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Thank you, Melody !

 Trigonometry

\(\begin{array}{|rcll|} \hline \mathbf{\tan{\theta}} &=& \mathbf{2\ \tan(60^\circ-\theta)} \quad | \quad \lambda = 2,\ A=60^\circ \\\\ (2-1)\sin{60^\circ} &=& (2+1)\sin(2\theta-60^\circ) \\ \sin{60^\circ} &=& 3\sin(2\theta-60^\circ) \quad | \quad \sin(60^\circ)=\dfrac{\sqrt{3} }{2} \\ \dfrac{\sqrt{3} }{2} &=& 3\sin(2\theta-60^\circ) \\ \mathbf{ \sin(2\theta-60^\circ)} &=& \mathbf{\dfrac{\sqrt{3}}{6}} \\ 2\theta-60^\circ &=& \arcsin\left(\dfrac{\sqrt{3}}{6}\right) +360^\circ n,\ n\in\mathbb{Z} \\ 2\theta &=& 60^\circ +16.7786548810^\circ +360^\circ n \\ 2\theta &=& 76.7786548810^\circ +360^\circ n \\ \theta &=& 38.3893274405^\circ +180^\circ n \\ \\ \mathbf{\theta_1} &=& \mathbf{38.3893274405^\circ} \\\\ \theta_2 &=& 38.3893274405^\circ+180^\circ \\ \mathbf{\theta_2} &=& \mathbf{218.389327440^\circ} \\ \hline \sin(2\theta-60^\circ) &=& \sin\Big(180^\circ-(2\theta-60^\circ)\Big) \\ &=& \sin(180^\circ-2\theta+60^\circ) \\ &=& \sin(240^\circ-2\theta ) \\\\ \mathbf{ \sin(240^\circ-2\theta )} &=& \mathbf{ \dfrac{\sqrt{3}}{6} } \\ 240^\circ-2\theta &=& \arcsin\left(\dfrac{\sqrt{3}}{6}\right) +360^\circ n,\ n\in\mathbb{Z} \\ 2\theta &=& 240^\circ-16.7786548810^\circ - 360^\circ n \\ 2\theta &=& 223.221345119^\circ - 360^\circ n \\ \theta &=& 111.610672560^\circ - 180^\circ n \\ \\ \mathbf{\theta_3} &=& \mathbf{111.610672560^\circ} \\\\ \theta_4 &=& 111.610672560^\circ+180^\circ \\ \mathbf{\theta_4} &=& \mathbf{291.610672560^\circ} \\ \hline \end{array}\)

Very impressive Heureka    

Trigonometry

\(\begin{array}{|rcll|} \hline \tan{\theta} &=& \lambda\ \tan(A-\theta) \\ \boxed{\tan{\theta}=\dfrac{\sin(\theta)}{\cos(\theta)} \\ \tan(A-\theta) = \dfrac{\sin(A-\theta)}{\cos(A-\theta)} } \\ \dfrac{\sin(\theta)}{\cos(\theta)} &=& \lambda\ \left(\dfrac{\sin(A-\theta)}{\cos(A-\theta)} \right) \\ \boxed{ \sin(A-\theta) = \sin(A)\cos(\theta)-\cos(A)\sin(\theta) \\ \cos(A-\theta) = \cos(A)\cos(\theta)+\sin(A)\sin(\theta) } \\ \dfrac{\sin(\theta)}{\cos(\theta)} &=& \lambda\ \left(\dfrac{\sin(A)\cos(\theta)-\cos(A)\sin(\theta)} {\cos(A)\cos(\theta)+\sin(A)\sin(\theta)} \right) \\ \sin(\theta) \Big(\cos(A)\cos(\theta)+\sin(A)\sin(\theta) \Big) &=& \lambda\ \cos(\theta) \Big(\sin(A)\cos(\theta)-\cos(A)\sin(\theta) \Big) \\ \cos(A)\sin(\theta)\cos(\theta)+ \sin(A)\sin^2(\theta) &=& \lambda\ \cos^2(\theta)\sin(A)-\lambda\ \cos(A)\sin(\theta)\cos(\theta) \\ (\lambda+1)\cos(A)\sin(\theta)\cos(\theta) &=& \lambda\ \cos^2(\theta)\sin(A)-\sin(A)\sin^2(\theta) \\ \boxed{ \cos^2(\theta) = 1-\sin^2(\theta) } \\ (\lambda+1)\cos(A)\sin(\theta)\cos(\theta) &=& \lambda\ \Big(1-\sin^2(\theta) \Big) \sin(A)-\sin(A)\sin^2(\theta) \\ (\lambda+1)\cos(A)\sin(\theta)\cos(\theta) &=& \lambda\ \sin(A)-(\lambda+1) \sin(A)\sin^2(\theta) \\ (\lambda+1)\cos(A)\sin(\theta)\cos(\theta)+(\lambda+1) \sin(A)\sin^2(\theta) &=& \lambda\ \sin(A) \\ (\lambda+1)\Big( \cos(A)\sin(\theta)\cos(\theta)+\sin(A)\sin^2(\theta) \Big) &=& \lambda\ \sin(A) \\ \boxed{\sin(\theta)\cos(\theta)=\dfrac{\sin(2\theta)}{2} \\ \sin^2(\theta) = \dfrac{1-\cos(2\theta)}{2} } \\ (\lambda+1)\Big( \cos(A)\dfrac{\sin(2\theta)}{2}+\sin(A) \Big(\dfrac{1-\cos(2\theta)}{2}\Big) \Big) &=& \lambda\ \sin(A) \\ (\lambda+1)\Big( \cos(A)\sin(2\theta)+\sin(A) \Big( 1-\cos(2\theta)\Big) \Big) &=& 2\lambda\ \sin(A) \\ (\lambda+1)\Big( \cos(A)\sin(2\theta)-\sin(A)\cos(2\theta) + \sin(A) \Big) &=& 2\lambda\ \sin(A) \\ \boxed{ \cos(A)\sin(2\theta)-\sin(A)\cos(2\theta) = \sin(2\theta-A) } \\ (\lambda+1)\Big( \sin(2\theta-A) + \sin(A) \Big) &=& 2\lambda\ \sin(A) \\ (\lambda+1)\sin(2\theta-A) +(\lambda+1)\sin(A) &=& 2\lambda\ \sin(A) \\ (\lambda+1)\sin(2\theta-A) &=& 2\lambda\ \sin(A)-(\lambda+1)\sin(A) \\ (\lambda+1)\sin(2\theta-A) &=& 2\lambda\ \sin(A)-\lambda\ \sin(A)- \sin(A) \\ (\lambda+1)\sin(2\theta-A) &=& \lambda\ \sin(A)- \sin(A) \\ (\lambda+1)\sin(2\theta-A) &=& (\lambda -1) \sin(A)\\ \mathbf{(\lambda -1) \sin(A)} &=& \mathbf{(\lambda+1)\sin(2\theta-A)} \\ \hline \end{array}\)

I used Mathcad to generate the result.  The two approaches are not related.  For the first, search for LambertW, for the second, search for Newton-Raphson (Wikipedia will give details for both).

Thank you! BTW, can you tell me, from which software is this screenshot? I found same answer (Lambert W formula) hidden somwhere in plot graph on Wolfram Alpha, but with no steps shown. I'curious how to go from one formula to second one.

Hi Hectictar and CalculatorUser,

Your suggestions sounded a little weird to me Hectictar but then I thought a little and understood how that could help.

When we do a task it only uses certain parts of our brain allowing other bits to wander off leading to day dreams etc. If less relevant  parts of the brain are occupied with certain types of things, such a drawing, then it may allow the bit that is trying to concentrate on the maths to concentrate better. 

For me, i can often concentrate better on maths if i have quiet music playing or even if I am in a food court in a shopping centre and have all the associated noise around me. (So long as no one is being specifically annoying) 

I used to have a very intelligent high achieving friend who could only study when he was watching TV at the same time! I always found that incredible .... it would never work for me.

Anyway, the idea of allowing part of your brain to be chanelled into something such as doodling sounds reasonable.    

Of course you may have to defend your sketching to your teacher  

wat?

Show that from any five integers,

one can always choose three of these integers such that their sum is divisible by 3.

Five integers have a remainder of 0 or 1 or 2 if they become divisible by 3:

\(\begin{array}{|c|c|c|c|} \hline & & & \text{one choose three} \\ \text{remainder 0} & \text{remainder 1} & \text{remainder 2} & \text{of these is divisible by 3} \\ 0 \pmod{3} & 1 \pmod{3} & 2 \pmod{3} & \text{sum of the remainder} \\ \hline 0 & 0 & 5 & 2+2+2 \\ 0 & 5 & 0 & 1+1+1 \\ 0 & 4 & 1 & 1+1+1 \\ 0 & 3 & 2 & 1+1+1 \\ 0 & 2 & 3 & 2+2+2 \\ 0 & 1 & 4 & 2+2+2 \\ \hline 1 & 0 & 4 & 2+2+2 \\ 1 & 4 & 0 & 1+1+1 \\ 1 & 3 & 1 & 1+1+1 \\ 1 & 2 & 2 & 0+1+2 \\ 1 & 1 & 3 & 2+2+2 \\ \hline 2 & 0 & 3 & 2+2+2 \\ 2 & 3 & 0 & 1+1+1 \\ 2 & 2 & 1 & 0+1+2 \\ 2 & 1 & 2 & 0+1+2 \\ \hline 3 & 0 & 2 & 0+0+0 \\ 3 & 2 & 0 & 0+0+0 \\ 3 & 1 & 1 & 0+0+0 \\ \hline 4 & 0 & 1 & 0+0+0 \\ 4 & 1 & 0 & 0+0+0 \\ \hline 5 & 0 & 0 & 0+0+0 \\ \hline \end{array}\)

Find the value of 

\(x = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}.\)

\(\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}} } \\\\ x-1 &=& \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}} \\\\ \dfrac{1}{x-1} &=& 2+ \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}} \\\\ \dfrac{1}{x-1} &=& 2+ (x-1) \\\\ 1 &=& (x-1)(2+x-1) \\ 1 &=& (x-1)(x+1) \\ 1 &=& x^2-1 \\ x^2 &=& 2 \\ \mathbf{x} &=& \mathbf{\sqrt{2}} \\ \hline \end{array} \)

We don't use (x,y,z) for spherical coordinates.  It just causes confusion.

\(\text{I'm going to assume that }(x,y,z) \equiv (\rho, \theta, \phi)\\ \begin{align*} -3 &= \rho \sin(\theta)\cos(\phi)\\ 4 &= \rho\sin(\theta)\sin(\phi)\\ -12 &= \rho\cos(\theta) \end{align*}\)

\(\theta= \arccos\left(\dfrac{-12}{\sqrt{(-3)^2+4^2+(-12)^2}}\right) = \arccos\left(\dfrac{-12}{13}\right)\\ \tan(\theta)=-\dfrac{\sqrt{13^2-12^2}}{12} = -\dfrac{5}{12}\)

\(\phi= \arctan\left(\dfrac{4}{-3}\right)\\ \tan(\phi)= -\dfrac 4 3\)

A)

Let \(f(x)\) be a polynomial of degree \(4\)  with rational coefficients which has \(1+2\sqrt{3}\) and \(3-\sqrt{2}\)
as roots, and such that \(f(0) = -154\).
Find \(f(1)\).

Vieta:

\(\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{ax^4+bx^3+cx^2+dx+e} \quad | \quad \text{polynomial of degree 4} \\\\ \dfrac{e}{a} &=& x_1x_2x_3x_4 \quad | \quad \text{Vieta-theorem},\ f(0) =e \\\\ \dfrac{f(0)}{a} &=& x_1x_2x_3x_4 \\\\ \mathbf{a} &=& \mathbf{\dfrac{f(0)}{x_1x_2x_3x_4}} \\\\ && \boxed{ f(0)=-154\\ x_1 = 1+2\sqrt{3}\\ x_2 = 1-2\sqrt{3}\\ x_3 = 3-\sqrt{2}\\ x_4 = 3+\sqrt{2} } \\\\ a &=& \dfrac{-154}{(1+2\sqrt{3})(1-2\sqrt{3})(3-\sqrt{2})(3+\sqrt{2})} \\\\ &=& \dfrac{-154}{(1-4\cdot 3)(9-2)} \\\\ &=& \dfrac{-154}{(-11)\cdot 7} \\\\ &=& \dfrac{ 154}{77} \\\\ \mathbf{a} &=& \mathbf{2} \\ \hline \end{array}\)

\(\mathbf{f(1)=\ ?}\)

\(\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{ a(x-x_1)(x-x_2)(x-x_3)(x-x_4) } \\\\ f(1) &=& 2\left(1-(1+2\sqrt{3})\right) \left(1-(1-2\sqrt{3})\right) \left(1-(3-\sqrt{2})\right) \left(1-(3+\sqrt{2})\right) \\ &=& 2(2\sqrt{3}) (-2\sqrt{3}) (-2+\sqrt{2}) (-2-\sqrt{2})) \\ &=& 2(2\sqrt{3}) (-2\sqrt{3}) \left((-2)^2-2\right) \\ &=& 2(2\sqrt{3}) (-2\sqrt{3}) \cdot 2\\ &=& -16\cdot 3 \\ \mathbf{f(1)} &=& \mathbf{-48} \\ \hline \end{array}\)

\(\text{let }a = \dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\ddots}}}\)

\(a = \dfrac{1}{2+a}\\ a^2+2a-1=0\\ a = \dfrac{-2\pm \sqrt{4+4}}{2} = -1 \pm \sqrt{2}\\ \text{we can rule out the negative solution since everything is positive}\\ a = -1 + \sqrt{2}\)

\(x= 1+a = \sqrt{2}\)

Add 1 to both sides     and we have that

                                   1                                

x + 1  =    2  +       __________

                             2 +  1

                                  ________

                                    2 +  .......

So we have that

x  =   1 +    1

                ______          multiply  through by   x + 1

                 x + 1

x(x + 1)  = (x + 1)  + 1

x^2 + 1x =  1x + 2         subtract 1x from both sides

x^2    =  2          take the positive square root.....since the original right side is positive

x = √2

The center is   (h, k)  = ( -1,3)

One of the asymptotic lines- the one with a positive slope - passes through  ( 1, 6)  [ and the center]

So.....we can write an equation of this line.....the slope  =  [ 6 - 3 ] / [ 1 - - 1 ]  =  3/2

So.....we can write the eqation of this line as

y  =  (3/2)(x - -1) + k   =  (3/2) ( x + 1) + 3   =    (b/a) ( x - h) + k

So.....

a = 2    b   = 3    h  =  -1    and k  =  3        and their sum  = 7

Here's a graph : https://www.desmos.com/calculator/lxpgxphby3

Haha! I use the title "Help!" all the time! I feel like it's just easier.

Have you tried doodling on your paper? Here are some ideas: If you have some paper with some text, you can draw stick figures climbing or falling off of the text. If you have notebook paper with the 3 holes, you can try using those holes as circles in part of a drawing, like balloons or the center of a flower. You could draw houses. You could draw bubble or block letters. These are some things I used to do all the time.

There should be a filter for 'troll' and all the posts deemed as troll posts will pop up lol

The focal points will lie on the y axis.....and the circle will be centered at the origin

kx^2 + y^2  = 1    we can write

   x^2                 y^2

  _____   +       _____  =    1

 (1/√k)^2              1

The  focal distance will be  =  √[ 1 - (1/√k)^2]  =  √[ 1 - 1/k ]

The radius of the circle will be the square of this

So  ... since the circle is tangent to the ellipse at two points on the x axis.....the point  (  √ (1 - 1/k) , 0 )  will be on the ellipse...so 

[√ (1 - 1/k)]^2                0

__________  +         ___    =     1         simplifying, we have

    (1/ √k)^2                  1

1 - 1/k  =  1/k

1 = 2/k

k = 2

Here's a graph :  https://www.desmos.com/calculator/nlcn74kcb7

You can utilize

1) Your own brain

2) someone elses brain

3) A calculator

4) Google (brains, learning sites and calculators)

5) Resouce texts.

.....

The roots are  (1 + √12) ( 1-√12) (3 - √2)  and (3 + √2)

So........the polynomial is

a [ x - (1+√12) ] [ x - (1 - √12) ] [ x - ( 3 - √2) ] [ x- (3 + √2)]

Simplifying....we have that......

a [x^4 - 8 x^3 + 8 x^2 + 52 x - 77]  = 

And since f(0)  = -154

Then  -77a  = -154

So....a  = 2

So....the polynomial is

2x^4 - 16x^3 + 16x^2 + 104x - 154

And f(1)  =   2 - 16 + 16 + 104 - 154  =  -48

Thank youuuu!

F = [G * M * m] / d^2, solve for M
M = (d^2 F)/(G m) 
M = (3.8E8^2 * 2.046E20) / (6.67428E-11 * 5.974E24)
M =~ 7.4 x 10^22 Kg - The mass of the Moon.
Note: You have the gravitational force between the Earth and the Moon listed as 20.46 x 10^25 N, which is wrong. The actual force is about: 2 x 10^20 N.

x = 2y^2 -8y + 3

x - 3  = 2(y^2 - 4y)         complete the square on y

x - 3 + 8  = 2(y^2 - 4y + 4)       simplify

(x + 5)  = 2(y - 2)^2 

(1/2)(x + 5)  = (y - 2)^2        

Vertex  =  ( -5, 2)

We have the form   (4p)(x + 5) = (y - 2)^2

4p = 1/2

p = 1/8

This parabola opens to the right

The axis of symmetry is y = 2

The focus  =  ( -5 + p , 2)  = (-5 + 1/8, 2)  = ( -39/8, 2)

The directrix is    x = -5 - p  =  -5 - 1/8  =  -41/8

Here's the graph :