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\(|a+bi|^2 =( \overline{a+bi}) \cdot (a+bi)\), and since \(|w| = 1\), we know \(|w|^2 = |w| \cdot |w| = 1 \cdot 1 = 1 =|w|\), giving us 

\(\begin{align*} |w| &= 1 \\ |w| &= |w|^2 \\ |w| &= \overline{w}w \\ \overline{w}w &= 1 \\ \overline{w} &= \frac1w. \end{align*}\)

we can get \(\overline{z} = \frac1z\) through the same process. we make \(x = \frac{w+z}{1+wz}\), which means \(\overline{x} = \frac{\overline{w}+\overline{z}}{1+\overline{w}\cdot\overline{z}}. \)we plug in our values for the congujates of \(w\) and \(z\), which gives us

\(\begin{align*} \overline{x} &= \frac{\frac{1}{w} + \frac{1}{z}}{1+ \frac{1}{w} \cdot \frac{1}{z}} \\ &= \frac{\frac{z}{wz} + \frac{w}{wz}}{\frac{wz}{wz}+\frac{1}{wz}} \\ &=\frac{\frac{w+z}{wz}}{\frac{1+wz}{wz}} \\ &= \frac{w+z}{1+wz} \\ &= x \end{align*}\)

we get \(\overline{x} = x\), which is only true when \(x\) is real.

Find the area  of the shaded green square in the blue isosceles right triangle,
whose legs have length 6 cm.

\(\begin{array}{|rcll|} \hline 3x &=& 6\sqrt{2} \quad | \quad : 3 \\ x &=& 2\sqrt{2} \\ x^2 &=& 4\cdot 2 \\ \mathbf{x^2} &=& \mathbf{8} \\ \hline \end{array}\)

The area  of the shaded green square is \(\mathbf{8}\)

Hint:

Let the radius equal a set value. You can choose any number or numeral.    

Now work out the dimensions of the sides

etc

I'll help more but I want evidence that you are listening.

Show us what you can do by yourself.

Dragon may have seen an easier way ... 

Split green square in half; we have 2 identical rectangles. Rectangle's diagonal is the radius of the semi circle.

Ratio of rectangle's short side to long side is  1 : 2

The angle between the radius and the longer rectange side is     tan(A) = 1 / 2 = 26.56505118°

Let the radius  r = 1

Green square side    a = cos(A) = 0.894427191

Green  area          A_g = a² = 0.8 u²

Blue square diagonal is the radius and it is also  r = 1

Blue square side      b = sqrt(1/2) = 0.707106781

Blue  area              A_b = 2 * b² = 1 u²   

The ratio of the  green  area to the  blue  area is  4 : 5  

Have you read the "Sum of two squares theorem" in your  Math Textbook?? If no, read about here: 

https://en.wikipedia.org/wiki/Sum_of_two_squares_theorem

also your anwser for the first question is wrong :(

im confused how you solved the second one

If  PN = 4 than AN = 12        ( see diagram )

If  AN = 12  than AM = 36

If  AM = 36  than  the side  AB = AM / sin(60°) = 41.56921938

41.56921938 / sqrt(3) = 24

The side length of the equilateral triangle is     24*sqrt(3)   

Both years would be cool but i made a typo, its 10 years

I dont understand how to do it.

I only need the answer but working out would be amazing as well

Do you understand the problem? And if so, how much work have you done on it?

you didnt give it in an interval

The centers of the four smaller circles are the vertices of a square.

The distance from one vertex to its adjacent vertex is 2.

This means that the distance across the square is 2·sqrt(2).

This makes the diameter of the circle 2 + 2·sqrt(2) and its radius 1 + sqrt(2).

There's no way to get 7 from the problem as it is stated.  Check to make sure you've copied it correctly. 

_.

Assume that its for 10 years

50 000  =  10 000x - 200 x^2

0 = -200x^2 + 10 000x - 50 000       Quadratic formula shows x = $ 5.635    or  44.635

You can only determine which option is "best" if you specify the term of the investment. In other words, how long do you intend to invest your money for. 1 Year? 5 years? 10 years?.....etc., since you have a large up-front deposit of $60,000.

I think you will have this many terms - somebody should correct me if I'm wrong.

[26 letters of the alphabet]^2

The number of terms of this expansion is calculated as follows:(The exponent of the polynomial)=2) + (number of terms - 1) nCr (the exponent of the polynomial) =[2+26 - 1] nCr 2 =27 nCr 2 =351 terms.

1)     Quadrilateral  PQRS is an  equilateral quadrilateral  or rhombus.

Tringles  PQR and QSP are identical.

Angle  QSR = 30°    

2)  In the figure below,  AB=BC=CD=DE=EF=FG=GA

What's the question here??? 

A, A, B, B, C, C =6! / 2!.2!.2! = 90 permutations.

Each permutation will begin with one of letters in: [90 / 6] x 2 =30 cases for each of the 3 letters. And in 4/6 of the cases at least two of the same letters will be next to each other. And in 1/3 of the cases, they will not be. So, for each permutation beginning with A, you will have:

1/3 x 30 = 10 permutations when they will NOT be next to each other. 
So, you will have a total of: 10 x 3 = 30 permutations where the 3 letters will be separated from each other. Here are just 10 of them beginning with the letter A:

{A, B, A, C, B, C}, {A, B, C, A, B, C}, {A, B, C, A, C, B}, {A, B, C, B, A, C}, {A, B, C, B, C, A},{A, C, A, B, C, B}, {A, C, B, A, B, C}, {A, C, B, A, C, B}, {A, C, B, C, A, B}, {A, C, B, C, B, A} = 10 permutations.

That is it, unless I made mistake somewhere!.

Thanks, Lucky!