CPhill

Note that angle BAD is acute

Note that angle DCB  is supplemental to angle BAD.....so cos DCB  =  - cosBAD

And by the Law of Cosines we have that

DB^2  = CB^2 + CD^2 - 2(CB)(CD)(-cos)(BAD)

DB^2  = BA^2 +  AD^2 - 2(BA)(AD) cos(BAD)     substituting we have that

DB^2  = 2^2 + 3^2  +  2(3)(2) cos BAD

DB^2  =  1^2 + 4^2 - 2(1)(4) cos BAD        simplify

DB^2 =  13 - 12cosBAD   ⇒ [ DB^2 - 13] /  [-12  ]  = cosBAD    (1)

DB^2  = 17 - 8 cosBAD   (2)

Sub (1) into (2)   and we have that

DB^2 = 17 - 8 [ DB^2 - 13] / [-12]

DB^2 - 17  = -8[DB^2 - 13] / [-12]

DB^2 - 17  = (2/3)[DB^2 - 13 ]

DB^2 - 17  = (2/3)DB^2 - 26/3                subtract  (2/3)DB^2 from both sides

(1/3)DB^2 - 17 = - 26/3             multiply through by 3

DB^2 - 51 = - 26        add 51 to both sides

DB^2  = 25      take the positive root

DB  = 5

Maybe not the most simple way....but

Let A(--2, 4)    B = (2, 4)    C  = (2, 0)    D = (-2, 0)

The height of the equilateral triangle will  be    side/ 2 * sqrt (3)  = 4/2 * sqrt (3) = 2 sqrt (3)

So point E will be located at  (0, 4 - 2sqrt (3) )

And using points B and E we can write the equation for the line containg the segment BE

The slope is    [ 4 -(4- 2sqrt (3)) ] / [ 2 - 0]  =  2sqrt(3) / 2  =  sqrt (3)

So....the equation of this line  is

y =  sqrt (3) ( x - 2) + 4

y = sqrt (3)x - 2sqrt(3) + 4

y = sqrt (3)x + (4 - 2sqrt (3) )       (1)

And we can find the equation of the line that joins AC

The slope is   [ 4 - 0 ] / [ -2 - 2 ]  =   4/-4 = - 1

So....the equation of this line is

y = -(x - 0) + 2

y = -x + 2       (2)

Set (1)  = (2) to find the x coordinate of P

sqrt (3) x + (4 - 2sqrt (3) )  = - x + 2        rearrange

x ( sqrt (3) + 1)  =  2 - 4 + 2sqrt (3)

x ( sqrt (3) + 1)  =  2sqrt (3) - 2

x  =  [ 2sqrt ( 3) - 2 ]

       _____________        rationalize the denominator

            sqrt (3) + 1

        [ 2sqrt (3) - 2 ] [ sqrt (3) - 1]             6 - 2sqrt (3) - 2sqrt (3) + 2

x =  ________________________  =   _______________________  =

        [sqrt (3)+ 1] [ sqrt (3) - 1 ]                    3 -  1

8 - 4sqrt (3)                4 - 2sqrt (3)

_________       =

    2

So the y coordinate of P  =

y = -( 4- 2sqrt (3))+ 2  =   2sqrt (3) - 2

So  the  area  of triangle APE  =   Area of triangle ABE - Area of triangle APB

Area of triangle  AEB  = (1/2)base * (height)  =  (1/2) (4) (  2sqrt (3))  = 2 (2sqrt (3))  = 

4sqrt (3)  units^2

And the area of triangle  APB  = (1/2) base (height) = (1/2)(4) [4 - (2sqrt (3) - 2 ) ]  =

2 ( 6 - 2sqrt (3) )  =  12 - 4sqrt (3)   units^2

So...the area of triangle APE  =  [ 4sqrt (3)]  - [ 12 - 4sqrt (3) ]  = [8sqrt (3) - 12] units^2 =

4 [ 2sqrt (3) - 3 ]  units^2  

x^3 - 3x^2 + 8  = 0

We have the form

ax^3 - bx^2+ cx + d  = 0

Using Vieta's Theorem, we have that

The sum of the roots  =  -( -b)/a =  r1 + r2 + r3  = --(-3)/1  = 3

And the product of the roots r1 * r2 * r3  = -d/a =  -8/1  =  - 8

So....the sum of the new roots

2r1 + 2r2 + 2r3 =   2 (r1 + r2 + r3) = 2(3)  =  6 ⇒  -b/1 ⇒  b = -6

And the product of the new roots  = 2r1 * 2r2 * 2r3  = 8(r1*r2*r3)  = 8(-8)  = -64  ⇒ -d / 1 ⇒ d = 64  

So the new polynomial  is

x^2 - 6x^2 + 64

1.  A possible formula is

P = 400(2.5)^(N )     where P is the population after N weeks

[ Note  150%  actually means that the population increases by 2.5 times every week ]

2.  After 10 weeks we have

P  = 400(2.5)^ (10)  =  400 (2.5)^10  ≈ 3,814,697

3.  We need to solve this

1,000,000,000  = 400 (2.5)^(N)     divide both sides by 400

2,500,000 = (2,5)^(N )         take the log of both sides

log 2500000 = log 2.5^(N )      and by a log property we can write

log 2500000 = (N ) log2.5        divide both sides by log 2.5

16.077 = N ≈ 16 weeks       

2)  What is the average of all positive integers that have four digits when written in base 3, but two digits when written in base 6? Write your answer in base 10.

The integers in base 3 can range from

1000₃ to  2222₃   =   [27  to 80] base10

The integers in base 6 can range from

10₆  to  55₆  = [6 to 35] base 10

So  the common integers are      27, 28, 29 , 30 , 31, 32, 33 , 34 and 35

Their average  =   [ 4(62) + 31]  / 9  =  31

50(20)  = 1000 

When each data point is doubled we have 1000 * 2  =  2000

So

[ 1000 + 2000] / 50  =

3000 / 100  =  

30 =  new average

I think it's supposed to be    sec^2 x - 1

To see why.....the left hand side  can be factored as

tan^2x  (sin^2 x + cos^2 x)  =

tan^2 x ( 1)  =

tan^2 x

But 

tan^2x =  sec^2 x - 1      add 1 to both sides

tan^2 x + 1  =  sec^2 x      which is an identity

x^3 + ax^2 + 11x + 6 = 0  (1)

x^3 + bx^2 + 14x + 8  = 0   (2)

By the Rational Roots Theorem, the possible  zeroes  for the first polynomial are ±  [ 1, 2 , 3 , 6]

And for the second polynomial they are ±  [ 1,2,3,4] 

Subtract (1) from (2)  and we get that

(b - a) x^2 + 3x + 2  = 0    

Two possible shared roots  of  -1 and - 2  can be found if we let  b - a  = 1

So we have that

x^2 + 3x+ 2  = 0

(x + 1) ( x + 2)  = 0

x = - 1  and x = -2

Now   let x  = -1   as a shared root

Then  (-1)^3 + a(-1)^2 + 11(-1) + 6  = 0

-1 + a - 11 + 6  = 0

And a =6

Also

(-1)^3 + b(-1)^2 + 14(-1) +8 = 0

-1 + b -14 + 8  = 0

And b = 7

So testing (-2) as a root we have that

(-2)^3 + 6(-2)^2 + 11(-2) + 6  =

-8 + 24 - 22 + 6  =

0

And 

(-2)^3 + 7(-2)^2 + 14(-2)  + 8 =

-8 + 28 -28 + 8  =

0

So

(a,b ) =  (6, 7)

Here is the graph to show that this is true :https://www.desmos.com/calculator/y69o2yts2n