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Thank you!

Thank you so much!

Thank you !

The best way to do this problem is to find the equation of the line in slope intercept form.

Find the slope first:

\(\frac{7-11}{1-3}=\frac{-4}{-2}=2\)

Now substitute 2 for the slope in y=mx+b and put one of the points in for x and y, I'll use (1,7)

7=2(1)+b

Solve for b, the y intercept

7=2+b subtract 2

5=b so the y-intercept is the point (0,5)

I found three different possible values of  d  using this graph:

Thanks :)

Good job  !!!!

THX, Pushy....I left out one....edit made!!!!!

1/m  + 1/n  = 1/4

[ m + n]  / mn  = 1/4         which implies that

mn / [ m + n ]  = 4

mn  =  4 [ m + n]

mn  =  4m + 4n

mn - 4n  = 4m

n [ m - 4 ]  =  4m

n  =       4m

            _____

             m - 4

When

m = 5   n  =  20

m = 6   n  =  12

m = 8   n  = 8

m = 12  n  = 6

m = 20  n = 5

So   (m, n)  = (5, 20)  (6,12) (8, 8)  (12, 6)  and (20, 5)

So, we can plug the variables into the equation. 

\(1- i = (1+3i) I.\)

We divide 1 + 3i on both sides to get:

\(\frac{1-i}{1+3i} = I\)

Then we multiply 1-3i on the numerator and the denominator of the left side to get:

\(-\frac12-i\).

So, \(I = -\frac12-i\)

So 

i^1 = i,

i ^ 2 = -1,

i^3 = -i,

i^4= 1,

i ^ 5 = i,

and it keeps repeating between the four (i,-1,-i,i).

So, to find i^11, we find 11mod4 (or the remainder when dividing 11 and 4) to get 3. So, i ^11 is -i.

Then you do the rest.

i^16 = 1

i^21 = i

i^26 = -1

i^31 = -i.

So, all of those added together is -i + 1 + i - 1 - i which is -i.

THX, Pushy.....I have a feeling that at least one of our members, heureka, can present this in a WAY better manner than I can....you might want to check back later....he may give his method.....

Note 

i^2  = -1     

And -1 raised to an even power = 1

And -1 raised to an odd power  =  - 1

i^11 =  I^10 * i     =   (i^2)^5 * i      =  (-1)^5 * i   = (-1) * i  =  - i

i^16  = (i^2)^8   = (-1)^8  =  1

i^21  = i^20 * i   =  (i^2)^10 * i  =  (-1)^10 * i   = 1 * i   =   i

i^26  = (i^2)^13  =  (-1)^13  = - 1

i^31  =  i^30 * i  =  (i^2)^15 * i   = (-1)^15 * i  =  -1 * i  = - i

So....adding the 5  results we get

-i + 1+ i - 1 - i  =

 - i

EDIT MADE....THX Pushy!!

Wow... that's a great solution! Thanks :)

\( log x - log y = a, log y - log z = 15, log z - log x = -7\)

log z - log x = -7

log y - log z  = 15        add these and we get

logy - log x  =  8       multiply through by -1

log x - log y  =  -8 =  "a" 

Some of our members may be able to do this in a better manner....but...here goes....!!!!

(a + bi)(a - bi)^3  +  (a - bi)(a + bi)^3  = 350

(a + bi)(a - bi)(a - bi)^2 + (a - bi)(a + bi) ( a + bi)^2  = 350

(a^2 + b^2) (a - bi)^2 + (a^2 + b^2)( a + bi)^2  = 350

(a^2 + b^2) [ (a - bi)^2 + (a+ bi)^2 ] = 350

(a^2 + b^2) [ a^2 - 2abi - b^2  + a^2 +2abi - b^2 ]  = 350

(a^2 + b^2 ) [ 2a^2 - 2b^2]  = 350

2 (a^2 + b^2)(a^2 - b^2) = 350

(a^2 + b^2)(a^2 - b^2)  = 175

(4^2 + 3^2) (4^2 - 3^2)  = 175

(4^2 + 3^2) (4 + 3) (4 - 3)  = 175

(25) (7) (1) = 175

So..by a little trial and error...let the first  complex number be  4+ 3i     and its conjugate be  4 - 3i

And let the second be -4 - 3i    and its conjugate be  -4 + 3i

And we have that

(4 + 3i)(4 - 3i)^3   + (-4 + 3i)(-4 - 3i)^3  =

(4 + 3i)(4-3i)(4 - 3i)^2  + (-4 + 3i) (-4 -3i)(-4 - 3i)^2  =

(16 + 9) ( 16 - 24i - 9) + ( 16 + 9) (16 + 24i -9)  =

(16 + 9)  [ 7 - 24i + 7 + 24i ]  =

(25) [ 14]  = 350

So.....the four  points are    4 + 3i  , 4 - 3i , -4 -3i  and  - 4 + 3i

We can make the  translation to Cartesian co-ordinates   (4, 3) (4 - 3) (-4, - 3) (-4, 3)

See the graph here : 

These points form an 8 x 6  rectangle....so....the area  = 48 units^2

I need this drug... but the answer is

B:  The new drug appears to reduce insomnia since the rerandomized mean differences are close to the mean difference found between the two groups studied.

Find the smallest possible value of a.

x^3 + ax^2 + ax + 1  = 0

First....notice that no matter what "a" might be,   -1  is a root because

(-1)^3 + a(-1)^2 + a(-1) + 1  =

-1  + a - a + 1   =   0

Using synthetic division

-1    [    1         a          a            1   ]

                      -1         1 - a        -1

         _________________________

            1      a - 1        1           0

So....the  remaining polynomial  is    x^2 + (a - 1)x  + 1

The discriminant of this  must be ≥ 0  for there to be remaining real roots....so.....

(a - 1)^2   - 4(1)(1)  ≥  0

a^2 - 2a + 1  - 4   ≥  0

a^2 - 2a  - 3 ≥  0

(a - 3) ( a + 1) ≥ 0

This will be true  on these intervals   (-inf, -1]   and  [3, inf)

So.....since a is positive......then its smallest value   =     3

a = 4?

It's a giant calculator which you can ask questions on. Like you did here. People will answer to them and... yeah :)