CPhill

B  =  (5, 5)

A  = (2, 1)

Subtract the x coordinates  =  (5 - 2) =  3 ...  square this  = 9

Subtract the y coordinates  = ( 5 - 1)  = 4 .....  square this   =  16

Add  the two results   9 + 16   =  25

Take the square root of this  = 5 units  =  the distance between  A and B

I like your way better, hectictar  !!!!

2. There is a polynomial which, when multiplied by $x^2 + 2x + 3$, gives $2x^5 + 3x^4 + 8x^3 + 8x^2 + 18x + 9$. What is that polynomial?

We can find this with division

                         2x^3  - x^2   + 4x  + 3

x^2 + 2x + 3  [  2x^5 + 3x^4 + 8x^3 +   8x^2 + 18x   + 9  ]

                        2x^5  + 4x^4 + 6x^3

                       ________________________

                                     -x^4  + 2x^3  + 8x^2

                                     -x^4  - 2x^3  -  3x^2

                                   ______________________

                                               4x^3  + 11x^2 +18x

                                               4x^3   + 8x^2 + 12x   + 9

                                              _____________________

                                                             3x^2 + 6x  +  9

                                                             3x^2  +6x  +  9

                                                            _____________

The polynomial we need is in red

1. \(f(x) = x^4-3x^2 + 2 and g(x) = 2x^4 - 6x^2 + 2x -1 \)

Let a  be a constant

What  is the highest pssible degree of

\(f(x) + a\cdot g(x) \)

Multiplying g(x)  by a non-zero constant  will not change the degree of g(x)

As long as "a"  is  not equal to   -1/2,    the degree  of  f(x) + a*g(x)  will  be  4

Call  the other intersection of BT and the circle, U

So  angles

RAB  + RAS   + SAU  =  180

RAB + 74 + SAU  =  180

RAB + SAU  = 106

SAU  =  106 - RAB

And we have that

m RTB  =  (1/2)  (minor arc RB - minor arc SU)

And since RAB and SAU  are central angles, their measures = the measure of their minor arcs...so..

28  = (1/2) ( RAB - SAU )

56  =  ( RAB - SAU)

56 = (RAB- (106 - RAB) )

56   =   2RAB - 106

162  = 2RAB

81°  =  RAB   =     minor arc  RB  =  minor arc BR

f(x)  = x^7 - 3x^3  + 2

g(x)  = f(x + 1)   =  (x + 1)^7 - 3(x+ 1)^3 + 2

The sum of the the coefficients  of g(x)  =   

Sum of the entries in Row 7 of Pascal^s Triangle  =  2^7 -

3 * Sum of the entries in Row 3  of Pascal's Triangle  = 3*2^3  +

2 =

2^7 - 3*2^3  +  2   =   106

(2z^2 + 5z - 6)(3z^3 - 2z + 1)

2z^2(3z^3 - 2z + 1)  + 5z(3z^3 - 2z + 1)  - 6(3z^3 - 2z + 1)

6z^5 - 4z^3 + 2z^2  + 15z^4 - 10z^2 + 5z  - 18z^3 + 12z - 6    combine like terms

6z^5 + 15z^4 - 22z^3 - 8z^2 + 17z - 6

Thanks, tertre   !!!!

More miles were biked initially than the tenth week.

(400 miles initially....320 miles in the 10th week )

The graph is decreasing between the first and seventh week.

f(x)  =  -2x^4  +  3x^2

If even, then    f(x)  =  f(-x)

f(-x)  =  -2(-x)^4  + 3(-x)^2   =   -2x^4  + 3x^2   =   f(x)    ⇒   even

f(x)   =  7

f(x)  = x^2 + 2x  - 8

Set these equal

x^2 + 2x  - 8  = 7   subtract 7 from both sides

x^2 + 2x  - 15  =  0        factor

(x + 5) ( x - 3)  =  0

Set each factor to 0

x +  5  =  0              x -  3   =  0

x  = -5                   x   =  3 

C  = (m, 4)

D   = (k, 0 )

Slope of line containing CD

(4 - 0)

_____   =  1

 m  - k

4  =  m  - k

m - k  = 4     (1)  

Height of trapezoid, H,  = 4

Area  =  (1/2)H * (sum of the bases)

34 = (1/2)* 4  (sum of the bases)

34   = 2(sum of the bases)

17  =  sum of the bases

But   the length of the bottom base  is    k + 1  and the length of the top base is m

So

(k + 1) +  m  =  17

m + k  = 16     (2)

And using (1)  and (2)   and we have that

m - k  =  4

m + k  =  16    add  these

2m  =  20

m  =  10

So....k  = 6

Note....the equation of the perpendicular line is y  = x - 6 

Here's a pic :