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

step-by-step explanation:

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl