Mathematics, 19.08.2019 13:30 1kafran1

What does x+2/5=12 equal and i would love to seehow you worked it out

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You said x + 2/5 = 12

It might be easier to handle if we use the decimal for 2/5.

x + 0.4 = 12

Subtract 0.4 from each side: x = 11.6

That's all there is to it. I'm sorry I can't make it
more complicated for you.

Answer from: Queen9336

X + $\frac{2}{5}$ = 12

First, subtract $\frac{2}{5}$ from both sides. / Your problem should look like: x = 12 - $\frac{2}{5}$
Second, simplify 12 - $\frac{2}{5}$ to $\frac{58}{5}$ / Your problem should look like: x = $\frac{58}{5}$

Answer as fraction: x = $\frac{58}{5}$
Answer as decimal: x = 11.6

Answer from: Quest

x= -3 and $x=\frac{-3+/-3i\sqrt{3} }{2}$

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

$(x^3-a^3)=$ $(x-a)(x^2+ax+a^2)$

where a is the cube root of $a^3$ .

here this is 27 and the cube root of 27 is 3. so we write using the form:

$(x-3)(x^2+(3)x+3^2)\\(x-3)(x^2+3x+9)$ .

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

$x=\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

where a = 1, b=3, and c=9

$x=\frac{-3+/-\sqrt{3^2-4(1)(9)} }{2(1)} \\x=\frac{-3+/-\sqrt{9-36)} }{2} \\x=\frac{-3+/-\sqrt{-27)} }{2}\\x=\frac{-3+/-3i\sqrt{3} }{2}$

this cubic has two complex roots.

Answer from: Quest

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.

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