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

i think da answer is

d.square root of 3

d

step-by-step explanation:

the two points from this line are   (0,4)   & (2,-2)

we can call

now, lets find the slope (m):

now finding b (the y intercept). this is the point at the y-axis where the line cuts the axis. looking at the graph, it is y = 4. so the point is 4. hence b = 4

the slope-intercept form is given by  

putting the points, we have:

d is the correct answer.