Iwill give brainliest and 10 points if answered ! the radioisotope cobalt-60 is used in cancer therapy. the half-life of this isotope is 5.27 years. which equation determines the percent of an initial amount of the isotope remaining after t years? a=100(12)5.27ta=5.27(12)100ta=100(1 2)t5.27a=5.27(12)t100

A=100(12)t5.27

Step-by-step explanation:

Assume / and ^ got lost from eqns.

A(t) = initial × (1/2)^(t/halflife)

So initial = 100, halflife = 5.27,

A=100(12)t5.27 means A=100(1/2)^(t/5.27)

so that A is 50 when t is 5.27

if a = 1,   and b = 2, then the answer is 47.99999999. if you round that to the nearest whole number, the answer is 48.