CPhill

1. 

We can start by cross-multiplying the two expressions:

(2x^2 + x + 3)(x + 1) = (2x + 1)(x^2 + x + 1)

2x^3 + x^2 + 3x + 2x^2 + x + 3  =  2x^3 + 2x^2 + 2x + x^2 + x + 1

2x^3 + 3x^2 + 4x + 3  =  2x^3 + 3x^2 + 3x + 1

4x + 3  = 3x + 1

x =  1 - 3

x = -2

Using the quadratic formula, the difference  in the  roots will simplify to  sqrt (k^2 -20)

So....we have

sqrt ( k^2 - 20) =  2sqrt (31)

k^2 - 20  = 4 ( 31 )

k^2  = 124  + 20

k^2  = 144

k = 12, -12 

If  k  = 12     we have

x^2 + 12x  + 5  = 0

And   sqrt (12^2 - 20) =  sqrt (124) =  sqrt ( 31 * 4) =  2sqrt (31)

So 12 is the largest possible k

Powers of 13 (starting with 13^1) end in

3, 9, 7, 1    a cycle of  4

So

13^19  ends in 7

Powers of 19  (starting with 19^1) end in

9, 1      a cycle of 2

So

19^13  ends in 9

So

13^19 * 19^13  ends in

7  * 9  =     ...3

10, 15, 20, 25, 30  = 100

y =ax^2 + bx  - 6

a^2 = 4

If this lies  completely below the   x axis, then a must  be negative because the  parabola will  turn  downward

So....a^2 =4  and  a  = -2

Since this lies  below the  x axis it  will  have no  real  roots....therefore, the discriminant  must  be < 0

b^2 - 4(-2)(-6)  < 0 

b^2  -48 < 0

b^2 < 48

b < 7  →     the largest value of b  =  6

Simplify the polynomial as

x^2  + 6x + 5

This can be factored as   (x+5) (x +1)

A,B  = 1

AB = (1)(1)  =  1

The height of  one  of these triangles = (1/2) diagonal of the square =  (1/2) 10 *sqrt (2) =  5sqrt 2

Half the base length =  5 sqrt (2) / sqrt (3) ) ...so the whole base length = 10sqrt (2) / sqrt (3)

Total area of both triangles =  2 (1/2) ( 5 sqrt (2)) (10 sqrt (2) / sqrt (3) = 

50 * 2 / sqrt (3)  = 

100 /sqrt (3) ≈ 57.735

We have a binomial probabiity  problem....3 of the  dice  must  show odd  and each one has a probability of (1/2) of that....the other 7 each have a probability of (1/2) of showing even.....we need to  choose any  3 of the  10 to show  odd...so we have

C(10,3) (1/2)^3 (1/2)^7  =  15 /128 ≈ .117

AB = 1

The interior angle of the  octagon = 135

The exterior angle PAB  =360 / 8  =  45

Angle ABP  = (360 -135) / 2  = 112.5

Angle APB = 180- 45 -112.5 = 22.5

Using the Law of Sines

PB / sin 45 = AB / sin 22.5

(sin 22.5) = sqrt [ ( 1 -sin 45 ) / 2  ]  =   sqrt [ (1 - sqrt (2)/2) /2 ]  =  sqrt [( 2 - sqrt (2)) / 4 ] = sqrt [ 2 -sqrt (2)] / 2  

PB / (1/sqrt (2)) = 1  / ( sqrt [2 -sqrt (2) ] / 2) =   

PB  =  2 ( 1/sqrt (2)) /  sqrt [2 -sqrt (2) ]  =  sqrt (2) / sqrt [ 2 -sqrt (2) ]  =

sqrt [(2 ) * sqrt( 2 + sqrt(2)] / sqrt [ 4 -2 ] =  sqrt (2) * sqrt (2 + sqrt (2) / sqrt (2)=

sqrt [ 2 + sqrt (2)]

The first function is defined  for  all x since  no real number makes the  denominator = 0

So....it's only the  second  function that we need to  consider

The quantity under the  root cannot be less than 0....so.......x cannot  be  less than 4

So  the domain is  [ 4, infinty )