All Answers 355975 Answers

E  is the midpoint of  AC,  so  AE  is half the length of  AC,  or   AE  =  1/2 * AC

One way to find the scale factor from  △ABC  to  △AFE  is to find  AE / AC

AE / AC   =   (1/2 * AC) / AC   =   1/2

So the scale factor is 1/2

(Remember the scale factor is the number which each side of the starting triangle must be multiplied by in order to make it the same size as the "goal" triangle. In this case, each side of △ABC must be multiplied by 1/2 to make it the same size as △AFE.)

Way off

Nice work hectictar, heureka and Melody....

I don't know much about these.....maybe I'll learn something  by  looking at each  solution  !!!!

We have an isosceles right triangle..... legs  = 4sqrt (2)    hypotenuse = 8

Perimeter  =   8  + 8sqrt (2)  =  8 ( 1 + sqrt (2) )

Let B = (0,0)

Let A  = (0, 4sqrt(2) )

C = (4sqrt(2) ,0)

Formula  to find the incenter  =

x = (  side opposite A * x coordinate of A   + side opp B * x coordinate of B + side opp C * x coordinate C) / perimeter

So we have

x =  ( 4sqrt (2) * 0  + 8 * 0  + 4sqrt (2)*4sqrt (2) )/ [ 8 (1 + sqrt (2))]  =

(32)  / [ 8 (1 + sqrt (2)]  =     

4/ ( 1 + sqrt (2) ) =         ( multiply by   1 -sqrt (2) on top/bottom  and simplify ]

4   ( 1 - sqrt (2) ) / (-1)   =

4 (sqrt (2)  - 1  ) =  4sqrt (2)  - 4   =    sqrt  (32) - 4

y = ( side opposite A *y coordinate of A  + side opp B * y coordinate of B + side opp C * y coordinate of C) /perimeter

If you do this correctly  you will find that  x  = y  =      sqrt  (32)  - 4

BI  =    sqrt    ( [ sqrt (32)  - 4 - 0 ]^2  + [ sqrt (32) - 4 - 0]^2 )  =

sqrt  [   32 - 8sqrt (32) + 16   +  32  - 8sqrt (2) + 16  ]  =

sqrt  [ 96  - 16 sqrt (32) ] =

sqrt [ 16  (6 - sqrt (32) ]  =

4 sqrt [ 6 - 4sqrt (2 ]   ≈  2.343  units

See the following image :

i is the intersection of the angle bisectors......

Sorry,

I started mine hours ago and got called away.

I did not know that Heureka and Hectictar had already answered it.

Lets see

\(z^2 + z + 1 = 0\\~\\ z=\frac{-1\pm \sqrt{1-4}}{2}\\ z=\frac{-1\pm \sqrt{-3}}{2}\\ z^2=\frac{1+-3\mp2\sqrt{-3}}{4}\\ z^2=\frac{-2\mp2\sqrt{-3}}{4}\\ z^2=\frac{-1\mp\sqrt{-3}}{2}\\ z^2=\bar z \)

\(z^3=z^2*z=\bar z*z=\frac{1}{4}-\frac{-3}{4}=1\\ so\\ z^{3n}=1\qquad \text{where n is a positive integer} so\\ z^{48}=z^{51}=1\)

\(z^{49}=z\\ z^{50}=\bar z\\ z^{51}=1\\ z^{52}=z\\ z^{53}=\bar z\\~\\ \text{Add them together and get}\\ 1+2(z+\bar z)\\ =1+2(\frac{-1}{2}+\frac{-1}{2})\\ =1+2(-1)\\ =1-2\\ =-1 \)

LaTex:

z^2 + z + 1 = 0\\~\\
z=\frac{-1\pm \sqrt{1-4}}{2}\\
z=\frac{-1\pm \sqrt{-3}}{2}\\
z^2=\frac{1+-3\mp2\sqrt{-3}}{4}\\
z^2=\frac{-2\mp2\sqrt{-3}}{4}\\
z^2=\frac{-1\mp\sqrt{-3}}{2}\\
z^2=\bar z

z^3=z^2*z=\bar z*z=\frac{1}{4}-\frac{-3}{4}=1\\
so\\
z^{3n}=1\qquad \text{where n is a positive integer}
so\\
z^{48}=z^{51}=1

z^{49}=z\\
z^{50}=\bar z\\
z^{51}=1\\
z^{52}=z\\
z^{53}=\bar z\\~\\
\text{Add them together and get}\\
1+2(z+\bar z)\\
=1+2(\frac{-1}{2}+\frac{-1}{2})\\
=1+2(-1)\\
=1-2\\
=-1

If \(z^2 + z + 1 = 0\), find
\(z^{49} + z^{50} + z^{51} + z^{52} + z^{53}\).

1)

\(\begin{array}{|rcll|} \hline && \mathbf{z^{49} + z^{50} + z^{51} + z^{52} + z^{53}} \\ &=& \left(z^{49} + z^{50} + z^{51} \right) + z^{52} + z^{53} \\ &=& z^{49}\left(1+ z + z^2 \right) + z^{52} + z^{53} \quad | \quad \mathbf{z^2 + z + 1 = 0} \\ &=& z^{49}\cdot 0 + z^{52} + z^{53} \\ &=& \mathbf{z^{52} + z^{53}} \\ \hline \end{array}\)

2)

\(\begin{array}{|rcll|} \hline && \mathbf{z^{49} + z^{50} + z^{51} + z^{52} + z^{53}} \\ &=& z^{49} + \left(z^{50} + z^{51} + z^{52}\right) + z^{53} \\ &=& z^{49}+ z^{50}\left(1+ z + z^2 \right) + z^{53} \quad | \quad \mathbf{z^2 + z + 1 = 0} \\ &=& z^{49}+ z^{50}\cdot 0 + z^{53} \\ &=& \mathbf{z^{49} + z^{53}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline z^{52} + z^{53} &=& z^{49} + z^{53} \\ z^{52} &=& z^{49} \quad | \quad :z^{49}\\ \dfrac{z^{52}}{z^{49}} &=& 1 \\ z^{52-49} &=& 1 \\ \mathbf{z^{3}} &=& \mathbf{1} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{z^{49} + z^{50} + z^{51} + z^{52} + z^{53}} &=& \mathbf{z^{49} + z^{53}} \\ &=& z^{3\cdot16+1} + z^{3\cdot17+2} \\ &=& z^{3\cdot16}z + z^{3\cdot17}z^2 \\ &=& \left(z^3\right)^{16}z + \left(z^3\right)^{17}z^2 \quad | \quad \mathbf{z^{3} = 1 } \\ &=& 1^{16}z + 1^{17}z^2 \\ &=& z + z^2 \\ && \boxed{z^2 + z + 1 = 0 \\ z^2 + z = -1 } \\ \mathbf{z^{49} + z^{50} + z^{51} + z^{52} + z^{53}} &=& \mathbf{-1} \\ \hline \end{array}\)

Ron, if you hadn’t responded to this, I’d never have known it was you. LOL

The post I trolled is definitely BB esque –it’s exactly the kind of post that at least two of the forum’s BBs would make (I can give examples).  Even though you’ve claimed ownership, the only corroborating evidence is that you included a copy of the question in your post. The BBs rarely do this. 

IDK if this is the best way...but we can first simplify it like this:

z⁴⁹ + z⁵⁰ + z⁵¹ + z⁵² + z⁵³

                                                        Factor  z⁴⁹  out of the first three terms and  z⁵¹ out of the last two terms

=  z⁴⁹(1 + z + z²)  +  z⁵¹(z + z²)

                                                        Since  z² + z + 1 = 0,   1 + z + z² =  0   and   z + z²  =  -1

=  z⁴⁹( 0 )  +  z⁵¹( -1 )

=  -z⁵¹

Now by the quadratic formula,  \(z\ =\ \frac{-1\pm\sqrt{1^2-4(1)(1)}}{2(1)}\ =\ \frac{-1\pm\sqrt{-3}}{2}\ =\ -\frac12\pm \frac{\sqrt3}{2}i\)

Let's pick  \(z\ =\ -\frac12 + \frac{\sqrt3}{2}i\)     (If we picked  \(z\ =\ -\frac12 - \frac{\sqrt3}{2}i\)   we would get the same answer)

Now let's re-express  z  to be in the form  \(r(\cos\theta+i\sin\theta)\)  so that we can use DeMoivre's Theorem.

By the Pythagorean Theorem,

\(r^2\ =\ (-\frac12)^2+(\frac{\sqrt3}{2})^2\ =\ 1\)   so taking the positive sqrt, we get     \(r \ =\ 1\)

An angle which has a cos of  \(-\frac12\)  and a sin of  \(\frac{\sqrt{3}}{2}\)  is  \(\frac{2\pi}{3}\),  so let  \(\theta\ =\ \frac{2\pi}{3}\)

And so...

\(z\ =\ 1(\cos(\frac{2\pi}{3})+i\sin(\frac{2\pi}{3}))\)          (we can check in a calculator that this does equal \( -\frac12 + \frac{\sqrt3}{2}i\) )

Then by DeMoivre's Theorem,

\(z^{51}\ =\ (1)^{51}(\cos(51\cdot\frac{2\pi}{3})+i\sin(51\cdot\frac{2\pi}{3}))\\~\\ z^{51}\ =\ (1)^{51}((1)+i(0))\\~\\ z^{51}\ =\ (1)^{51}\\~\\ z^{51}\ =\ 1\)

And so...

\(-z^{51}\ =\ -1\)

If \(z^2+z+1=0\),  find  \(z^{49} + z^{50} + z^{51} + z^{52} + z^{53}. \)

Hello Guest!

\(z^2+z+1=0\\ z=-\frac{1}{2}\pm\sqrt{\frac{1}{4}-1}\\ z=-\frac{1}{2}\pm\sqrt{-\frac{3}{4}}\\ z=-\frac{1}{2}\pm \frac{1}{2}\sqrt{3}\cdot i\\ \color{blue}z=-\frac{1}{2}\pm sin(\frac{\pi}{3})\cdot i\)

  !

Don't know.... maybe it depends on what universe reality you are in   

OMG! You’re right!

I can’t completely untangle myself from this Quantum Entangled Spooky Dumbness at a Distance BS.  I need a doctor!

The only doctors who could fix Quantum Entanglement problems are the time-traveling doctors: Dr. Tud and Dr. Who.

Dr. Tud is lost in time, and Dr. Who is on First and seems stuck there.

I may be doomed ... I wonder how long it takes for this to wear off....

GA

That's nice, but I like my method too.

Let's denote a side of a square with a letter  a

Tangent of angle BAC = 1.33333333

Tangent of angle BCA = 0.75

So, we have     (a / 1.3333333) + a + (a / 0.75) = 5  ==>   a = 1.621621622     or     60 / 37

Using the triangle inequality that says the  sum of any two side lengths of a triangle is greater than the remaining side....

1)          11 + 19  .....the missing side  must not exceed 29

2 )        11  + missing  side  > 19    .....the missing side must be 9 (or greater)

Putting these together, we have that

9 ≤ missing side ≤ 29

Number of possible lengths  =   29 - 9 + 1  =  21

Thx, GA   !!!!

I needed to take a  break......my quest for the Roman zero was more arduous than I thought....!!!

Call the side of the square  = s

Triangle ABC is similar to triangle AXW

So 

AB/BC  = AX /XW

3/ 4   =  AX / s

AX = (3/4)s

Triangle  ABC  is also similar to triangle ZYC

So

AB/ BC =  ZY/YC

3/4 = s/ YC

YC = (4/3)s

And

AX  + XY  + YC   =  5

(3/4)s +   s + (4/3)s   =  5

(9/12)s  + (12/12) + (16/12)s = 5

(37/12)s  =  5

s = 5(12/37)  =  60/37

YES  IT IS  !    

The approximate answer is   35.354%    

This is an interesting answer: 

Using similar triangles twice, and  calling the length of the side of the square x,

in the triangle CYZ,  CY/x = 4/3, so CY = 4x/3,

in the triangle WXA,  x/XA = 3/4, so XA = 3x/4.

The hypotenuse of the triangle is

CY + YX + XA  = 4x/3 + x + 3x/4 = 37x/12 = 5,

so x = 60/37.

(Solved by Guest -  Nov 13, 2015)

It’s great to have you back, CPhill.

Sisyphus’ bolder was lonely and listless. 

No one here has the strength to budge it, let alone push it up the hill.  

3. 

Multiply both sides by abc to get rid of all division parts
−a^2b−a^2c−ab^2+abx−ac^2+acx−b^2c−bc^2+bcx=3abc

"Simplify"
x(ab+ac+bc) = a^2b+a^2c+ab^2+3abc+ac^2+b^2c+bc^2

Factor out x
x(ab+ac+bc) = a^2b+a^2c+ab^2+3abc+ac^2+b^2c+bc^2

Divide both sides by ab+ac+bc

x = (a^2b+a^2c+ab^2+3abc+ac^2+b^2c+bc2)/ab+ac+bc

Hope it helped.

By the way all exponents are alone, meaning they are NOT being multiplied by a variable or being added with other numbers. Sorry if it is confusing.

LOL!!..GA...Freudian typo  !!!  (since fixed...it's a family show....)

Thank you so much for the answers!

2. A cowboy agreed to work for one year for 19,200 dollars plus a horse. He quit after 7 months and was paid 10,450 dollars plus a horse. How many dollars was the horse worth, if the paid salary reflects the fair proportion of his yearly salary?

(7/12)  * 19200  =   11200  =  pay for 7 months

11200  - 10450  =   $750  =  horse's worth