{1000}+0,006906755=a*2\\\\-0,005293305+0,006906755=2*a\\a = 0,00161345 / 2 = 0,000806725$

Rearranging, we can model the number of infected troops (I) as

$-\frac{ln(1000/I-1)}{1000}+0,006906755=0,000806725*t\\\\-\frac{ln(1000/I-1)}{1000}=0,000806725*t-0,006906755\\-ln(1000/I-1)=0,806725*t-0.6906755\\\\\frac{1000}{I}-1=exp^{0,806725*t-0.6906755} \\\\\frac{1000}{I}=exp^{0,806725*t-0.6906755}+1\\\\I=\frac{1000}{exp^{0,806725*t-0.6906755}+1}$

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3 years ago

V=1/2at^2 solve for t

malfutka [58]

$\dfrac{1}{2}at^2=V\qquad\text{multiply both sides by 2}\\\\at^2=2V\qquad\text{divide both sides by}\ a\neq0\\\\t^2=\dfrac{2V}{a}\to\boxed{t=\sqrt{\dfrac{2V}{a}}}$

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3 years ago

-3(1-2x)-3=-20-8x what is the work and answer

larisa86 [58]

You have to keep order of operations in mind.

1. Take care of the parentheses; distribute.
-3+6x-3=-20-8x
Simplified: 6x-6=-20-8x

2. Isolate the x variable.
Add 8x to both sides and add 6 to both sides.
14x=-14

3. Solve for x.
Divide by 14 on both sides.
x=-1

4 0

3 years ago