A person invested $200 in an account growing at a rate allowing the money to double every 13 years. How much money would be in the account after 15 years to the nearest dollar

$431

Step-by-step explanation:

Given parameters:

Principal  = $200

Time given for the amount to double  = 13yrs

Unknown:

Amount in the account after 15yrs = ?

Solution:

To solve this problem, we need to find the interest rate on this investment.

    I =

I = interest

P = principal

R = Rate

T = time

 Now,

  if the amount should double,

     Interest earned  = $200

So;

         200  =  

         200  = 26 x R

           R = 7.7%

Now, in 15yrs, this amount will be;

 let us find the interest;

           I  =  

           I = $231

The amount in the account will be:

  Amount  = Principal + Interest

  Amount = $200 + $231  = $431

false

step-by-step explanation:

4y - 6 + 4y^2 - 9

before simplifying this contains 4 terms

after simplifying

4y -15 + 4y^2

it contains 3 terms

(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