Alan

(4+6+8+...+2n-4)+(4+6+...+2n-6)+...+4 = (n^3 - 3n^2 - 4n + 12)/3.      n >= 4

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Here's the first one:

\(f\left( x\right)=2+x^3(7\ln(x)-9) \\ \\ f'(x)=3x^2(7\ln(x)-9)+\frac{7x^3}{x}\rightarrow x^2(21\ln(x)-20) \\ \\ f'(1)=1^2(21\ln(1)-20)\rightarrow -20 \\ \)

"Why does 0.999 equal 1."

As written it doesn't!  However, if you mean the 9's go on forever then:

Let x = 0.999999...... etc.

  10x = 9.999999..... etc.

Subtract the first from the second to get:   9x = 9  or x = 1

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If your equations are:

2y = x + 2               (1)

x - 3y = -5              (2)

Then:

Rearrange (2) to get:   x = -5 + 3y     (3)

Replace x in (1) using (3):   2y = -5 + 3y + 2  

Rearrange: -y = -3  so   y = 3

Put this into (3):    x = -5 +3*3   so  x = 4

(x, y) = (4, 3)

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N5 + N8 = 30000

0.08*N8 - 0.05*N5 = 840. Replace N8 using the above:

0.08*(30000 - N5)  - 0.05*N5 = 840

0.13*N5 = 30000 - 840

N5 = (30000 - 849)/0.13

N5 = $12000

I think you forgot a 5x^2 in your subtraction Max!

Here's an alternative approach:

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Here's a graph:

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