What is the amount of a $4,000.00 annuity due at 12 percent compounded semiannually for 3 years?

$ 5674.076

Step-by-step explanation:

The question is on compound interest

The formulae = A= P(1+ r/n) ^nt where P is the principal amount, r is the rate of interest in decimal, n is number of compoundings per year and t is the total number of years.

Given; P= $4,000.00 , r=12/100=0.12, n=2 and t=3

Substituting values in the equation  A= P(1+ r/n) ^nt

A= 4000 ( 1+0.12/2)^2×3

A=4000(1.06)^6

A=$ 5674.08

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0

an

y = 6x -3y -12

y +3y +12 = 6x

=> . x = { 2y / 3 +2 } this is equation of line in terms of x

=> . if line meets the x-axis the co-ordinates of meeting point would be ( x , 0 )

=> . x = 2 ( 0 ) / 3 + 2

=> . x = 2

so, the point is ( 2 , 0 )