Given that Begin equation . . . begin fraction . . . 2 over 3 . . . end fraction . . . times a equals 16 . . . end equation, which of the following illustrates how the multiplicative inverse property can be used to find the value of a? A Begin equation . . . 1 times the quantity . . . begin fraction . . . 2 over 3 . . . end fraction . . . times a . . . equals 1 times 16 . . . end equation B Begin equation . . . begin fraction . . . 2 over 3 . . . end fraction . . . times a minus . . . begin fraction . . . 2 over 3 . . . end fraction . . . minus a equals 16 minus . . . begin fraction . . . 2 over 3 . . . end fraction . . . times a . . . end equation C Begin equation . . . 0 times the quantity . . . begin fraction . . . 2 over 3 . . . end fraction . . . times a . . . equals 0 times 16 . . . end equation D Begin equation . . . begin fraction . . . 3 over 2 . . . end fraction . . . times the quantity . . . begin fraction . . . 2 over 3 . . . end fraction . . . times a . . . equals . . . begin fraction . . . 3 over 2 . . . end fraction . . . times 16 . . . end equation

the homeowners insurance cost for a 115,000 wood frame house is 80,500.

step-by-step explanation:

it is given that the coverage rate for wood frame house is 0.7.

the value of wood frame house is 115,000.

.

homeowners insurance cost for a 115,000 wood frame house is

therefore the homeowners insurance cost for a 115,000 wood frame house is 80,500.

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.