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A series is given by
\(\dfrac{1}{7^1}+\dfrac{2}{7^2}+\dfrac{3}{7^3}+\dfrac{4}{7^4}+\dfrac{5}{7^5}+\cdots \).

This is a infinite arithmetico-geometric sequence.

In general: \({\color{red}a}{\color{blue}b }+(a+d){\color{blue}br }+{\color{red}(a+2d)}{\color{blue}br^2 }+{\color{red}(a+3d)}{\color{blue}br^3 }+\cdots + {\color{red}\Big(a+(n-1)d \Big)}{\color{blue}br^{n-1} } + \cdots \)

\(\begin{array}{|lrcll|} \hline \text{arithmetical sequence:}\quad 1+2+3+4+5+\ldots \\ {\color{red}1}+({\color{red}1}+1*{\color{orange}1})+({\color{red}1}+2*{\color{orange}1})+({\color{red}1}+3*{\color{orange}1})+\dotsb+\Big({\color{red}1}+(n-1)*{\color{orange}1}\Big)+\dotsb \\ \boxed{a={\color{red}1},\ d={\color{orange}1} \text{ is the common difference} } \\\\ \text{geometric series:}\quad \frac{1}{7^1}+\frac{1}{7^2}+ \frac{1}{7^3}+ \frac{1}{7^4}+ \frac{1}{7^5}+\cdots \\ {\color{blue}\frac{1}{7} } +{\color{blue}\frac{1}{7} }\left({\color{green}\frac{1}{7}}\right)^1 +{\color{blue}\frac{1}{7} }\left({\color{green}\frac{1}{7}}\right)^2 +{\color{blue}\frac{1}{7} }\left({\color{green}\frac{1}{7}}\right)^3 +\dotsb +{\color{blue}\frac{1}{7} }\left({\color{green}\frac{1}{7}}\right)^{n-1} +\dotsb\\ \boxed{ b={\color{blue}\frac{1}{7} },\ r={\color{green}\frac{1}{7}}\ \text{ is the common ratio} } \\ \hline \end{array} \)

Formula:
sum of a infinite arithmetico-geometric sequence

\(\begin{array}{|rcll|} \hline \mathbf{s} &=& \mathbf{\left(\dfrac{b}{1-r}\right) \Bigg( a+d\left( \dfrac{r}{1-r} \right) \Bigg)} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{s} &=& \mathbf{\left(\dfrac{b}{1-r}\right) \Bigg( a+d\left( \dfrac{r}{1-r} \right) \Bigg)} \\\\ && \boxed{a={\color{red}1},\ d={\color{orange}1} \text{ is the common difference} } \\ &&\boxed{ b={\color{blue}\frac{1}{7} },\ r={\color{green}\frac{1}{7}}\ \text{ is the common ratio} } \\\\ s &=& \left(\dfrac{{\color{blue}\frac{1}{7}} }{1-{ \color{green}\frac{1}{7}} }\right) \Bigg( {\color{red}1} + { \color{orange}1} \left( \dfrac{ {\color{green}\frac{1}{7} }}{1-{\color{green}\frac{1}{7}} } \right) \Bigg) \\\\ s &=& \frac{1}{7}*\frac{7}{6} \left( {\color{red}1} + \frac{1}{7}*\frac{7}{6} \right) \\\\ s &=& \frac{1}{6} \left( {\color{red}1} + \frac{1}{6} \right) \\\\ s &=& \frac{1}{6} *\frac{7}{6} \\\\ \mathbf{s} &=& \mathbf{\frac{7}{36}} \\ \hline \end{array}\)

\(\mathbf{\dfrac{1}{7^1}+\dfrac{2}{7^2}+\dfrac{3}{7^3}+\dfrac{4}{7^4}+\dfrac{5}{7^5}+\cdots} = \mathbf{\dfrac{7}{36}}\)

does r=1/7?

What is the value of sin^-1 (−√3/2) ?

Hello Guest!

\(sin^{-1}(-\sqrt{\frac{3}{2}}\ )= \frac{1}{sin(- \sqrt{\frac{3}{2}}\cdot \frac{180^o}{\pi}\ )}=\frac{1}{\color{blue}sin(-70.17^o)}=-1.063\)

  !

Look  at the graph here   :   https://www.desmos.com/calculator/a6sxjsqyr4

Note that

There is no y intercept.....so  (1)  is false

From the graph....(2)  is true

There are no x intercepts.....(3)  is false

The last is true........csc  = r /y  .....    at   0  + n*pi    where n  is an integer,  y  = 0....and since we can't divide by  0, the  csc is not defined at these  values

We  just  need  to  find  the  LCD  of  6  and 9 .....   this  =  18

So.....if he buys     18/ 9    =  2 package of sausages    and    18 / 6  =   3 packets of  rolls  he will have the same number

Proof   

2 * 9    =18

And

3 * 6   =18

We  can solve this    as follows....

We  can find  his initial   distance from  the  tree, D,   thusly

tan (68°)  =  80  / D      rearrange as

D =  80   / tan (68°)  ≈  32.3  ft

Now....let  x  be  the  additional distance that he steps back and we  have that

tan  ( 41°)  = 80  /  ( 32.3  + x)     rearrange as

(32.3 + x)  tan (41°)   = 80

32.3*tan(41°)  +  x *  tan (41°)  =  80

x * tan (41°)  =  80  -  32.3 * tan (41°)

x =    [  80  - 32.3 * tan (41°)  ]   / tan (41°)  ≈    59.73  ft 

The second part  of this problem might be better solved with the  cotangent  because

cot (41°)  =  (32.3  + x)   / 80      is a little easier to  solve  !!!!

Each successive term is smaller than the preceding one......the series wil converge  ( to  7/36)

Ah I see thanks! I made a teeny tinsy mistake in the equation (it's supposed ot be 220.2 instead of 22.2). Does it still mean the same then?

We are asking ourselves.....where  does  the sin  = -√3/2  ?

This  is  a  60°  angle  in  the  4th quadrant    =   - pi / 3   (or   -60°  )

2. does not pass horizontal line test

9. [0,π]

[ Because  of  the nasty roots that we will find in this problem....I leaned heavily on WolframAlpha for  most of the computations ] 

Note that   AB    = BX   so.....AX   = 28

Using the Law of Cosines    we have

BC^2  = AB^2  + AC^2 - 2(AC*AB) cos CAX

15^2  =  14^2  +13^2  - 2(13 * 14) cos CAX

Solving for the cos CAX  we  get  that  cos (CAX)  = 5/13

So....the sin of CAX =  12/13 

And we can find   CX  using the Law of Cosines  again

CX^2  = AC^2  + AX^2  - 2(AC * AX) cos CAX

CX^2  = 13^2  + 28^2  - 2(13 * 28) (5/13)

Solving this for  CX  gives us  CX  =   √673

And again  we can find the cosine of angle  BCX  as

BX^2  =  BC^2  + CX^2  - 2(BC * CX) cos BCX

14^2  = 15^2  + 673  -2 ( 15 * √673) cosBCX 

Solving this  for  cos BCX gives  us   117/ [ 5 √673 ]

And sin BCX   =  √  [  1   - [ 117/ [ 5 √673 ] ]^2 ]    =   56 / [ 5√673]

Connect  BD

Since  quadrilateral ABDC  is inscribed in a circle....angles  CAB  and   CDB are supplemental....so the have the same sines  =12/13

So.....using the Law of Sines

BC / sin  CDB  =   BD  /sin BCX

15 / (12/13)  = BD  / [  56 / [ 5√673] ] 

Solving this gives that  BD   =  182  / √673

Finally.....using the Law of Cosines we  can  find   CD  as folllows

BD^2  =   BC^2  + CD^2   -  2 (BC * CD) cos BCX

Let  CD   = x   and we have that

182^2 / 673   =   15^2 + x^2  -  2 (15 * x)*(117/ [5sqrt (673)] )

Solving this for x  = CD  gives us two  values

CD  = 281 / √673  ≈  10.83   or      

CD  =  421 / √673  ≈  16.23

But     BC  =  15   and is clearly  greater than CD

So   CD =   281 / √673   ≈  10.83   is  the  correct solution

A short computer code could only find one such n:

When n=1. 1!  +  2!  +  3! = 9

Sqrt(9) = 3 - which is a perfect square.

How many numbers are in the list \(3\dfrac23,\, 4\dfrac13,\, 5,\, 5\dfrac23,\, \ldots,\, 26\dfrac13,\, 27\) ?

\(\begin{array}{|rcll|} \hline && 3\dfrac23,\, 4\dfrac13,\, 5,\, 5\dfrac23,\, \ldots,\, 26\dfrac13,\, 27 \\\\ &=& \dfrac{11}3,\, \dfrac{13}3,\, \dfrac{15}3,\, \dfrac{17}3,\, \ldots,\, \dfrac{79}3,\, \dfrac{81}3 \\\\ \text{arithmetic progression, formula: }\quad \mathbf{a_n} &=& \mathbf{a_1+(n-1)d} \\\\ && a_1 = \dfrac{11}3,\qquad a_n = \dfrac{81}3,\qquad d = \dfrac{2}3 \\\\ \dfrac{81}3 &=& \dfrac{11}3+(n-1)\dfrac{2}3 \\\\ \dfrac{81}3- \dfrac{11}3 &=& (n-1)\dfrac{2}3 \\\\ \dfrac{70}3 &=& (n-1)\dfrac{2}3 \quad | \quad *3 \\\\ 70 &=& (n-1)*2 \quad | \quad :2 \\\\ 35 &=& n-1 \\\\ n &=& 35+1 \\\\ \mathbf{n} &=& \mathbf{36} \\ \hline \end{array}\)

See the graph  for this region : https://www.desmos.com/calculator/q9jn22a8q4

Because we are rotating about the x axis,  the area will be a function of  y.....so.... we  need  to  write y  =x^3   as    x  = y^1/3

So  we  have

      8

2pi ∫  radius  *  height   dy      =

      0

       8

2pi  ∫  y  * y^(1/3)  dy   =       

       0

      8

2pi ∫   y ^ (4/3)   dy    =

     0

                                 8

2pi   * [ ( 3/7) y^(7/3) ]        =

                                 0

(6 pi / 7 )  *  [  ( 8^(7/3)   -  0^(7/3) ]  =

(6 pi /7) * [  8^(1/3) }^7   =

(6pi /7)  *  [ 2^7 ]  =

(6 pi / 7) * 128     ≈     344.68  units^3

Note  ...using the change-of-base theorem we can write

P  = log₂ 3  *  log ₃ 4 *  log ₄ 5 *   .......  *log ₂₀₀₇ 2008  * log ₂₀₀₈ 2009        as

                  log 3    *   log 4   *    log 5                        log 2008   *  log  2009

P =          _____     _____       _____    *  ......     *   _______      _________    =

                  log 2        log 3         log 4   .                    log 2007       log  2008

                    1                   log 2009     

                 ____    *         ________                     =

                  log 2                   1

                          log  2009

                         ________  =       log ₂ 2009  =   log _a b

                              log 2

a =  2    b   = 2009

a  +  b  = 

2011

Note  that   1323  factors as  3^3 * 7^2

So   1323 * 100 =    132,300  =  3^3 * 7^2 * 100  =   3^3 * 7^2 * 10^2

So....we have

log (132300)  = 

log  ( 3^3  * 7^2  * 10^2)          using a log property  that  log (a * b * c)  = log a + log b + log c

log 3^3  +  log 7^2  +  log 10^2      and another  property says that   log a^b   = b log a

3log3 + 2 log 7  +  2* log 10   =          [ log 10  = 1 ]

3 (.477) + 2 (.845)  + 2(1)  ≈ 

5.121

https://web2.0calc.com/questions/determining-the-approx-value-of-this-logarithm 

This will help but not give you the answer. I myself don't know how to do this.

\(4\cdot2\cdot1\cdot2\cdot2=32.\)

I did this because the 4 is for the hairstyles: bald and the other 3. The 2's are for the eyebrows, ears and lips and the 1 (didn't have to be included) is for the googly eyes.

No it is combination maths. C for combination or C for choice

9C4 = 9C5     

The brackets that Rom used is just another way to display this.

There is a C function on any 'proper' calculator.

9C4 means how many different combinations of 4 items can I choose if I have 9 items altogether to choose from.

I’m sure Rom will carefully explain this to you.

K thnx!!!

\(\dbinom{10}{5}5! = 30240\)

The time  =      4  /    [ 17 -12  ]  =  4/5  seconds

2log ( a -2b)  =  log a  + log b

By log properties, we can write

log (a - 2b)^2  = log (ab)

This implies that

(a -2b)^2   = ab       simplify

a^2 - 4ab  + 4b^2   = ab     subtract  ab  from both sides

a^2  -5ab +  4b^2    = 0        factor as

(a - 4b) ( a - b)  =   0

So...either

a - 4b   = 0                        or                  a  - b  = 0

a  = 4b  ⇒ a/b  = 4                                    a  =  b                  

The second answer (in red)  would  mean that  b would have to be negative in log (a - 2b)......but this isn't defined  for log  b

So

a/b  = 4