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The vertex  is at x = 3

And the x coordinate of the vertex  = -b / [2a] = 3 →  b = -6a

The product of the roots =   c /a =  8 →   c = 8a

And we know that

a(1)^2 - 6a(1) + 8a   = 6

3a = 6    →

a = 2     b  = -12    c = 16

Here'a a graph : https://www.desmos.com/calculator/flafi02koq

sin (pi/6) = 0.5

acos(0.5) = pi/3    

[√6 - i √2 ]  /  [ i √3 ]       mult. top/ bottom by  i √3

[  i √18 - i^2 √6 ] / [ i^2 *3 ]  =

[3i√2 + √6 ] / [ - 3]  =

- [ 3 i√2 + √6 ] / 3

CPhil THANK YOU the formula 2, because we are beginning on the second day, to the power of n-1 is now clear to me. Thank you for your input.

~Amy Jane 

\(\dfrac{\sqrt6 - i\sqrt2}{i\sqrt3}\\ =\dfrac{\sqrt6}{i\sqrt3} - \dfrac{i\sqrt2}{i\sqrt3}\\ =\dfrac{\sqrt2}{i}-\dfrac{\sqrt2}{\sqrt3}\\ =-i\sqrt{2} - \dfrac{\sqrt6}{3}\)

Multiply top and bottom by isqrt 3

(isqrt18- i^2 sqrt 6 )  /  i^2  3   = ( i 3 sqrt 2 + sqrt 6 )  /  -3  =  -isqrt2 - (sqrt 6)/3     ???

The zeros are  2 and 4 so the equation now looks like

(x-2)(x-4)  =  x^2 -6x +8   

f(1) = 6?        1^2 -6 +8 =  3            which needs to double to equal 6

so   2x^2 - 12x +16  = 0    fits the bill!     Graph to follow .... 

Simplify the following:
((sqrt(6) - i) sqrt(2))/(i sqrt(3))

sqrt(2) (-i + sqrt(6)) = -i sqrt(2) + 2 sqrt(3):
(-i sqrt(2) + 2 sqrt(3))/(i sqrt(3))

Multiply numerator and denominator of (-(i sqrt(2)) + 2 sqrt(3))/(i sqrt(3)) by -i:
((-(i sqrt(2)) + 2 sqrt(3)) (-i))/(i sqrt(3) (-i))

i×i = -1:
(-(-(i sqrt(2)) + 2 sqrt(3))×i)/(--1 sqrt(3))

((-(i sqrt(2)) + 2 sqrt(3)) (-i))/(-sqrt(3) (-1)) = (-1)/(-1)×((-(i sqrt(2)) + 2 sqrt(3))×i)/(sqrt(3) (-1)) = ((-(i sqrt(2)) + 2 sqrt(3))×i)/(sqrt(3) (-1)):
((-(i sqrt(2)) + 2 sqrt(3))×i)/(-sqrt(3))

i (-i sqrt(2) + 2 sqrt(3)) = sqrt(2) + 2 i sqrt(3):
(sqrt(2) + 2 i sqrt(3))/(-sqrt(3))

Multiply numerator and denominator of (sqrt(2) + 2 i sqrt(3))/(sqrt(3) (-1)) by -1:
(-(sqrt(2) + 2 i sqrt(3)))/(sqrt(3))

Rationalize the denominator. (-(sqrt(2) + 2 i sqrt(3)))/(sqrt(3)) = (-(sqrt(2) + 2 i sqrt(3)))/(sqrt(3))×(sqrt(3))/(sqrt(3)) = (-(sqrt(2) + 2 i sqrt(3)) sqrt(3))/(3):
(-(sqrt(2) + 2 i sqrt(3)) sqrt(3))/(3)

sqrt(3) (sqrt(2) + 2 i sqrt(3)) = 6 i + sqrt(6):
Answer: |(-6 i + sqrt(6))/(3)

1)

a - At least one Head =63/64 =0.984375

b- Exactly 3 Heads =6C3 / 64 =20 / 64 =0.3125

c- More than 4 Heads =[6+1] / 64 =0.109375

I entered it in W/A engine and it gave this result. Check and see if it is my mistake in entering it,

or check to see if you made a mistake:

sum_(n=2)^∞ (log((1 + 1/n)^n (n + 1)))/(log(n^n) log^(n + 1)(n + 1))≈1.27891

I think that naswer may confuse.....

DECIMAL 10 in hexadecimal (base 16)  is 'A'

Hexadecimal 10   in  decimal (base 10) is  16          NOW  I am done.... LOL

The hexadecimal  number '10'  in the base of  16 is   16  

1 2 3 4 5 6 7 8 9 A B C D E F    10 = 16        I am not sure which version you are asking about....

-.- whyyyyyyyyy

XD I was gonna say potato but then when you said that I remembered, "IT'S 42"

yes it us

"A"  is hexadecimal 10...

The questions are a) sketch the line with the given equation.

b) what is the slope of the line? What does it represent? 

C) what is the F-intercept of the line? What does it represent?

" 42 "

What is  '...the given frequency..'   ???

If it is ONE year compounding     5000 (1+.03)^10 =  $ 6719.58  after ten years

it 5/8 or 8/5

9 (2d)  = 18d

(-5,-9)

We can reflect twice to find the answer

First...across x ......   (-5,-9) → (-5, 9)......

And (-5, 9) across y  →  ( 5, 9 )  .....and (5, 9) is the answer.....

Note.......any successive term divided by its predecessor  =  

a_{n + 1} / a_n      =  1/2  =  r

Example.....   -82 / -164   =  1/2

I looked at this earlier and I agree wiih EP......

Each side of the base  = 8

Thus....each face can be divided into two congruent right triangles..........and each of these would have a "bottom" leg  = 4.....and in each case, the lateral edge of 4 would serve as a hypotenuse....however......a leg can never = a hypotenuse.....