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(e_1, e_2) = (4,6).

(a) Let P be the intersection of AD and MN, and let Q be the intersection of AL and MN.  Then by Menelaus's Theorem, Q is the midpoint of PK.  Since AD and LK are parallel, triangles ADL and LKQ are similar.  And since Q is the midpoint of PK, QK = PK/2 = DL/2.  Therefore, from the similar triangles, AD/LK = 2.

(b) Since QK = DL/2 and QG = LG/2, triangles GDL and GKQ are similar.  Then KG = DG/2, so by part (a), triangles ADG and LKG are similar.

(c) In part (b), we found that triangle GDL and GKQ were similar.  Then angle QKG = angle LDG, so D, G, and K are collinear.  And by the similarity in part (b), DG/GK = 2.  Easy!

Let z = a + ib where a and b are real, so that conjugate of z = a - ib.

Put these in your equation, and multiply out the terms, then separately equate real and imaginary parts on both sides of the equation to get two simultaneous equations that you can solve to find a and b.

There is more to this “rude dude” than bad manners. 

I get a somewhat different answer from Guest #1:

On division by (x - 1)(x - 2) we would have

\(\displaystyle \frac{f(x)}{(x-1)(x-2)}=q(x)+\frac{r(x)}{(x-1)(x-2)},\)

where q(x) is of degree two less than f(x) and r(x) is linear,  ax + b say.

Multiply throughout by (x - 1)(x - 2) to get

\(\displaystyle f(x)=q(x)(x-1)(x-2)+ax + b\)

and now substitute x = 1 and x = 2.

That  gets you

\(\displaystyle f(1)=2=a+b,\\ f(2)=3=2a +b.\)

Solve those simultaneously to get a = 1 and b = 1.

"Find all values of x that satisfy 6(x + 1/x)^2 - 35(x + 1/x) + 50 = 0."

First let y = x + 1/x.  Then we have  6y^2 -35y + 50 = 0  Solve this using the quadratic equation, to get two values of y.

Then set x + 1/x equal to each value of y in turn, multiply through by x each time to get another quadratic (this time in x) and solve for x.  

You will get four values of x in total.

"A bacteria culture starts with 600 bacteria. After 1 hour there are 1000 bacteria. Assuming the size of the culture grows exponentially, find the time required for the population size to double. (You may leave ln,log, and (or) e in your answer)."

You could learn a thing or 2 about manners guest!  

CPhil's answer was spot on. A typo at the end is absolutely nothing. 

You should be assamed of yourself talking to CPhill like that after he went out of his way to answer YOUR question!

Let the numbers  be  X, Y  and Z  in that order

And we have that

X + 4  = Z    ⇒  Z  - 4  = X        (1)

Y = 3Z + 5      (2)

X + Y  + Z  = 16      (3)

Sub (1)  and (2) into (3)   and we have that

[Z - 4] + [3Z + 5]  + Z  = 16   simplify

5Z + 1  =16       subtract 1 from each side

5Z  = 15       divide both sides by 5

Z  =3

So....the first number, X  = Z - 4   = 3  - 4   = -1

Let the twins' current ages  =  A

Let the current age  of the other brother =  A^3

And next year  we have that

A^3 + 1   =  ( A + 1)^2

A^3 + 1  = A^2 + 2A + 1

A^3 - A^2  - 2A  =  0

A(A^2 - A - 2)  =  0

One solution is that  A  =0    (reject)

And   using the second factor we have that

A^2 - A  - 2  = 0

(A - 2) ( A + 1)  = 0

Setting both factors to 0 and solving for A  we have that

A = 2     or  A = -1   (reject)

So

A= 2        and the non-twin  = 2^3 =  8

So.....the sum of the ages  =

2 + 2  + 8  = 

12

a

Net change  =  r(12)  - r(6)  =

[ 6 - 24/12]  - [ 6 - 24/6]  =

[6 - 2 ]  - [ 6 - 4]  =

4  -  2   =

2

b

Avg change  =  [ r(12) - r(6) / [ 12 - 6]  =    2 / [ 12 - 6 ]   =   2 / 6   =   1 / 3

See the following image :

Let the radius of the large circle  = R

Due to  symmetry, QF = EF   = R - 7

And PF  = R - 10

And FB  =  R

And by the intersecting chord theorem in the smaller circle

(QF)(EF)  = (PF)(FB)

(R - 7) (R - 7)  =(R- 10)(R)

(R - 7) (R -7) = R^2 - 10R

R^2 - 14R + 49 = R^2 - 10r     rearrange  as

49  = 4R

R =  49/4  =  12.25 =  the  radius of the larger circle

C'mon, dude....I just made a typo......get a grip  !!!!

a.

Net change =  r(12)  - r(6)  =

[ 6 - 24*12  ]  - [ 6 - 24*6 ]  =

-24 ( 12)  +  24(6)   =    -144

b.

Avg rate of change =    Net Change  / [ 12  - 6]   =   -144/6   =  -24

THX, Melody....yep....just a typo on my part  !!!!!

The least  prime  is 11   and the greatest  is 43

And the possible primes are

11, 13, 17, 19, 23 , 29, 31, 37, 41 , 43

The possibilities are

          11     13    17    19    23    29   31    37   41    43

11               OK   OK  OK   OK  OK  OK   OK  OK  OK

13                       OK  OK   OK  OK  OK   OK 

17                              OK   OK  OK  

19                                      OK 

So   19  possible products  (if I counted correctly)