Eliza savage received a statement from her bank showing a checking account balance of $324.18 as of january 18. her own checkbook shows a balance of $487.38 as of january 29. the bank returned all of the cancelled checks but three. the amounts of these three checks are $15.00,$77.49, and $124.28. how much did eliza deposit in her account between january18 and january 29?

Eliza has now on her account  $ 487 + ( $15+77,49 + 124,28) =  $ 703,77 and this is the last amount  M

NOW  M = C + I, where, M is the final amount, C is the initial capital and I is the interest, of this formula  obtain the I, WHERE  I =  M-C ,  I = $ 703,77 - 324,18 = 379,59

NOW  I = C *i*n, WHERE, i , is the rate of interest, n is the time between the dates, ( 29 -19 = 11 days ), and with the formula you obtain i = I / C*n, i = 379,59/ ( 324,18* 11) = 0.1

then you must calculate the initial capital deposited in her account is calculated with

C = M / 1+(i*n) , C = 703,77/ 1 + (0,1*11) , 703,77/ 2.1 =  $335,12

687.09 dollars       happy now

step-by-step explanation: 6870

1857 times .63 equals 1169.91 subtract that from 1857 you get 687.09

(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