Alan

It is possible that this equation was meant to be:

$$\frac{14}{x^2-9}+\frac{6}{x-3}=\frac{8}{x+3}$$

in which case, multiply all terms by x² -9

$$14+\frac{6(x^2-9)}{x-3}=\frac{8(x^2-9))}{x+3}$$

Express x² - 9 as (x+3)(x-3) and cancel as appropriate

$$14+6(x+3)=8(x-3)$$

Expand bracketed terms and collect like terms on the same side

$$14+18+24=8x-6x$$

$$56=2x$$

$$x=28$$

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Here's another way of looking at it:

Let the distance betwen the cities be d miles

The speed, s, of the car is distance travelled divided by time taken or (d/6 miles)/(3/5 hours) = (d/6)*(5/3) miles per hour = 5d/18 miles per hour

At this speed the car will travel 5d/18 miles in one hour.

As a fraction of the distance between the cities this is (5d/18)/d or just 5/18

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September price = S

October price = 1.2S = $1.5  so September price S = $1.5/1.2 = $1.25

Money earned in September = 0.4*900*$1.25 = $450

Money earned in October = 0.4*700*$1.5 = $420

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Can tackle it like this:

$$2^{x+4}-2^{x+1}=3$$

$$2^x2^4-2^x2^1=3$$

$$2^x(2^4-2^1)=3$$

$$2^x(16-2)=3$$

$$2^x\times14=3$$

$$2^x=\frac{3}{14}$$

$$\ln{2^x}=\ln{\frac{3}{14}}$$

$$x\times \ln2=\ln3-\ln14$$

$$x=\frac{\ln3-\ln14}{\ln2}$$

$${\mathtt{x}} = {\frac{\left({ln}{\left({\mathtt{3}}\right)}{\mathtt{\,-\,}}{ln}{\left({\mathtt{14}}\right)}\right)}{{ln}{\left({\mathtt{2}}\right)}}} \Rightarrow {\mathtt{x}} = -{\mathtt{2.222\: \!392\: \!421\: \!336\: \!447\: \!9}}$$

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Time consuming:  Not really.  Think about the symmetry.  The off-axis ones are simply all the positive and negative combinations of 1/2 and √3/2.  The on-axis ones are simply, 1, -1, i and -i.

As for proving there are exactly n roots to an n'th order polynomial, you'll have to go to Carl Friedrich Gauss (I think he was the first) for this!

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How do you fold this into a square pyramid?  Even when the triangles are folded flat the combined distance of two of them is only 6 inches, so even when flat thay don't span a 10 inch base!

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They are the solutions to the 12th order polynomial x¹² = 1

A 12th order polynomial has exactly 12 roots.

They can be pictured as equally spaced points on the unit circle in the complex plane.

Each root is separated from the next by an angle of 360/12 = 30 degrees.

Starting with the obvious root x = 1 + 0i  on the x-axis, the next one (in a clockwise direction) has coordinates x = cos(30) and y = sin(30) (because the radius of the unit circle is 1, by definition!).  cos(30) = (√3)/2 and sin(30) = 1/2 so this point can be represented by the complex value (√3)/2 + i(1/2).

Just work your way around the unit circle in a similar manner for the others.

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2x² - 16x = 20

Divide each term by 2

x² - 8x = 10

Subtract 10 from both sides

x² - 8x - 10 = 0

This doesn't factorize nicely, so use the quadratic formula to solve it.

$$x=\frac{-(-8)\pm \sqrt{(-8)^2-4\times1\times (-10)}}{2\times1}$$

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For the first solution we assume 3x is greater than 5 so the equation simplifies to:

4(3x - 5) - 5 = -1

12x -20 -5 = -1

12 x -25 = -1

12x = 24

x = 2

For the second solution we assume 3x is less than 5, so the equation simplifies to:

4(5 - 3x) - 5 = -1

20 - 12x - 5 = -1

15 - 12x = -1

12x = 16

x = 16/12

x = 4/3

So the sum of the two solutions is 2 + 4/3 = 10/3

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The gradient of a straight line that is perpendicular to y + 3 = 2x is -(1/2), as the gradient of the given line is +2.

So the equation of the perpendicular line is y = -(1/2)x + c, where c is a constant.  We can find the value of c because we know the line passes through (-5, 2), so at that point we must have:

2 = -(1/2)*(-5)+c

or

2 = 5/2 + c

so c = 2 - 5/2 

or c = -1/2

so the equation of the line becomes

y = -(1/2)x - (1/2)

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