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f)

$\dfrac{3x^3-3a^2x}{x^2-2ax+a^2}\cdot\dfrac{a-x}{a^3x+a^2x^2}\\ =\dfrac{3x(x-a)(x+a)}{(x-a)(x-a)}\cdot\left(-\dfrac{x-a}{a^2x(a+x)}\right)\\ =\dfrac{-3x(x-a)^2(x+a)}{a^2x(x-a)^2(x+a)}\\ =-\dfrac{3}{a^2}$

To help describe where it cuts off, it cuts off at the bottom of the rectangle designated for the calculator. If you could please find a solution to why it does this, that would be very helpful.

Here is an example:

I always have google chrome at 100%, my ocd says no to any other size XD

But thank you anyway for answering CPhill. You always come through for me!

Check your screen size.....if it's too large, it will "cut off"  these functions.......

Im using Google Chrome set at 125%.....the functions show up just fine, for me.....

how to work out the size of an exterior angle of a regular pentagon

interior angle of a regular pentagon = [180(5-2)]/5 = 36*3= 108 degrees

Exterior angle = 180-108 = 72 degrees

Hi Juriemagic,

Your factorisation on the right is not correct.

the first few lines are good but

$9x^2+1 \ne (3x+1)^2\\~\\ (3x+1)^2=9x^2+6x+1$

Other than that little error our answers are the same I think :)

Hi Melody,

the left side I also got, however this is what I did with the right side:

${18x^3} \over9x^3+x-27x^2-3$

$18x^3 \over(9x^3+x)+(-27x^2-3)$

$18x^3 \over x(9x^2+1)-3(9x^2+1)$

$18x^3 \over (x-3)(9x^2+1)$

$18x^2 \over (x-3)(3x+1)(3x+1)$

so now it's;

$-(3x+1)-{18x^3 \over(x-3)(3x+1)(3x+1)}$

and this is where it ends for me...yes, I do not know, either = or * would have worked nicely, but subtraction?..I don't know..

Melody, anyways, thank you kindly once again for your help...I think I'll let this one pass..

The others were equations .. this one is a subtraction.

Maybe it was meant to be an equals sign ?

I do not think this simplifies very nicely

${3x(2x-3)-(3-2x) \over 3-2x} - {18x^3 \over 9x^3-27x^2+x-3}\\ ={-3x(3-2x)-1(3-2x) \over 3-2x} - {18x^3 \over 9x^2(x-3)+1(x-3)}\\ ={(-3x-1)(3-2x) \over 3-2x} - {18x^3 \over (9x^2+1)(x-3)}\\ ={(-3x-1)(3-2x) \over 3-2x} - {18x^3 \over (9x^2+1)(x-3)}\\ ={-(3x+1)} - {18x^3 \over (9x^2+1)(x-3)}\\ ={-(3x+1) (9x^2+1)(x-3)-18x^3 \over (9x^2+1)(x-3)}\\$

You can expand the numerator and denominator out if you want but nothing will cancel.

I should have added that x cannot equal 3 or 3/2

7^{x^2 - 9}  = 1    ...... note  1  =  7⁰

 So

7^{x^2 - 9}  =  7⁰     since the bases are the same, solve for the exponrents

x^2  - 9  = 0

(x + 3) (x - 3)  = 0

Set each factor to 0   and solve for x  ...so   ....x = 3   or  x = - 3

Heureka,

thank you kindly...something so small that i just do not think of...I do appreciate your help!

The second one for today is also about solving

$\mathbf{7^{x^2-9}=1}$

$\begin{array}{|rcll|} \hline 7^{x^2-9} &=& 1 \quad & | \quad \cdot 7 \\ 7^{x^2-9}\cdot 7 &=& 7 \\ 7^{x^2-9}\cdot 7^1 &=& 7 \quad & | \quad a^b * a^c = a^{b+c} \\ 7^{x^2-9+1} &=& 7 \\ 7^{x^2-8} &=& 7^1 \\ x^2-8 &=& 1 \\ x^2 &=& 9 \quad & | \quad \text{square root both sides} \\ x &=& \pm 3 \\ \hline \end{array}$

Thank you CPhill,

I would never have thought of that..your help is greatly appreciated!

3.2^x  =  24    take the log of both sides

log  3.2^x  =  log 24       and we can write

x * log 3.2  = log 24       divide both sides by  log 3.2

x  =  log 24  / log 3.2  ≈   2.732

determining the sum of this arithmetic series?

5 + 8 + 11 + ... + 53

arithmetic series:

$\begin{array}{lrrrrrrrrrrrrrrrrr} & {\color{red}d_0 = 5} && 8 && 11 && 14 && 17 && \cdots && 53 \\ \text{1. Difference } && {\color{red}d_1 = 3} && 3 && 3 && 3 && \cdots && 3 \\ \end{array}$

Formula:

$\begin{array}{|rcl|} \hline a_n &=& \binom{n-1}{0}\cdot {\color{red}d_0 } + \binom{n-1}{1}\cdot {\color{red}d_1 } \\ s &=& \binom{n}{1}\cdot {\color{red}d_0 } + \binom{n}{2}\cdot {\color{red}d_1 } \\ \hline \end{array}$

n = ?

$\begin{array}{|rcl|} \hline a_n &=& \binom{n-1}{0}\cdot {\color{red}d_0 } + \binom{n-1}{1}\cdot {\color{red}d_1 } \quad & | \quad a_n = 53 \quad {\color{red}d_0 } = 5 \quad {\color{red}d_1 } = 3 \\\\ 53 &=& \binom{n-1}{0}\cdot {\color{red}5 } + \binom{n-1}{1}\cdot {\color{red}3 } \\ 53 &=& 5+(n-1)\cdot 3 \\ 53 &=& 2 +3n \quad & | \quad -2\\ 51 &=& 3n \quad & | \quad :3 \\ 17 &=& n \\ \mathbf{n} &\mathbf{=}& \mathbf{17} \\ \hline \end{array}$

sum (s) = ?

$\begin{array}{|rcl|} \hline s &=& \binom{n}{1}\cdot {\color{red}d_0 } + \binom{n}{2}\cdot {\color{red}d_1 } \quad & | \quad n = 17 \quad {\color{red}d_0 } = 5 \quad {\color{red}d_1 } = 3 \\\\ s &=& \binom{17}{1}\cdot {\color{red}5 } + \binom{17}{2}\cdot {\color{red}3 } \\ s &=& 17\cdot 5 + \frac{17}{2} \cdot \frac{16}{1}\cdot 3 \\ s &=& 17\cdot \left(5+\frac{16}{2}\cdot 3 \right) \\ s &=& 17\cdot(5+ 24) \\ s &=& 17\cdot 29 \\ \mathbf{s} &\mathbf{=}& \mathbf{493} \\ \hline \end{array}$

The sum is 493

Here's the   Calculus solution for this one...it's a little difficult  !!!

 (x - 3)^2 + (y - 3)^2 = 6

(y - 3)^2  =  6 -  [ x^2 - 6x + 9]

( y - 3)^2  =  6x- x^2 - 3

y - 3  = √ [ 6x- x^2 - 3] 

y = √ [ 6x- x^2 - 3]  + 3

So......call y/x  =  z  ....so we have

z =  y/x  =   [ √ [ 6x- x^2 - 3]  + 3]  / x

The derivative  of this is messy, but we have

z'  =   - 3 ( x  + √ [ 6x- x^2 - 3] - 1 )  /   [ x^2 * √ [ 6x- x^2 - 3] ]

We  can find a solution that minimizes  x   by  solving   this :

( x  + √ [ 6x- x^2 - 3] - 1 )  =  0

x - 1  =   -√ [ 6x- x^2 - 3]        square both sides

x^2 - 2x + 1  =  6x- x^2 - 3

2x^2 - 8x + 4  =  0

x^2 - 4x + 2  = 0

The solution that minimizes x  is      2 - √ 2

And using    (x - 3)^2 + (y - 3)^2 = 6    

When  x =  2 - √ 2     we have

( 2 - √ 2 - 3 )^2 +  (y - 3)^2 = 6

( 1 + √ 2)^2 +  (y - 3)^2 = 6

1 + 2 + 2√ 2 + (y - 3)^2 = 6

 (y - 3)^2 = 3 -  2√ 2

y =  √ [ 3 -  2√ 2] + 3    ⇒ √  [ ( 1 -√2  ) ^2 ]  + 3   or √  [ ( √2 - 1 ) ^2 ]  + 3  ⇒   

y =  4 - √ 2     or   y =   2 + √ 2

And   

[  2 + √ 2 ] / [  2 - √ 2 ]   maximizes  y / x

N =((L - F) / D) + 1, where N=Number of terms, L=Last term, F=First term, D=Common difference

N =((53 - 5) / 3) + 1

N = 17 Number of terms.

S =(F+L) / 2  x  N  , where F=First term, L=Last term. N=Number of terms, S=The Sum.

S =(53+5)/2  x  17 

S =58/2  x  17 

S =29   x  17 =493

Let's arrange the series in a slightly different manner

53 +  [ 50 + 5 ]  +  [ 47 + 8 ] +  [ 44 + 11]  , etc.

So....we have some "n"  pairs each summing to 55  plus the last term

Finding the number of pairs is a little tricky....but we can use this "formula"

n  =  (  [ last even term - first term ] / 3   + 1 )  / 2

So....the number of pairs is

 ([ 50 - 5 ] / 3  +   1)  / 2  =     

(45/ 3  + 1) / 2  =

[15 + 1 ] / 2 =

16 / 2  =

8

So......the sum is``    53  + (  8 * 55 )  =    493

Assuming that each kid got at least one piece of candy [ we don't want any riots....!!! ]

The number of ways is 

C  [ k -1 , n - 1 ]

Where =  10   and n  = 4

So we have

C [ 9 , 3  ]  =  84  ways

[ This is analogous to the problem of distributing k identical balls into n distinguishable boxes]

[ x^2 + 1 ] / [ x - 2]

You would need Calculus to find the actual range

The graph here shows the approximate range :

https://www.desmos.com/calculator/gxvksotyo6

The range is  ( - inf, -.472 ] U  [ ≈ 8.472, inf )

[6x + 2 ] / [ x + 2 ]

The  function  will tend toward   y = [ 6x / x  ]  =   6   at both ends

See the graph, here :    https://www.desmos.com/calculator/mebyec89wp

Alternatively.....if you haven't had Calculus....you could solve this directly for x

6 / [ x + 1 ] =  9 / [ 14 - x ] 

Cross-multiply

6 [ 14 - x ]  = 9 [ x + 1 ]

84 - 6x  =   9x + 9

75  = 15x

 x  = 5       =  c  =   (5, 0 )    

Let the point  c on the x axis be  (x, 0)

ac + cb  can be represented as D =

D  =  sqrt [ (x + 1)^2 + 6^2 ]  + sqrt [ (14- x)^2 + 9^2]

D = [  x^2 + 2x  + 37 ]^(1/2) + [ x^2  - 28x + 277 ]^(1/2)

Take the derivative and set to 0

D'  =  [ 2x + 2] / ( 2  [  x^2 + 2x  + 37 ]^(1/2))  +  [ 2x - 28] / (2   [ x^2  - 28x + 277 ]^(1/2) )  = 0

[ x + 1 ] /   [  x^2 + 2x  + 37 ]^(1/2)  +  [ x - 14] / [ x^2  - 28x + 277 ]^(1/2)  = 0

[ x + 1 ] /   [  x^2 + 2x  + 37 ]^(1/2)  =   [ 14 - x ] /  [ x^2  - 28x + 277 ]^(1/2)

Square both sides

[ x^2 + 2x + 1 ] /  [  x^2 + 2x  + 37 ] =  [ x^2 - 28x + 196] /  [  x^2 -28x + 277 ]

Cross - multiply

[ x^2 + 2x + 1 ] [  x^2 -28x + 277 ]  =  [ x^2 - 28x + 196]  [  x^2 + 2x  + 37 ]

Simplify

x^4 - 26 x^3 + 222 x^2 + 526 x + 277  =  x^4 - 26 x^3 + 177 x^2 - 644 x + 7252

45x^2  + 1170x  - 6975  =  0

9x^2  + 234x -  1395  =  0 

Solving this for  x produces   x = 5  or x = -31  [ reject the second solution ]

So.....the distance is minimized when   c  =   (5, 0 )

We can prove this if

arctan  [  6/[5- -1] ]  =  arctan [ 9 / [ 14 - 5] ]   is true

arctan  [  6/6]  =  arctan [ 9/9]

arctan 1  =   arctan 1         and it is true

And the minimum distance  ac + cb  is

sqrt [ (5 + 1)^2 + 6^2 ]  + sqrt [ (14- 5)^2 + 9^2] =

√ 72  +  √ 162  =

√ 2  [ 6 + 9]  =

15√ 2  units  

Thank CPhill!