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We try the values y= 1, 2, 3, 4, 5, 6,...and so on.

For the function to have an upper bound, the coefficients of the polynomial when the original polynomial divided by (y) has to be positive. I recommend using Sythetic Division to divide the polynomial with the number.

Trying y=1, gives us a polynomial that alternates in signs. Same with y=2. Yet, y=3 gives us a polynomial with all positive coefficients. y=3: Answer

does this example help with what your suppposed to do with it? can you help me on it now? i am super confused

i believe it is -3

I think I saw this earlier.  Check out  https://web2.0calc.com/questions/help-please_7890 

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where is the picture? 

SO much information is missing!

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

Welcome back logic nice to see you again...

...but sadley i dont have an answer...

i made a dumb mistake :( 

thank you for clearing everything up!! 

Solve the following system:
{3 x + 5 y = -3 | (equation 1)
x + 2 y = -4/3 | (equation 2)

Subtract 1/3 × (equation 1) from equation 2:
{3 x + 5 y = -3 | (equation 1)
0 x+y/3 = (-1)/3 | (equation 2)

Multiply equation 2 by 3:
{3 x + 5 y = -3 | (equation 1)
0 x+y = -1 | (equation 2)

Subtract 5 × (equation 2) from equation 1:
{3 x+0 y = 2 | (equation 1)
0 x+y = -1 | (equation 2)

Divide equation 1 by 3:
{x+0 y = 2/3 | (equation 1)
0 x+y = -1 | (equation 2)

Collect results:
x = 2/3         and          y = -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

yay!! :)) 

thnaks foryour help on the second one, and the result is just the answer

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

\(b)~a_n = a_0 d^n,~n \in \{0\}\cup \mathbb{N},~a_0, d\in \mathbb{R}\\ \text{If we let $a_0 = \left(\dfrac 1 d\right)^{r-1},~\dfrac 1 d \in \mathbb{N}$}\\ \text{we end up with exactly $r$ integers in the sequence}\)

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

Credit to Cphill for answer

\(a)~a_n = a_0 + n d,~n \in \mathbb{Z^+},~d,a_0 \in \mathbb{R}\\ \text{Suppose there are 2 or more integers in the sequence, at say $k_1,~k_2$}\\ a_0 + k_1 d \in \mathbb{Z} \text{ and }a_0 + k_2 d \in \mathbb{Z}\\ (k_1-k_2)d \in \mathbb{Z}\\ m(k_1 - k_2)d \in \mathbb{Z},~m \in \mathbb{Z}\\ a_0 + k_1 d + m(k_1-k_2)d \in \mathbb{Z}\\ \text{This corresponds to index }\\ (m+1)k_1 - mk_2\\ \text{and thus there are infinitely many integers in the sequence as $m \in \mathbb{Z}$}\)

\(\text{On the other hand we can easily form a sequence with 1 integer by having $\\a_0 \in \mathbb{Z},~d \not \in \mathbb{Q}$}\)

This question is flawed....to see why....

If the squares overlapped to a width  of 10....it would mean that each square would have to contribute a  width of 10.....but.....the width of  each square is only 7, so this is impossible

Mathematically.....let  x  be the width of  each  of the unshaded rectangles

Then....the width of the overlap  must just be 7 - x   

And this equals 10

So

7 - x   = 10

-x  = 3

x  = -3        which is impossible

1. the range of x is going to be like a ray, with a end point while goes to infinity in another direction.

\(f(x)=\sqrt{{\color{red}3}x+{\color{red}6}}-7\)

You do \((-1)*\frac{{\color{red}6}}{{\color{red}3}}\Rightarrow-2\)

This means the minimum of x is -2, while the maximum goes to infinity.

I couldn't solve it, but I can get you started on this problem

After thinking of this for an eternity, I only came up with that the largest triangle you can make into a circle is an inscribed equilateral triangle with points XYZ.

So

If you have that inscribed equilateral triangle XYZ that is inscribed in a circle

to fit into the restrictions of the problem.

you will have to deform the equilateral triangle to do so.

You have to have vertice X as close to A (vertice of square) as possible this makes A your new point on of the triangle instead of X. Then you have to make sure vertices Y and Z are collinear with C (vertice of square)

Y and Z have to be as far as apart as possible while it has to be as close to edge CD of square as possible while remaining in circle w.

That was with my imagination, I am not sure if this is entirely correct.

Thank you, thats similar what my math teacher said!