You go to the doctor and he gives you 14 milligrams of radioactive dye. After 16 minutes, 5.25 milligrams of dye remain in your system. To leave the doctor's office, you must pass through a radiation detector without sounding the alarm. If the detector will sound the alarm if more than 2 milligrams of the dye are in your system, how long will your visit to the doctor take, assuming you were given the dye as soon as you arrived

31.7 min

Step-by-step explanation:

Given that;

N(t) = Noe^kt

No = amount of radioactive isotope initially present

N(t) = amount of radioactive isotope present at time t

t = time taken

k = decay constant

5.25 = 14e^-k16

5.25/14 = e^-k16

0.375 = e^=k16

ln(0.375) = =-16k

k = ln(0.375)/-16

k = 0.0613 min^-1

So time taken for 2 mg to remain;

2 = 14e^-0.0613t

2/14 = e^-0.0613t

0.1429 = e^-0.0613t

ln(0.1429) = -0.0613t

t = ln(0.1429)/-0.0613

t = 31.7 min

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !

don't give

take it once

try and try and you will not

happy to

pls mark as brainliest