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# How can you predict the value of a quantity given the rate of change?

• Réponse publiée par: elaineeee
Given:
vi=80 gal
cin=2 lb/gal
qin=2 gal/min
qout=2 gal/min
(since qin=qout, the volume is constant)
assume xo(original amount)=0
cout=x/v=x/80

dx/dt=cinqin-coutqout

dx/dt=(2)(/80)(2)

dx/dt=2[(160-x)/80]

dx/dt=(160-x)/40 (separable de)

dx/(160-x)=dt/40 (integrate)

-ln(160-x)=(1/40)t+c (general solution)

initial condition: xo=0,t=0
-ln(160-0)=(1/40)(0)+c
c=-ln(160)=-5.0752

-ln(160-x)=(1/40)t-5.0752 (particular solution)

a.) x at any time t x(t)

-ln(160-x)=(1/40)t-5.0752

ln[(160-x)^-1]=(0.025)t-5.0752

e^{ln[(160-x)^-1]}=e^(0.025t-5.0752)

(160-x)^-1=e^(0.025t-5.0752)

1/(160-x)=e^(0.025t-5.0752) (isolate x)

1/[e^(0.025t-5.0752)]=160-x

x(t)=160-{1/[e^(0.025t-5.0752)]}

b. if c=1 lb/gal, t=?

from c=x/v

1=x/80
x=80

-ln(160-x)=0.025t-5.0752

-ln(160-80)=0.025t-5.0752
-ln(80)=0.025t-5.0752

0.025t=5.0752-ln(80)

0.025t=0.6932

t=27.728 mins
• Réponse publiée par: joyce5512

intindihin mo lang tulad ng pag intindi sa jowa mo siz

• Réponse publiée par: elaineeee
12 a.m.

Step-by-step explanation:

Just find the LCM or least common multiple of the minute-intervals of their barking. Then, add it to 11 p.m. to know the answer.

Finding the time when the dogs bark together:

4(ban): 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 603(tag): 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 605(black): 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60

They bark together every 60 minutes (1 hour).

11 p.m. + 1 hour = 12 a.m.

• Réponse publiée par: kambalpandesal23

0.8km

Step-by-step explanation:

let A=arthur

let M=merlin

A=7.6 kph

M=6.8kph

in one hour of running, A ran 7.6 km and M ran 6.8km

to find the distance between each person, subtract M from A

7.6km-6.8km=0.8km

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How can you predict the value of a quantity given the rate of change?...