A radioactive substance decays at a continuous rate of 12% per year, and 90 mg of the substance is present in the year 2000. (a) Write a formula for the amount present, A (in mg), t years after 2000. MathPAD Response msnViewer_resp_str_c04q_go_2_006_1_ svg ViewEdit (b) How much will be present in the year 2010

A(t) = 90e^0.12t ;

299 mg

Step-by-step explanation:

General Continous growth rate equation :

A = Pe^rt

A = amount present, t years after year of initial amount.

Here,

P = 90 mg, amount in year 2000

r = rate = 12% = 0.12

t = years after 2000

Therefore,

A is written as ;

A(t) = 90e^0.12t

Amount present in year 2010 ;

t = 2010 - 2000 = 10

A(10) = 90e^0.12(10)

A(10) = 90 * e^1.2

= 90 * 3.3201169

= 298.81052

= 299 mg

answer: x

step-by-step explanation:

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3