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The answer to your question is 40/7.  I hope this helps!

So n = 9 is the smallest that works.

The solution was actually 13/4.

The solution was actually 13/4.

-1/2

x^2+x^2-3x-2x+5+1=2x^2-5x+6

we divide that by six so the constant term is the same:1

(2x^2-5x+6)/6=1/3x^2-5/6x+1

so now we know A=1/3, B=-5/6

A+B=1/3-5/6=-1/2

|z|=1

z=1 or -1

if z=1,

|1+1|+|1-1+1^2|=2+1=3

if z=-1,

|1-1|+|1+1+(-1)^2|=0+3=3

it can only be 3

proof:

|z|=1

z^2=1

|1±z|+|1±z+1|=1±z+1±z+1=3±z±z, one has the be -1 so 3

A line and a circle intersect at A and B as shown below. Find the distance between A and B.     
The circle is x^2 + y^2 = 1, and the line is y = x.     

x² + y² = 1 is a circle, centered on the origin, and radius 1.    

y = x is a straight line, that passes through the origin.    

Since the straight line passes through the center of the circle,    

it creates a diameter through the intersection points A and B.    

The radius of the circle is 1, so its diameter is twice that.    

So, the distance between A & B ( i.e., the diameter ) is 2.    

_.     

\( BC = 3, AC = 19\ and\ AB = 13.\\ BC<(AC-AB)\\ \color{blue }The\ requirements\ are\ pointless.\)

  !

What is the largest value of k such that the equation 6x-x^2=k has at least one real solution?     

                         6x – x²  =  k     

                         6x – x²  – k  =  0     

times –1             x² – 6x + k  =  0     

discriminant        (–6)² – (4)(1)(k)  must be > 0 to have real solution(s)     

                            36 – 4k  >  0     

                                      k  <  9     

_.     

How do you solve this equation     

Find all values of c such that 3*(2c + 1) = 28*(3*c) - 9.  If you find more than one value of c, then list your values in increasing order, separated by commas.     

Just remember PEMDAS.            3 • (2c + 1)  =  28 • (3 • c) – 9     

                                                            6c + 3   =  84c – 9     

                                                                   12  =  78c     

                                                                      c  =  12 / 78     

reduces to                                                     c  =  6 / 39     

_.     

If I give my brother 5 dollars, then we will have the same amount of money. If instead he gives me 24 dollars, then I'll have twice as much money as he will have. How much money does my brother currently have (in dollars)?     

Call B how much my brother has     

Call M how much I have     

(M – 5)  =  (B + 5)           ——>>  M  =  B + 10     (equation 1)     

(M + 24)  =  2 • (B – 24)                                        (equation 2)     

Substitute (B+10) in place     

of M in (equation 2)                     (B+10) + 24  =  2 • (B – 24)     

                                                             B + 34  =  2B – 48     

                                                                    82  =  B     

                                                                      B  =  82     

check answer     

from (eq 1)   M = 82 + 10          from (eq 2)   92+24  =?  2 • (82–24)     

                     M = 92                                           116  =?  116     true     

_.     

Find the number of positive integers that satisfy both the following conditions:     

Each digit is a 1 or a 2     

The sum of the digits is 3     

     1 1 1     

     1 2     

     2 1          I found 3 integers that satisfy the conditions.     

_.     

Let a and b be real numbers such that a - b = 1 and a^3 - b^3 = 1.     

                  a   –  b    =  1     

                  a³  –  b³  =  1     

            c.   When a = 0, then b = –1      

                  When b = 0, then a = +1      

            a.   ab is either [0 • –1] or [0 • +1]  ...  0 either way     

            b.   a+b is either [0+(–1)] or [0+(+1)]   ...  –1 or +1      

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