The total bear population in a certain area is represented by the function p=120(1.016)^t , where t is time in years. how could this function be rewritten to identify the weekly growth rate of the population? what is the weekly growth rate? drag and drop the choices into the boxes to correctly complete the table. if a value does not match, do not drag it to the table. function: weekly growth rate: option box: p=120(1.016^1/52)^52t p=120(1.016)^t/52 p=120(1.016^52)^t 0.03% 0.019% 1.0003%

Step-by-step explanation:

The function could be rewritten a couple of ways. We could leave t defined as the number of years, and rewrite the function as ...

  P = 120(1.016^(52/52t)) = 120(1.016^(1/52))^(52t) . . . matches one answer choice

Or, we could redefine t to be time in weeks and rewrite the function as ...

  P = 120(1.016^(t/52)) . . . . matches another answer choice

Either way, the weekly growth factor gets evaluated to be ...

  1.016^(1/52) ≈  1.0003053

This corresponds to a weekly growth rate of 0.03053%

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18