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It says...

"The diagram below shows the initial position of square DEFG,  an intermediate position, and the final position."

Where is the diagram?

Evaluate $1+\frac i2-\frac 14-\frac i8+\frac 1{16}+\frac i{32}-\cdots$ (where $i$ is the imaginary unit).  

Express your answer in the form a+bi, where a and b are real.

\(\begin{array}{|rcll|} \hline && 1 + \dfrac{i}{2} - \dfrac{1}{4} - \dfrac{i}{8} +\dfrac{1}{16} + \dfrac{i}{32} - \dfrac{1}{64} - \dfrac{i}{128} + \cdots \\ &=& \underbrace{ 1 - \dfrac{1}{4} +\dfrac{1}{16} - \dfrac{1}{64} + \cdots}_{=\text{geom. sequence: } \\ a_1 = 1,\ r=-\dfrac{1}{4}} + \underbrace{ \dfrac{i}{2} - \dfrac{i}{8} + \dfrac{i}{32} - \dfrac{i}{128} + \cdots}_{=\text{geom. sequence: } \\ a_1 = \dfrac{i}{2},\ r=-\dfrac{1}{4}} \\ &=& 1\cdot \dfrac{1}{1-(-\dfrac{1}{4})} + \dfrac{i}{2} \cdot \dfrac{1}{1-(-\dfrac{1}{4})} \\ &=& 1\cdot \dfrac{4}{5} + \dfrac{i}{2} \cdot \dfrac{4}{5} \\ &=& \mathbf{\dfrac{4}{5} + \dfrac{2}{5} i} \\ \hline \end{array}\)

1)   Where the values in the vertical columns of f(x) and g(x) are equal is the values of x that make them so

              Look where x=3     they are both  -9     so x = 3 is an answer

         and look at where x = 7   they are both 2  so x = 7 is an answer

2) the graph of g(x) is a parabola open UPWARD (like a bowl)  ....becaues the x^2 has a positive co-efficient (+2)

       so the second graph is the one to look at.....

        THEN, notice where the lines of the graphs cross....those are where the functions are equal valued.....then

           note the 'x' values of those points....there are two....  at x = 0   and x= 1

             (the actual POINTS are   0, -4     and      1, - 2   )

Cool ?  

Happy Anniversary!!!

answered.

B<4

B>-10

This is the union of the 2.  It includes the whole number line,

You just need to add the arrows at the ends.  And name the number line B

Suppose the line tangent to the graph of f at x = 2 is y = 4x + 1

This tells you that the point (2,9) is on the f graph

\(f(2)=9\)

The Equation of tangent to the f funtion    at x=2   is    y=4x+1

and suppose the line tangent to the graph of g at x = 2 has slope 3 and passes through (0, -2).

At   f'(2) = 3

And this tangent line, the one at f(2)  passes through (0,-2)

For  this line we are told (0,-2) is a point and the gradient is 3.    We need to find the find the y value for x=2.

\(m=\frac{rise}{run}\\ 3=\frac{y_1--2}{2-0}\\ 6=y_1+2\\ y_1=4\\ g(2)=4 \)

The Equation of tangent to the g funtion    at x=2   is    y=3x-2

a. y = f(x) + g(x) 

f'(2)+g'(2)=4+3=7

f(2)+g(2)= 9+4=13

So the equation of the tangent at x=2 is

y=7x+13

b. y = f(x) - 2g(x)

2g(x) the gradient and the y values will be doubled    so gradient of tangent=6 at (2,8)

f'(2)+2g'(2)=4+6=10

f(2)+g(2)= 9+8 = 17

y=10x-17

c. y = 4f(x)   

tangent of 4f(x)=4(4x+1) at x=2     is   y = 16x+4

I think that is right.

You can try convincing yourself (one way or the other) by drawing some graphs.

Thank you!

This is a convex mirror.   The focal point is MIDWAY betwen the center of curvature and the lens surface and is NEGATIVE

So for this mirror  f = -1.75  (1/2 of the RADIUS)    It will produce a virtual image on the same side as the focal point (opposite the side of the OBJECT ....the child's face) so the distance result will be negative thusly:

1/f  =  1/do + 1/di          do = distance to the object = 16cm    f = - 1.75     solve for di

-1/1.75 = 1/16 + 1/di

-1/1.75 - 1/16 = 1/di

di = - 1.58 cm

I do its fine man: What goes up and never comes down?

Wow! This inspires me to see the light of day, so I will keep chugging along, albeit slowly.

These examples should aid you with proper syntax.

Well, X².....by 2042, I'll probably be dead....so....you may catch me yet  !!!

Question is topic

Thanks for the salute! 

According to my profile information, I have 795 answers attached to my name currently. I have not yet reached my one-year anniversary. You are on your fourth, yet you have produced more than 30 times the amount of constructive responses than I have. Assuming that I continue answering questions at the same rate, I will reach the incredibly competitive benchmark of 23965 questions answered in 2042, which is 24 years from now!

Who deserves the salute now? 

f(x)  = 4x^(1/2)   - x

f ' (x)   =  (1/2*4x^(-1/2)  - 1  =    2x^(-1/2)  - 1

The tangent line will be horizontal if  the slope  = 0

So.....set the derivative equal to 0 and solve for x

2x^(-1/2)  -  1  = 0

2x^(-1/2)  = 1 

x^(-1/2)  = 1/2

We can write this as

x^(1/2)  = 2        square both sides 

And  x  =  4

When  x  =  4 ,   y  = 4√4  - 4   =  8 - 4  =  4

So.....the point where the tangent line is horizontal is  (4,4)

To find where the slope is  -1/2....we have

2x^(-1/2)  -  1  = -1/2

2x^(-1/2)  =  1/2

x^(-1/2)  =  1/4

And we can write this as

x^(1/2)  =  4     square both sides

x  = 16 

When  x  = 16 ,   y  = 4√16  - 16   =  0

So....the point where the tangent line  = -1/2  = (16, 0 )

If we have a multiplicity of 2 at  (-1,0)  and the point (7,0) is also on the graph, we must have that

P(x) =

a (x + 1)^2 (x - 7)      

a (x^2 + 2x + 1) (x - 7)  

a(x^3 + 2x^2 + x -7x^2 -14x - 7)  

a(x^3 - 5x^2 - 13x - 7)  

And if   (0, -14)  is on the graph.....we have that

-14  = a ( 0^3 - 5(0)^2 - 13(0)  - 7)

-14 = -7a

a  = 2

Here's a graph : https://www.desmos.com/calculator/wbejrdzi2x

2.  We  can find  P  with the tangent inverse  (other inverse functions are also possible )

arctan (12/5)  = P = 67.38° 

3.  C  = arctan (22/120) = 10.39°

4.  We know the adjacent side to L  and the hypotenuse.......the cosine inverse is the way to go, here

arccos ( 18/60)  = L  = 72.54°

5.   We know an angle an opposite side and we are looking for the hypotenuse

So.....the sine function is what we need

sin (62)  =  45 / hypotenuse     rearrange as

hypotenuse  = board length  =  45 / sin (62)   =  51 in