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Compute the sum

\(\frac{2}{1 \cdot 2 \cdot 3} + \frac{2}{2 \cdot 3 \cdot 4} + \frac{2}{3 \cdot 4 \cdot 5} + \cdots\)

The equation of the line through $2+3i$ which is perpendicular to the line through $0$ and $2+3i$ has equation \[az + b\overline{z} = 26,\]

where $a$ and $b$ are constant complex numbers.

Find the product $ab$ in rectangular form.

\(\begin{array}{|rcll|} \hline az + b\overline{z} &=& 26 \\ &\boxed{ z = b} \\ & \boxed{\overline{z} = a} \\ ab + ba &=& 26 \\ 2ab &=& 26 \\ \mathbf{ab} &\mathbf{=}& \mathbf{13} \\ \hline \end{array}\)

The equation of the line through $2 + 3i$ and $0$ can be written as \[az + b \overline{z} = 0\]for some complex numbers $a$ and $b$.

Find the quotient $b/a$ in rectangular form.

\(\begin{array}{|rcll|} \hline az + b \overline{z} &=& 0 \\ &&\boxed{ z = 2+3i} \\ && \boxed{\overline{z} = 2-3i} \\ a(2+3i) + b (2-3i) &=& 0 \\ b (2-3i) &=& -a(2+3i) \\\\ \dfrac{b}{a}&=& \dfrac{-(2+3i)} {(2-3i)} \\\\ \dfrac{b}{a}&=& \dfrac{-(2+3i)} {(2-3i)}\cdot\dfrac{(2+3i)} {(2+3i)} \\\\ \dfrac{b}{a}&=& \dfrac{-(2+3i)(2+3i)} {(2-3i)(2+3i)} \\\\ \dfrac{b}{a}&=& \dfrac{-(4+12i+9i^2)} {4-9i^2} \quad & | \quad i^2=-1 \\\\ \dfrac{b}{a}&=& \dfrac{-(4+12i-9)} {4+9} \\\\ \dfrac{b}{a}&=& \dfrac{-(-5+12i)} {13} \\\\ \dfrac{b}{a}&=& \dfrac{5-12i} {13} \\\\ \mathbf{\dfrac{b}{a}} &\mathbf{=}& \mathbf{\dfrac{5} {13} -\dfrac{12} {13}i} \\ \hline \end{array}\)

Consider the line with equation

\( (2-i)z + (2+i)\overline{z} = 20. \)

 \[(2-i)z + (2+i)\overline{z} = 20.\]
Where does this line intersect the real axis?

On the real axis:  \(z = \overline z\)

\(\begin{array}{|rcll|} \hline (2-i)z + (2+i)\overline{z} &=& 20 \quad & | \quad z = \overline z \\ (2-i)z + (2+i)z &=& 20 \\ z\Big((2-i)+(2+i)\Big) &=& 20 \\ z(2-i+2+i) &=& 20 \\ 4z &=& 20 \\ \mathbf{ z } & \mathbf{=} & \mathbf{5} \\ \hline \end{array}\)

The line intersect the real axis at 5.

Let the long side of a rectangle = x

Let the short side   =  y

By the diagram....it must be that    3x = 4y   ⇒  x = (4/3)y

So  L =  3x  = 4y

And W = 2y + x   =  2y + (4/3)y  =  (2 + 4/3)y = (10/3)y

So

L / W   =    (4y) /  [ (10/3) y ]  =    4 / (10/3)  =   12 / 10  =   6 / 5

2^(2/3) =  (2^2)^(1/3)  = (4)^(1/3) 

2^(-5/4)

Note that this is  (2/1)^(-5/4)

We can "flip" the fraction and make the exponent positive.....so we have

(1/2)^(5/4)  =   [(1/2)^5 ] ^(1/4)   =  [ 1/32 ]^(1/4)   

The last answer might also be simplified as    [ (1/16) (1/2) ]  ^(1/4)  =  

(1/16)^(1/4) * (1/2)^(1/4) =

(1/2)* (1/2)^(1/4)

Thanks Alan :)

No it wouldn't

I think this is your actual question.......

Alan is telling you the question that your friend has effectively answered, which is a different question.  Some confusion here. :)

Anyway....

\((2x^3y)^6\\ =(2\times x^3\times y)^6\\ =2^6\times(x^3)^6\times y^6\\ =64\times x^{18}\times y^6\\ =64 x^{18} y^6\\ \)

\(\sum \limits_{n=0}^\infty \dfrac{n!}{(n+1+\delta)!} = \dfrac{1}{\delta \cdot \delta!}\\ 2\sum \limits_{n=0}^\infty \dfrac{n!}{(n+3)!} = \dfrac{2}{2\cdot 2!} = \dfrac 1 2\)

If CD = DA, then AB = BC  [ not AB = CD ]  since a kite has two distinct [different ]  pairs of equal adjacent sides 

[If AB = CD, we would have a square or a rhombus ]   

Let the diagonals be AC and BD

And let their intersection = E

Since P is a centroid, then EP = 1/3  of EB

And similarly, EQ = 1/3 of ED

So.....the area of the kite =   AC * EB / 2   +  AC * ED / 2  =  (1/2) [ AC * EB + AC * ED ]

And the area of triangle ABC = AC * (1/3)EB / 2  =  AC * EB / 6

And the area of triangle ADC = AC * (1/3)ED / 2  = AC * EB / 6

So...the area of   PCQA =   area of triangle ABC + area of triangle ADC  =  (1/6) [ AC * EB + AC * ED] =

(1/3) (1/2) [ AC * EB + AC * ED ]  =

(1/3) area of kite  =   20

divide by 100 on top and bottom to get 620/8, then it is 310/4, then 155/2, so it is 77.5.

HOPE THIS HELPED!

Thanks! I wasn’t exactly sure what bisects means. 

Proof of this

Height * Width = Area

x * y   = 4

x * z =  6

w * z = 5

w * y = ???

y = 4 /x

w = 5/z

w * y   =     4/ x  *  5 /z     =   20 / xz  = 20 / 6   =   10 / 3

thanks

The products of diagonal areas are equal

Thus

4*5 = 6 *x  

20 = 6 * x       divide both sides by  6

20 / 6  =  x  =   10 / 3   =  missing area

Factors of 2   =  1 , 2

Factors of 6 =  1, 2, 3 , 6

The greatest in common is 2

Mmmmmm.....the question seems to have changed....LOL!!!

We need  8 *90  = 720 points on the 8 tests

So

720 -  5(87)  = 285 remaining

So....we need to average

285 /3   =   95  on the last three

y = 3.4 (5) + 5  =    17 + 5  = $22

360°

To see this....draw a diagonal connecting any pair of opposite vertices

This will split the trapezoid into two triangles....

The interior angles of each triangle will sum to 180°

So  ...2 * 180°  =  360°