Mathematics, 16.09.2020 08:01, leahphi

What is 1/6d+2/3 = 1/4(d-2)

Answers: 1

Show answers

Answers

1/6d + 2/3 = 1/4(d - 2)

First, simplify $\frac{1}{6} d$ to $\frac{d}{6}$ / Your problem should look like: $\frac{d}{6}$ + $\frac{2}{3}$ = $\frac{1}{4}$ (d - 2)
Second, simplify $\frac{1}{4} (d - 2)$ to $\frac{d-2}{4}$ / Your problem should look like: $\frac{d}{6}$ + $\frac{2}{3}$ = $\frac{d-2}{4}$
Third, multiply both sides by 12 (the LCM of 6,4) / Your problem should look like: 2d + 8 = 3(d - 2)
Fourth, expand. / Your problem should look like: 2d + 8 = 3d - 6
Fifth, subtract 2d from both sides. / Your problem should look like: 8 = 3d - 6 - 2d
Sixth, simplify 3d - 6 - 2d to d - 6 / Your problem should look like: 8 = d - 6
Seventh,add 6 to both sides. / Your problem should look like: 8 + 6 = d
Eighth, simplify 8 + 6 to 14 / Your problem should look like:14 = d
Ninth, switch sides. / Your problem should look like: d = 14

d = 14

answered: Guest

x² + 11x + 18 < 0

[-11 ± sqrt(121 - 72)]/2 = (-11 ± 7)/2

solution: (-9, -2)

step-by-step explanation:

answered: Guest

Answer: f(x) = -2x^3 +3x^2 +11x -6see attachedan infinite number. since the magnitude of the leading coefficient is not specified, it may be any negative number. (we have chosen the smallest magnitude integer that makes all coefficients be integers.)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.