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how did u get 8 choose 2 as the denominator 

You should do the MULT FIRST then the SUBTR

6 - 56  = -50

Then you can home in further:

4/ (3 (sqrt8)) === 4 / (3 sqrt(4x2)) =  4 / ((3x2)(sqrt2)) = 2/(3sqrt2) = 2 sqrt2 / (3 sqrt2 sqrt2) = 2 sqrt2 /6  = sqrt(2) / 3

 'f(x)=1/2*(x^3-3x^2+6)

So we have

(1/2) (x^3 - 3x^2 + 6)  = 0        divide through by 1/2

x^3 - 3x^2 + 6  = 0      the possible rational zeroes are ±1, ± 2, ± 3 and ± 6

Unfortunately........none of these produce a "zero"

If you know Calculus, you could adopt an iterative approach [ Newton's Method] to get an approximate answer or answers

Another alternative is to just graph the function and see where the zeroes lie.......this is guaranteed to have one "real" zero

Here's the graph: https://www.desmos.com/calculator/dx17hd5ze5

WolframAlpha  gives the real solution at  x ≈ -1.1598

Here is the Wolframalpha answer and graph : https://www.wolframalpha.com/input/?i=x%5E3+-+3x%5E2+%2B+6+%C2%A0%3D+0

There are also two non-real solutions of  x ≈ 2.0979 + 0.7850 i    and  x ≈ 2.0979 - 0.7850 i

Using a graphing calc I get   -1.196  as a zero

00062 → 6.2*10^(-4)

6.2 x 10^(-4)

If it's just the real zero you are interested in, start by plotting a graph:

(4^4) × (4^4)      ....we are using the rule that says :    a^m * a^n   =  a^(m + n).........so we have

4^(4 + 4)   =  4^8  =  65536

4096

8.5×10^5          7.05×10^8

8.5 x10^5 =     850000

7.05 x 10^8 =  705000000

And

850000   < 705000000

So

8.5 x 10^5  <   7.05 x 10^8

7,6*10^-4m*1,5*10^7m=7.6*1,5*10^-4m*10^7m =11,4*10^(-4m+7m)=11,4*10^3m

29,5 h * 8,25 $/h = 243,375 $

Omi67,

Why is the focal length a half of the radius ?

I understand very little of this type of physical problem but if you could give me a reason that would be great.  ://

Thanks :)

y-2 = -1/2( x+1)  |+2

   y = -1/2( x+1) + 2

   y = -1/2x - 1/2 + 2

   y = -1/2x + 3/2

Number of digits given by floor(log₁₀(3^1000)) + 1. or floor(1000*log₁₀(3)) + 1 → 478

Is the screen measure in both cases a diagonal? Or by length do you mean width?

If the 29, 719 etc were meant to be half-lives then the original equations are indeed incorrect (they are missing the ln(2)).  However, I don't see that they are explicitly called half-lives anywhere.