All Answers 358108 Answers

$${\frac{\left({\mathtt{19}}{\mathtt{\,-\,}}{\mathtt{4}}\right){\mathtt{\,\times\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{12}}}{{\mathtt{6}}}} = {\mathtt{150}}$$

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This is the forth or fifth time this has been posted.  It has been answered at least 3 times in full !

I first answered it a few days ago.  I think I even put it in my End of Day Wrap!

Are you losing track of your posts.  

If you become a member they will be stored automatically in your watchlist!!

This exact same question has been answered twice elsewhere.

One time was today by geno (I think)

The other was by me a couple of days ago.

Please do not keep repeating the same question!

Thanks Archy - I like people putting up puzzles  

Thanks for playing TakakiroMaeda   

6(19+4)+12=150

$$\left({\mathtt{19}}{\mathtt{\,\times\,}}{\mathtt{5}}\right){\mathtt{\,\small\textbf+\,}}\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{12}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{6}} = {\mathtt{149}}$$

I answered this same question some time back.  

You don't need to apologise.  I like being 'needed'  

Nullstellen von x2-1,6+0,2=0

Gesucht sind die Nullstellen von

$$x^2-1.6+0.2 = 0 \\
x^2 - 1.4 = 0 \quad | \quad +1.4\\
x^2 = 1.4 \quad | \quad \pm\sqrt{} \\
x_{1,2}= \pm\sqrt{1.4} \\
x_{1,2}= \pm 1.18321595662 \\
\boxed{x_1 = 1.18321595662 \ oder \ x_2 = -1.18321595662}$$

Die beiden Nullstellen von $$x^2-1.6+0.2 = 0$$ lauten $$x_1 = 1.18321595662$$ und $$x_2 = -1.18321595662$$

To find x take logs of both sides:

ln(0.000061) = ln(0.5^x)

Using the fact that ln(a^b) = b*ln(a) you have:

ln(0.000061)=x*ln(0.5)

or x = ln(0.000061)/ln(0.5)

$${\mathtt{x}} = {\frac{{ln}{\left({\mathtt{0.000\: \!061}}\right)}}{{ln}{\left({\mathtt{0.5}}\right)}}} \Rightarrow {\mathtt{x}} = {\mathtt{14.000\: \!831\: \!231\: \!761\: \!287\: \!9}}$$

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If you have the coordinates of the end points simply average each of them. i.e. if the end points are (x1,y1) and (x2,y2) then the midpoint is ( (x1+x2)/2, (y1+y2)/2 )

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if f(x) = c (a constant) then

c = 2ln(x-3)

Divide both sides by 2

(c/2) = ln(x-3)

Raise both sides to the power e

e^c/2 = e ^ln(x-3)

But, e^ln(a) = a so

e^c/2 = x-3

Add 3 to both sides

x = 3 + e^c/2 

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If you have something like 

$$\frac{\pi}{x}=2$$

then multiply both sides by x

$$\pi=2*x$$

then divide both sides by 2

$$\frac{\pi}{2}=x$$

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Alternatively, if you have y = ax² + bx +c, then the vertex is at x = -b/(2a)

Plug this value of x back into the equation to find the corresponding value of y.

Put two other values of x into the equation to find two other values of y for the other points.

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If  you mean a series expansion of something like ln(1 + x) then you can use a Taylor series to get:

$$ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...$$

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$$\sum_{n=0}^{n=\infty}n^3z^{2n}=\frac{z(z^2+4z+1)}{(z-1)^4}$$

This has a singularity at z = 5, so I guess the radius of convergence of your expression is when x²/5 = 1 or x = ±√5

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Edit: Just noticed that Chris has answered the same question elsewhere!

cos^2(x)=2cos(x)   suntract 2cos(x) from both sides

cos^2(x) - 2cos(x) =0    factor

cos(x) [ cos(x) - 2] = 0

Setting the first factor to 0, we have that cos(x) = 0    and this happens at pi/2 + n(pi) ..  (where n is an integer)

Setting the second factor to 0, we have ... cos(x) - 2 = 0   or  (adding 2 to both sides)  cos(x) = 2   which is impossible

Heureka's method is from Euclid.....

It says that, if a - b = c  and c / b, then c / a   and c is the greatest common factor of a and b

So

   1135620

 - 1134218

-------------

        1420 

And 1420 / 1134218  →    so,  1420 / 1135620     and is the greatest common factor of both numbers......

what is the greatest common factor of 1134218 and 1135620

Use the euklid algorithmn:

$$a= 1135620 \ and \ b= 1134218 \ and \ a> b\\ \\
\begin{array}{c|c|c|c|c|c}
\hline
a & b & q & r \\
\hline
1135620 & 1134218 & \frac{a}{b}=\textcolor[rgb]{0,1,0}{1} & \textcolor[rgb]{0,0,1}{1402} & 1135620 = \textcolor[rgb]{0,1,0}{1} * 1134218 + \textcolor[rgb]{0,0,1}{1402} \\
1134218 (old\ b) & 1402(old\ r) & \frac{a}{b}=\textcolor[rgb]{0,1,0}{809} & \textcolor[rgb]{0,0,1}{0} &1134218 = \textcolor[rgb]{0,1,0}{809} * 1134218 + \textcolor[rgb]{0,0,1}{0}& r= 0 \ \textcolor[rgb]{1,0,0}{Stop }\\
\hline
\end{array}$$

if  r = 0 then in the line before r = 1402 is the greatest common factor

Asumiing we have a function

N = ab^t    where N is the bacteria count at some time t........note.....when t  = 10, N = 800....and when t = 40, N = 1800...so we have

800 = ab¹⁰     and solving for "a," we have     800/b¹⁰ = a

And we also have

1800 = ab⁴⁰      and substituting, 800/b¹⁰  for a, we have

1800 = (800/b¹⁰)b⁴⁰ →   1800 = 800b³⁰

And dividing both sides by 800  we have

1800/800 = b³⁰  →  9/4 =  b³⁰

And taking the (positive) 30th root of each side, we have that b ≈ 1.0274

And a  = 800/(1.0274)¹⁰  ≈ 611    And this was the initial culture size  (when t = 0, b⁰ = 1)

A window or 2.

it is 1.58333333333333333333333333333333333333333333...

p.s. why did you do that anon #29?

Press the buttons on the calculater by clicking on them.

We can use the sine for this

sine (Θ) = opp / hyp  =  5 / 10   = 1 / 2

And this is easily evaluated, since the sine of 30° is 1/2

April's Rate ...(3/4) mi / (1/8) hr = 6 mph

Tim's Rate ....( 2/3)mi / (2/9) hr = 3 mph