Atumor is injected with 0.4 grams of iodine-125, which has a decay rate of 1.15% per day. to the nearest day, how long will it take for half of the iodine-125 to decay?

The Time period required for decay of Iodine-125 to half of its value is 60 days   .

Step-by-step explanation:

Given as :

The initial quantity of iodine-125 = 0.4 gram

The rate of decay = 1.15 %

Let The time period for decay = x day

The finial quantity after decay = half of initial quantity

I.e The finial quantity after decay = 0.2 gram

Now ,

The final quantity after decay = initial quantity ×

Or, 0.2 gm = 0.4 gm ×

or, =

Or, 0.5 =  

Or, = 0.9885

Taking log both side

log ( ) = Log 0.9885

or, × log 0.5 = - 0.0050233

or,   × ( - 0.301029 ) = - 0.0050233

or, x =  

∴ x = 59.92  ≈ 60 days

Hence The Time period required for decay of Iodine-125 to half of its value is 60 days   . Answer

v = 1/3 a^2h

step-by-step explanation:

if the height is 57.7cm and the height of the slant edge is 18.2cm, then the volume is 1/3 a^2h.

b, i believe, but not sure.