Alan

Have a look at https://en.wikipedia.org/wiki/Cubic_function for the solution to this cubic equation.

CH₄ + 2O₂ = CO₂ + 2H₂O

Molecular weight of methane = 16

Molecular weight of carbon dioxide = 44

Mass of carbon dioxide = mass of methane×44/16  →  6.30×10−3×44/16  → 0.017 g

Here's another way to approach this question:

The equation can be written as 3x = e^-x

e^-x can be written as a series: 1 - x + x²/2 - x³/6 + x⁴/24 - x⁵/120 + ...

So 3x = e^-x can be written as 3x = 1 - x + x²/2 - x³/6 + x⁴/24 - x⁵/120 + ...

Subtract 3x from both sides to get  0 = 1 - 4x + x²/2 - x³/6 + x⁴/24 - x⁵/120 + ...

Now approximate this by  0 ≈ 1 - 4x₀  so that x₀ ≈ 0.25  

Take the next term to get   0 ≈ 1 - 4x₁ + x₁²/2   Solve this to get x₁ ≈ 0.258  (ignoring the other solution which is bigger than 1, so the series would diverge, and is clearly not a solution to the original equation).

The true solution lies between 0.25 and 0.258  (the question allows for an interval as a solution!).

It is easy enough to take higher order approximations numerically to tighhten the interval if required.

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(x^3-x+1)-(3x-1) 

This is just an expression that simplifies to x^3 -4x + 2

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If you have the quadratic equation  \(ax^2+bx+c=0\)

where a, b and c are constants, then the two solutions are given by the quadratic formula, which is:

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Ok, you've managed to correct it.

You've slipped up somewhere Spawn.  Guest#1 has the right results (albeit with a rather long solution method!)

There are an infinite number of ordered pairs for y = -2x + 6

For example:   

(x,   y)

(0,  6)

(1,  4)

(2,  2)

(3,  0)

First find the components in the horizontal and vertical directions.

Horizontal:  Fx = 9 - 14*cos(65°)    or  Fx = 3.083   

Vertical:   Fy = 14*sin(65°)  or Fy = 12.688

The resultant has magnitude sqrt(Fx² + Fy²)  = 13.058

It has an angle to the horizontal axis of  tan^-1(Fy/Fx)  =  76.34°  

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Well, pi*diameter is the circumference of a circle, so pi*radius would be the arc length of a semicircle.