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Can you elaborate please...

do you know the answer?

Notice, they the appear the same number of times in each group of 100 from 0 - 399

4 14 24 34 40 41 42 43 44 45 46 47 48 49 54 64 74 84 94

8 18 28 38 48 58 68 78 80 81 82 83 84 85 86 87 88 89 98

They each appear 20 times in each group of 100

But starting at page 400 to page 488, the 4's appear

89 + 19 times   = 108 times

But the 8's only  appear 18 times  from pages 400 - 488

So the 4's appear   108  - 18    = 100 more times

What's an ellipse?

sec x  cot x  =  -2

(1/cosx) (cos x/ sin x)  =  -2

(1/sin x )  =  - 2            which also implies  that

sin x = -1/2

And this is  true when x  =  7pi/ 6     and  11pi/6

(x , y)  =  (2, -3)

r  =  sqrt  [ 2^2 + (-3)^2]  = sqrt [ 4 +9]  =  sqrt (13)

csc  θ  =  r / y     =  sqrt (13)  / -3     =      - sqrt (13)  /  3

question 3 and 4 is done but it mentions them in the question so I included the.  The question is Provide an example of two complex numbers in the form c + di and , where c, d, e, and f are positive real numbers such that their product lies in the other possible quadrant. Support your example by determining its product.

Join the centers  of each  of  the small circles

This will create  a square  with a side  of  2  and  a diagonal  of   2√2

The area  of the square  = 2^2  =  4    (1)

Then  the  diameter  of  the  large circle  =  2 + 2√2

So  the radius  =   1 + √2

So  the area of  the  larger circle  =  pi (1 + √2)^2  =  pi  ( 3 + 2√2)    (2)

And we will  have  4 additional  white  areas  that  each  have  the area  of 3/4 of a circle with a radius of 1

So.....the total  of  the areas  =  4 (3/4) pi (1)^2  =  3 pi      (3)

So....the  grey  area  =   (2)  -  (1)  - (3)  =

pi ( 3 + 2√2)  - 4  - 3pi  =

2√2 pi  -  4   ≈   

4.89  units^2   (rounded to 3 sig digits)

Each successive  term  is larger  than the previous  term.....this series will diverge

never mind 

1 revolution is 1 circumference

C  =  2·pi·radius  =  2·pi·40 ft  =  80·pi ft  =  253.3274123 ft

                                                =  253.3274123 ft  x  12 in / ft  =  3015.928947 in

10 min  =  10 min  x  60 sec / min  =  600 sec

4 revolutions / 10 min  =  4 x 3015.928947 in / 600 sec  =  5.0 in/sec

I'm not sure what you are asking --

In problem 3, are you asking for an example whose product ends in QI and another example whose

product ends in QIV?

If so, (8 + 7i)(6 - 5i)  =  83 + 2i    (QI)

         (3 + 4i)(1 - 2i)  =  11 - 2i     (QIV)

In problem 4, are you asking how to solve the problem?

If so:  4sqrt(3) - 4i

          r  =  sqrt( ( 4sqrt(3) )² + ( -4 )² )  =  sqrt( 48 + 16 )  =  sqrt( 64 )  =  8

          theta  =  tan^-1( -4 / (4sqrt(3) )  =  tan^-1( - 1 / sqrt(3) )  =  (-1/6)·pi

          --->     4sqrt(3) - 4i  =  8·cis( (-1/6)·pi )

         sqrt(2) + sqrt(2)·i

         r  =  sqrt( ( sqrt(2) )² + ( sqrt(2) )² )  =  sqrt( 2 + 2 )  =  sqrt( 4 )  =  2

          theta  =  tan^-1( sqrt(2) / sqrt(2) )  =  tan^-1( 1 )  =  pi/4

          --->     sqrt(2) + sqrt(2)·i  =  2·cis( pi/4 )

Multiplying them together:  8·cis( (-1/6)·pi ) x 2·cis( pi/4 )  =  16·cis( 1/12·pi )

haha I'm also trying to do this problem last-minute right now

but i advise u to read the message board instead. there are a lot of really useful tips there

We can use something  called Pick's  Theorem  to answer  this

Area  =

Number of boundary points with integer coordinates  / 2   + Number of  Interior points with integer  coordinates    -1

Number of boubary points with integer coordinates  =   6

Number of interior points with integer coordinates  = 31

So....the area  =

6/2  +  31  -  1  =

3  +  30  =

33 units^2

I misread the text !!!  

Phill's answer's correct!!! 

Not  to  hard  to  find  this....T is  (2/3)  of  the  distance  from D to F

The coordinates  of T  can be found as

[ 1 + (2/3)(7 - 1)  ,  4 + (2/3)(1 - 4) ]  =

[ 1 + (2/3)(6) , 4 + (2/3)(-3) ]  =

[ 1 + 4.   4  - 2 ]  =

( 5 , 2)

The  octagon will  be formed  8 congruent  isosceles triangles  with  equal sides of  5  and  an included angle  of 45°

The area of the octagon  is

8 (1/2) (5^2) sin (45°)  =

100  ( sqrt (2)  / 2)   =

50 sqrt (2)   units^2  

See heureka's excellent answer, here :

https://web2.0calc.com/questions/geometry_136

Evaluate:  ( 5 + 5·sqrt(3)·i )⁷

1)  Write  5 + 5·sqrt(3)·i  in r·cis( theta )  form:

      If you have  a + bi  use  r  =  sqrt(a² + b²)  and  theta  =  tan^-1( b/a ):

      r  =  sqrt( ( 5 )² + ( 5·sqrt(3) )²  =  sqrt( 25 + 75  =  sqrt( 100 )  =  10

      theta  =  tan^-1( 5·sqrt(3) / 5 )  =  60^o   [If you are to use radians, use pi/3.]

          [You need to check which quadrant you should use; since both 5 and 5·sqrt(3) are

           positive, the angle is in the first quadrant; so 60^o is the correct angle.]

2)  You now have:  5 + 5·sqrt(3)·i   =   10·cis( 60^o )

     To raise to a power, raise the constant to that power and multiply the angle by the power.

     [ 10·cis( 60^o ) ]⁷  =  10⁷ · cis( 60^o · 7 )  =  10⁷ · cis( 420^o )  =  10⁷ · cis( 60^o )

3)  To write this in rectangular form, use:  x  =  r·cos( theta )     and     y  =  r·sin( theta )

      x  =  10⁷·cos( 60^o )  =  10⁷· (1/2)  =  5·10⁶

      y  =  10⁷·sin 60^o )  =  10⁷· (sqrt(3)/2)  =  5·sqrt(3)·10⁶

      --->     5·10⁶  +  5·sqrt(3)·10⁶·i

If MB  =  6.5    then  BC  =  2(6.5)   =  13

Using  the Law  of  Cosines  we  have  that

AB^2  = AM^2  + MB^2  -  2(AM*MB) cos (AMB)

5^2  =  (6.5)^2  + (6.5)^2  - 2(6.5)^2 * cos (AMB)

[ 25  - 6.5^2  - 6.5^2]  / [ - 2* 6.5^2  ]  =  cos (AMB)

-59.5 / -84.5    =  119/169

Then the sin of AMB   =  sqrt [ 169^2  - 119^2]   / 169   =  120/169

Using  the Law  of Sines

sin(AMB)  /AB = sin (ABM) / 6.5

(120/169) /5  =  sin (AMB)  /6.5

6.5 ( 120/169) / 5 =  sin AMB  =  12/13

Then the area  of  ABC  =

(1/2) AB * BC  sin (AMB)  =

(1/2) (5)(13) * (12/13)  =

(1/2) 60  =

30 units^2

We need to  first  find  out  how long  it  will take A and B working together to fill the tank....this  is easy

A  fills    1/6  of the tank in one hour

B fills  1/4  of  the tank in one hour

Add these fractions   =   (1/6) + (1/4)  =  10/24   =  5/12  =  portion of tank filled by both working together in one hour

Time  for  A  and B  ti fill the tank  =  12/5  hrs

So.....C  fills   1 / (12/5)  =  5/12 of the tank in one hr

So....if all  three  work together    (5/12)  + (5/12)  =  10/12  =  5/6  of the tank is filled in one hour

do you have the answer