CPhill

Thanks, tertre....!!!

Good to see you ......!!!

Not a tough as it looks

Let's suppose that  the slope of  some line  =  a/b

This theorem says that   another line will be perpendicular  if it  has  a negative reciprocal slope

The negative reciprocal of a/b   = -b/a

Farther.....if we multiply these two slope together and the result  = -1.....the lines are perpendicular

So

(a/b) (-b/a)  =

- (a/b) (b/a)  =

-  (ab) / (ba)  =

- (ab) / (ab)  =

- 1        ,,,,so....we see that they are truly perpendicular lines  !!!!!  

Divide each term by 3

1, -7/3, 5/3, -1

Remove the first term    and change any minuses to pluses

7/3, 5/3/ 1

1) take the largest value and add 1 to it  →  7/3 + 1  = 10/3

2) Sum the values =  5   .....compare this with 1 and take the larger value  = 5

Take the smaller  of  (1), (2)

The upper bound is 10/3

See this graph  : https://www.desmos.com/calculator/1mf8fqbard

However....I  think that you are using the syntethic division test to find the greatest integer that is an upper bound.....this should be 3

Test 2

2 [  3  -7   0  5  -3]

           6

  _______________

     3  -1                              we get a negative so we can stop

Test  3

3 [  3 - 7   0  5   -3]

           9   6 18   69

   _______________

     3   2    6  23  66       we ge all positives....so 3 is the largest integer upper bound 

I think the following image is what we need

This figure is an isosceles trapezoid....PQ is parallel to RS  and PS = QR

And the diagonals of an isosceles trapezoid are equal

And SR = SR

And PS = QR

Therefore...by SSS....triangle SPR is congruent to triangle RQS →   "B"

Floor (t)  =  2t + 3

First off.... "t"  cannot be positive

A graph  might give us a clue  .....here : https://www.desmos.com/calculator/ugdkn4zwei

Note that the graph  appears to pass through the t values of  -3, - 3.5  and -4

Check

Floor (-3)  = 2(-3) + 3      ????

-3   = -3       true

Floor (-3.5)  =  2(-3.5) + 3  ???

-4  = -7 + 3   ???

-4 = -4   true

Floor (-4)  = 2(-4) + 3   ???

-4  = -8 + 3   ???

-4  =  -5    not true

So.....the values of  "t"  that make this true are    

-3.5 , -3

We can solve this by the method of partial fractions

Note that the first denominator can be factored as  ( x -2) (x+ 1)

Multiply  everything through by   (x-2) (x + 1)   and we get that

x + 7  =  A( x + 1)   + B( x - 2)      simplify

x + 7  =  (A + B)x  + (A - 2B)        equating terms, we have that

A + B  =1      (1)  

A - 2B  =  7    →   -A + 2B  = -7     (2)

Add (1) and (2)  and we have that

3B = -6

B = -2

So   A -2  = 1

A = 3

(A,B)  =  ( 3, -2)

2. For integers a, b, c, and d, (x^2+ax+b)(x^2+cx+d)=x^4+x^3-2x^2+17x-5. What is the value of a+b+c+d?

(x^2 + ax + b) ( x^2 + cx +d)   =

x^4 + cx^3  + dx^2

+       ax^3   + (ac)x^2  + (ad)x

+                     bx^2      + (bc)x  + bd

___________________________________________

x^4  + (a + c)x^3  + (ac + d + b)x^2  + ( ad + bc)x  + bd    =  1x^4  + 1x^3 - 2x^2  + 17x -  5

Equating terms   it can be seen that   bd  =  -5

Suppose that   b = -1   and d  = 5

Then

ac + d  + b =   -2      →  ac + 4  =  - 2   →   ac  = -6      (1)

a + c =  1   →  c = 1- a     (2)

Sub (2)  into (1)

a ( 1 - a)  = -6

-a^2 + a  + 6  = 0

a^2 - a - 6  = 0

(a -3) (a + 2)  = 0

So  a  = 3   or a = -2

If we let a  = 3, then c = -2

Then  

ad + bc  = 17   ??

3(5) + (-1)(-2)  = 17

15 + 2  = 17    is true

Therfore   a  = 3, b = -1, c = -2  and  d  = 5      and their sum  =  5

BTW.....the same sum is achieved if we let b = 5 and d = -1

1.   x^2 - 11x + 4  = 0

Call the solutions a, b

First note, that from Vieta, the sum of the solutions will  =  11/1  = 11

So   a + b  = 11      (1)

Square both sides  of this to give that    a^2 + 2ab + b^2  =  121     

Rearrarange this  as   a^2 + b^2  =  121  - 2ab         (2)

And  the product of the roots  =   4/1   = 4

So

ab  =  4     (3)

And the sum of the cubes  =

a^3  + b^3     which we can factor as

(a + b) ( a^2 - ab + b^2)

(a + b) ( [ a^2 + b^2]  - ab )     (4)      

Sub (1) , (2) and (3)   into  4  and we have that

a^3 + b^3   =

(11) ( [121 - 2ab] - ab)

(11)  ( 121 - 2(4)  - 4 ]

(11) ( 121 - 12)

(11) ( 109)  =

1199

3.      √ [ x + 1]

        ___________

          (x - 2) (x - 4)

First note that   if x = 2   or x = 4......the denominator  = 0....so.....these two values are not in the domain

Also...under the radical.....x + 1  must be  ≥  0

So

x + 1  ≥ 0

x  ≥ -1

So   the domain is

[ -1, 2)  U (2, 4)  U (4, infinity)

See the graph here : https://www.desmos.com/calculator/a4regqy27f

2.       √[-2x + 7]

          ________

                x

First of all....note that x CANNOT  =  0   because this would mean that we are dividing by 0 which is undefined

Also....under the radical.....    -2x + 7  must be   ≥ 0

So

-2x + 7  ≥  0

-2x  ≥ - 7       divide both sides by -2   and reverse the inequality sign

x  ≤ 7/2

So....the domain is

(-infinity, 0)  U  ( 0, 7/2)

See the graph here to confirm this :  https://www.desmos.com/calculator/qg90mexgvo