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[28 - 1] / 3 + 1 = 10 terms for the sequence

1 + 4 +7 +......+ 28 =145

10x + 145 = 155

10x =155 - 145

x = 1

Hey Melody!  

  That is what I did (on paper) to find the other root.....I got the same numbers as you did for a and b.  I just didn't feel like transcribing it all on the keyboard when I was done !    (Maybe for a REGISTERED user who wanted derivation I would have...)

 The question states a and b are real.....well , yes they are, but 'b' is part of   b i   (which obviously is not real)...a bit misleading, but at least we both figured it out....

A 3-digit number is memorable if it is of the form 1B1 or A1C.  This gives us 10 + 10*10 = 110 3-digit memorable numbers.

1) Extend AD and BC to meet at P.  Then you get a 30-60-90 triangle and a 45-45-90 triangles, so by Pythagoras, CD = (6 - 2*sqrt(3)) + (4 + 2*sqrt(3)) = 10.

2) Using the area condition, the bases of the trapezoid work out to 8 and 16, so the median is (8 + 16)/2 = 12.

Using the distance conditions, a + bi works out to 4 + 8i, so a = 4 and b = 8.

There are two triangles with a base of sqrt(2), two triangles with a base of 2 sqrt(2), two triangles with a base of 3 sqrt(2), and three triangles with a base of 2, for a total of 2 + 2 + 2 + 3 = 9 triangles.
 

In the diagram, angles PQR and PST are right angles. If PR = 35, PT = 11, and QT = 10, then what is PS?

Let PS = x

\(\begin{array}{|rcll|} \hline \dfrac{x}{11} &=& \dfrac{21}{35} \\\\ x &=& \dfrac{11*21}{35} \\\\ \mathbf{x} &=& \mathbf{6.6} \\ \hline \end{array}\)

PS is 6.6

The circle x^2 + y^2 + Ax + By + C = 0 passes through the origin, and has the same center as the circle x^2 + y^2 - 4x + 6y - 3 = 0.  What is A + B + C?

Thank you, CPhill !

Find the numerical value of \(\sin(40^\circ)*\sin(80^\circ) + \sin(80^\circ)*\sin(160^\circ) + \sin(160^\circ)*\sin(320^\circ)\)

\(\begin{array}{|rcll|} \hline && \sin(40^\circ)\sin(80^\circ) + \sin(80^\circ)\sin(160^\circ) + \sin(160^\circ)\mathbf{\sin(320^\circ)} \\\\ && \boxed{ \mathbf{\sin(320^\circ)} = \sin(360^\circ-40^\circ) = -\sin(40^\circ) } \\\\ &=& \sin(40^\circ)\sin(80^\circ) + \sin(80^\circ)\sin(160^\circ) - \sin(160^\circ)\sin(40^\circ) \\ &=& \sin(40^\circ)\sin(120^\circ-40^\circ) + \sin(120^\circ-40^\circ)\sin(120^\circ+40^\circ) - \sin(120^\circ+40^\circ)\sin(40^\circ) \\ &=& \sin(40^\circ)\sin(120^\circ-40^\circ) - \sin(120^\circ+40^\circ)\sin(40^\circ)+ \sin(120^\circ-40^\circ)\sin(120^\circ+40^\circ) \\ &=& \sin(40^\circ)\Big( \underbrace{\sin(120^\circ-40^\circ) - \sin(120^\circ+40^\circ)}_{\boxed{\mathbf{\text{Formula:}~2\cos(x)\sin(x) = \sin(x+y)-\sin(x-y)} \\ \sin(120^\circ-40^\circ) - \sin(120^\circ+40^\circ) = -2\cos(120^\circ)\sin(40^\circ) }} \Big)+ \sin(120^\circ-40^\circ)\sin(120^\circ+40^\circ) \\\\ &=& \sin(40^\circ)\Big( -2\mathbf{\cos(120^\circ)}\sin(40^\circ)\Big)+ \sin(120^\circ-40^\circ)\sin(120^\circ+40^\circ) \\\\ && \boxed{ \mathbf{\cos(120^\circ)} = \cos(180^\circ-60^\circ) = -\cos(60^\circ)=-\dfrac{1}{2} } \\\\ &=& \sin(40^\circ)\Big( -2\left(-\dfrac{1}{2}\right)\sin(40^\circ)\Big)+ \sin(120^\circ-40^\circ)\sin(120^\circ+40^\circ) \\ &=& \sin^2(40^\circ)+\underbrace{\sin(120^\circ-40^\circ)\sin(120^\circ+40^\circ)}_{ \boxed{ \mathbf{\text{Formula:}~-2\sin(x-y)\sin(x+y) = \cos(2x)-\cos(2y)}\\ \sin(120^\circ-40^\circ)\sin(120^\circ+40^\circ)=-\dfrac{1}{2}\Big(\cos(240^\circ)-\cos(80^\circ) \Big) } } \\\\ &=& \sin^2(40^\circ)-\dfrac{1}{2}\Big(\mathbf{\cos(240^\circ)-\cos(80^\circ)} \Big) \\ && \boxed{ \mathbf{\cos(240^\circ)} = \cos(180^\circ+60^\circ) = -\cos(60^\circ)=-\dfrac{1}{2}\\ \cos(80^\circ) = \cos(2*40^\circ) = 2\cos^2(40^\circ)-1 } \\\\ &=& \sin^2(40^\circ)-\dfrac{1}{2} \Bigg( -\dfrac{1}{2} -\Big( 2\cos^2(40^\circ)-1\Big) \Bigg) \\ &=& \sin^2(40^\circ)-\dfrac{1}{2} \Big( -\dfrac{1}{2} - 2\cos^2(40^\circ)+1 \Big) \\ &=& \sin^2(40^\circ)+\dfrac{1}{4} + \cos^2(40^\circ)-\dfrac{1}{2} \\ &=& \underbrace{\sin^2(40^\circ) + \cos^2(40^\circ)}_{=1} +\dfrac{1}{4}-\dfrac{1}{2} \\ &=& 1 +\dfrac{1}{4}-\dfrac{1}{2} \\\\ &=& \dfrac{1}{4}+\dfrac{1}{2} \\\\ &=& \mathbf{ \dfrac{3}{4} } \\ \hline \end{array}\)

Here's a pic of the situation :

Since AD  is an angle bisector.....then  DB/ DC  = AB/AC

So

25/7  = AB/ AC

AB = (25/7)AC

And  since AB is a diamemter,  triangle ACB  is right....and BC  = BD + DC  =  25 + 7  = 32

So

BC^2  + AC^2  = AB^2

32^2  + AC^2  = [(25/7)AC]^2

1024  = [625/49  - 1 ] AC^2

1024  = [576/49]AC^2

1024 * [49 / 576]  = AC^2

AC = sqrt (1024 * 49/ 576 ]  =   32*7 / 24  = 28/3

And triangle ACD  is also right...with hypotenuse AD  and legs CD and AC....so....

AD^2 = CD^2  + AC^2

AD^2  = 7^2  + (28/3)^2

AD^2 = 49 + 784/ 9

AD^2 = [ 441 + 784 ] / 9

AD=  sqrt [ 1225/ 9]  = 35/3

And by the intersecting chord theorem

BD * DC = AD*DE

25 * 7  = (35/3)(DE)

175 = (35/3) (DE)

DE = 175 (3) / 35  = 15

And  triangle DEB is also right   with   hypotenuse DB = 25   and leg DE = 15

So

DB^2  = DE^2 + BE^2

25^2  = 15^2  + BE^2

25^2 - 15^2 = BE^2

625 - 225 = BE^2

400 = BE^2

20  = BE

Very nice, heureka   !!!!!

Rearrange as

x^2 - (1/2)x +  y^2 + y  =  1/2      complete the square on x and y

x^2 - (1/2)x + 1/16 + y^2 + y + 1/4  =  1/2 + 1/16 + 1/4

Factor  and simplify

(x - 1/4)^2  + (y + 1/2)^2  =  8/16 + 1/16 + 4/16

(x - 1/4)^2  + (y + 1/2)^2  =  13/16

Center  = (1/4 , -1/2)

Radius = √ (13/16 )  =  √13  / 4

Multiply  through by 2

2sin(40) sin(80)  + 2sin(80)sin(160)  + 2sin (160)sin(320)

Using ....[ 2sin AsinB ]   =  cos(A -B) - cos(A + B)

cos ( 40 - 80) - cos(40+ 80)  + cos(80 - 160) - cos(80 + 160)  + cos(160 - 320) - cos(160 + 320)  =

cos(-40) - cos (120)  + cos (-80) - cos(240)  + cos (160 - 320) - cos (480)  =

[cos(480) = cos(240)  =  cos (120)  = -1/2 ] 

cos (-40) - (-1/2)  + cos(-80)  - (-1/2) + cos (-160)  - (-1/2)   =

cos (40)  + cos(80)  + cos (200)  + 3/2   =

{Using....cos A + cos B  = 2[cos ((A + B)/2)] * [ cos( (A - B)/2) ]

2[ cos ((80 + 40)/2) ] * [ cos ( ((80 - 40)/2) ]   + cos(200) + 3/2  =

2cos (60) cos (20)  +  cos (200) +  3/2  =

2(1/2) cos (20) + cos(200)  + 3/2  =

cos (20 )  + cos(200) +  3/2   =

cos(20) + cos (180 + 20) + 3/2   =

cos(20)  + [ cos180cos20  - sin180sin20 ]  +  3/2  =

cos(20) + (-1)cos(20) - (0)sin(20) + 3/2  =

cos (20)  - cos(20)  + 3/2  =

3/2

But....since we multplied  the original identity by 2.....we must divide the result by 2  to get the correct answer  = 

3 / 4 

The spherical coordinates of (-3, 4, -12) are \((\rho, \theta, \varphi)\).
Find \(\tan \theta + \tan \varphi\).

I assume: the spherical coordinates are \((\text{radius}~ \rho, \text{inclination}~ \theta, \text{azimuth}~ \varphi)\)

\(\begin{array}{|lrcrl|} \hline (1) & x &=& -3 &= \rho \sin(\theta) \cos(\varphi) \\ (2) & y &=& 4 &= \rho \sin(\theta) \sin(\varphi) \\ (3) & z &=& -12 &= \rho \cos(\theta) \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline \dfrac{(2)}{(1)}: & \dfrac{\rho \sin(\theta) \sin(\varphi)}{\rho \sin(\theta) \cos(\varphi)} &=& \dfrac{4}{-3} \\ & \mathbf{\tan(\varphi)} &=& \mathbf{-\dfrac{4}{3}} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \dfrac{(2)}{(3)}: & \dfrac{\rho \sin(\theta) \sin(\varphi)} {\rho \cos(\theta)} &=& \dfrac{4}{ -12} \\ & \tan(\theta)\sin(\varphi) &=& -\dfrac{1}{3} \quad | \quad \text{square both sides} \\ & \mathbf{\tan^2(\theta)\sin^2(\varphi)} &=& \mathbf{\dfrac{1}{9}} \\ \hline \dfrac{(1)}{(3)}: & \dfrac{\rho \sin(\theta) \cos(\varphi)} {\rho \cos(\theta)} &=& \dfrac{-3}{ -12} \\ & \tan(\theta)\cos(\varphi) &=& \dfrac{1}{4} \quad | \quad \text{square both sides} \\ & \mathbf{\tan^2(\theta)\cos^2(\varphi)} &=& \mathbf{\dfrac{1}{16}} \\ \hline & \tan^2(\theta)\sin^2(\varphi)+\tan^2(\theta)\cos^2(\varphi) &=& \dfrac{1}{9}+\dfrac{1}{16} \\ & \tan^2(\theta)\left(\underbrace{ \sin^2(\varphi)+ \cos^2(\varphi)}_{=1} \right) &=& \dfrac{25}{9*16} \\ & \tan^2(\theta) &=& \dfrac{25}{9*16} \\ & \tan(\theta) &=& \dfrac{5}{3*4} \\ & \mathbf{\tan(\theta)} &=& \mathbf{\dfrac{5}{12}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \tan(\theta) + \tan(\varphi) &=& \dfrac{5}{12} -\dfrac{4}{3} \\\\ &=& \dfrac{15-48}{36} \\\\ &=& \dfrac{-33}{36} \\\\ \mathbf{\tan(\theta) + \tan(\varphi)} &=& \mathbf{-\dfrac{11}{12}} \\ \hline \end{array}\)

Lots try counting them systematically

1 x 2 ?

1 x 3 ?

1 x 4 ?

1 x 5 ?

2 x 3 ?

2 x 4 ?

2 x 5 ?

3 x 4 ?

3 x 5 ?

4 x 5 ?

that is it.

Please no one else interfer.  Asker guest can do this by themselves now.  How else will they learn? 

f t b l l         o o a

There are 3 ways to arrange the vowels with each other.

there are 5!/2! = 3*4*5 = 60 ways to arrange the consonants with respect to each other

There are 6 positions that the vowels can go

so that is

60*3*6 = 1080 ways     

That is what I get anyway. 

                    x^2  - 1           (x  -1) (x + 1)

f(x)  =         _______  =     ___________    =   - [  x + 1 ]

                       1 - x              -  ( x - 1)

f(2)  = - [ 2 + 1 ]=  -3

g(2)  = 2 + 2  = 4

So

f(g(2))  - g(f(2))  =

f ( 4)  - g (-3)  =

- [ 4 + 1 ]  - [ -3 + 2]   =

- 5  - [ -1 ]  =

-4

THX, hectictar  and heureka    !!!!!

Here goes  nothing  !!!

[ - 3 - ( a + h)^2  ]  -  [ -3 - (a -h)^2  ]

_____________________________   =

                  h

[ -3 - ( a^2 + 2ah + h)^2 ]  - [ -3 - (a^2 - 2ah + h^2)  ]

__________________________________________   =

                       h

[-3 - a^2 - 2ah - h^2    - [ -3 - a^2 + 2ah -  h^2  ]

________________________________________  =

                        h

[ -3 -a^2 -2ah - h^2   + 3 + a^2 - 2ah  + h^2 ]

_______________________________________   =

                         h

- 4ah

____   =

   h

-4a

 (cos x/2-sin x/2)^2 = 1-sin x

Work with the left side

[ cos (x/2)  - sin (x/2)] ^2  =

{cos (x/2) - sin(x/2) ] * [ cos(x/2) - sin (x/2) ]  =

cos^2 (x/2)  - 2sin(x/2)cos(x/2)  + sin^2 (x/2)

Let x/2   = θ     ......and we have that

cos^2 (θ)  -  2sin(θ)cos(θ)  + sin^2(θ)   =

sin^2(θ)  + cos^2(θ)  - 2sin(θ)cos(θ)  =

1   -  sin (2θ)    .....now.....back-substitute

1  - sin [ 2  (x/2) ]   =

1 - sin x         which =  the right side  !!!