Does anyone know the answer!!!!

Mathematics

2 answers:

Anna35 [415]3 years ago

5 0

Answer:

I believe is B or -3/5

Good luck!

egoroff_w [7]3 years ago

3 0

Answer:

Step-by-step explanation:

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Jason runs 440 yards in 75 seconds. At this rate, how many minutes does it take

Leya [2.2K]

Answer:

5 minutes

Step-by-step explanation:

We know that 1760 yards are in a mile. We can convert the given info into a proportion. 440/75 = 1760/x, where x is the number of seconds it takes to run a mile. Cross-multiply to get 440x = 75*1760. Divide both sides by 440 to get x = 75*4 = 300 seconds. The problem asks for how many minutes so you have to convert 300 seconds to minutes. To do this, we have to divide 300 by 60 to get 5 minutes.

3 0

3 years ago

Read 2 more answers

Please help me with question when your answering it make sure you explain it so I get it

IrinaK [193]

So plug in the number of passengers into each equation and see which answer matches Layton's. then from all the answers you find from plugging in the number of passengers, choose the cheapest one for B

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4 years ago

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

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