Horatio is solving the equation -3/4+2/5x=7/20x-1/2 Which equations represent possible ways to begin solving for x? Check all that apply. A. Two-fifths x = StartFraction 7 Over 20 EndFraction x + one-fourth

B. Two-fifths x = StartFraction 7 Over 20 EndFraction x + five-fourths

C. Negative three-fourths = negative StartFraction 1 Over 20 EndFraction x minus
one-half

D. Negative three-fourths = three-fourths x minus one-half

E. Negative three-fourths + StartFraction 1 Over 20 EndFraction x = negative one-half

c

step-by-step explanation:

there is vertical asymptote at x = 2 and x = 0

vertical asymptote occurs at the points where the denominator is 0. so we can say that

x = 2 occurs when there is a factor (x-2) in the denominator, and

x = 0 occurs when there is a factor "x" in the denominator

thus, the denominator of the equation would have "x * (x-2)", this is option c only. that's the correct answer.

step-by-step explanation:

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.