If a tank fills in 4hours and empties in 6hours how long does it take if the tank is emptying and filling at the same time, how long will it take to be filled? this is all the info the problem has if you can figure it out ill give 100 points and the thing: )

12 hours.

Step-by-step explanation:

In 1 hour if the tap filling the tank runs on its own it will fill 1/4 of the tank. But if the tap that empties the tank is run at the same time it empties 1/6 of the tank. So the net result in 1 hour is that 1/4 - 1/6 = 1/12 of the tank is filled.

In the next hour the same thing happens - another 1/12 is filled.

So the tank will be filled in 12 hours.

Every hour, the tank would fill 1/4 of the way, but it would be set back by 1/6 of it's original progress due to the emptying. So how much progress did it actually make in that hour? 1/4 - 1/6 = 0.08333333. It filled 0.0833...%

This would mean '1' = 100%, so we have the equation 0.0833... x ? = 1

Well, 0.083333 is equal to the fraction 1/12. Also, 0.083333 x 12 equals 1. You can take either of those two routes to find your answer, but we can tell either way that it takes 12 hours.

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).

17.08

correct me if i am wrong.