The December bank statement and cash T-account for Stewart Company follow: BANK STATEMENT Date Checks Deposits Other Balance Dec. 1 $ 55,200 6 $ 7,680 47,520 11 610 $ 28,000 74,910 17 16,900 58,010 23 37,000 95,010 26 610 94,400 30 23,400 19,000 NSF* $ 360 89,640 31 Interest earned 80 89,720 31 Service charge 210 89,510 * NSF check from J. Left, a customer. Cash (A) Dec. 1 Balance 55,200 Deposits Checks written during December: Dec. 11 28,000 7,680 23 37,000 610 30 19,000 16,900 31 14,000 610 210 23,400 5,400 Dec. 31 Balance 98,390 There were no deposits in transit or outstanding checks at November 30. Prepare a bank reconciliation for December.

tube or goog would be a great resource to with that question

answer: your answer is a

choose (5, 1.5) and (-1.5,-3.5)

step-by-step explanation:

5 - 1.5 =3.5

-1.5 - (-3.5) = 2

3.5 + 2 =5.5

hope this !

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.