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Let N be what he started with

After he gave1/3 of his coins to Clark  and 1/3 of the remaining coins to Lois  we have

After giving (1/3) to Clark, he had (2/3)N left

After giving (1/3) of these to Lois, we have

(2/3)[(2/3)N] = 144 

(4/9) N  = 144

N =  144 * 9 / 4   =  324

After giving (1/3) of these to Clark he had   216 left

He gave (1/3) of these to Lois =  216 / 3  =   72

OK....I see what you mean.....

The domain   is correct...( -inf, inf)

In the form

y = trig function (Bx + C)

Factor out B

y  = trig function B( x + C/B )

Set   x + C/B = 0   and solve for   X

Example

sin ( 3x - pi )

sin 3(x - pi/3)

Set

x + pi/3  = 0

x =  -pi/3          this is the phase shift

y= (1/2)x - 5/2 

No graph is shown.....

Let the number of coins that he started with= C
2/3C left - after he gave 1/3 to Clark
4/9C left - after he gave  1/3 to Lois
4/9C = 144, solve for C
C =144 x 9/4 =324 - Coins that Lex started with.
324 x  1/3 =108 Coins he gave to Clark
324 - 108 =216 Coins left
216 x 1/3 =72 Coins he gave to Lois.
216 -72=144 - Coins that he is left with.
                            

I don't believe you. Im sorry.

Let w be the width of the frame. 

The length of the shorter strip of frame is 6 + 2w cm

The length of the longer strip (excluding that on the shorter sides) is 11 cm 

The total area of the frame is then given by 2*(6+2w)*w + 2*11*w or 4w² + 34w cm

This must equal 18 cm², so:

4w² +34w = 18

Divide by 2 and rearrange:   2w² + 17w - 9 = 0

This factors as:  (2w - 1)(w + 9) = 0

Because the width must be positive the only sensible solution of this is 2w = 1, or w = 1/2 cm

.

As follows:

Edit:  Just noticed heureka beat me to it.  Luckily, we came up with the same result! 

Let $f(z)$ be a rotation by $\pi/2$ counterclockwise around $2+i$, as in the picture below:

\(\text{Let $z_0 = 2+i$}\)

\(\begin{array}{|rcll|} \hline f(z)-z_0 &=& (z-z_0)\left( \cos\left(\dfrac{\pi}{2}\right) + i\sin\left(\dfrac{\pi}{2} \right) \right) \\ f(z)-z_0 &=& (z-z_0)\left( 0 + i\cdot 1 \right) \\ f(z)-z_0 &=& (z-z_0)i \\ \mathbf{f(z)} & \mathbf{=}& \mathbf{z_0 + (z-z_0)i} \\\\ f(z) &=& z_0 + (z-z_0)i \quad | \quad z_0 = 2+i \\ f(z) &=& 2+i + \Big(z-(2+i)\Big)i \\ f(z) &=& 2+i + (z-2-i)i \\ f(z) &=& 2+i + iz-2i-i^2 \quad | \quad i^2 = -1 \\ f(z) &=& 2+i + iz-2i+1 \\ f(z) &=& iz+3-i \quad | \quad \text{compare with }f(z) = az + b \\ \\ \mathbf{(a,b)} & \mathbf{=} & \mathbf{(i,3-i)} \\ \hline \end{array} \)

The calculator here doesn't seem to like non-integer factorials.  It gets 0.5! wrong.  

Webcalc:   0.5! = 0.8777726925922196...

In fact:  0.5! = 0.88622692545275801365...

(Non-integer factorials are defined using the Gamma function:   \(n! = \Gamma(n+1)\)  ).

It is undefined  :)

Not so stupid really - questions such as this have been asked many times.

Thanks Alan :)

I did that, I just didn't think to finish it with a run. 

I was looking for some other way to do it. 

But I guess I could do it in EXCEL easily enough. :/

We can write:

a + (a+1) + (a+2) + ... + (a+n-1) = na + (n-1)n/2

Hence we have:  na + (n-1)n/2 = 100, which we can rewite as:  n² + (2a-1)n - 200 = 0 

Solve this for n in terms of a: \(n=\frac{\sqrt{4a^2-4a+801}-2a+1}{2}\)  (the other solution results in negative values for n).

Let a run from 1 to 100 to find the only integer solutions are:  (a,n) = (9,8) and (18,5) as found by Chris and Guest.

(There is also a = 100 and n = 1, but the question requires n>=2).

(Alternatively, solve for a in terms of n and run through n = 2 to 100, and look for integer solutions for a)