All Answers 358504 Answers

The sum is

6r(1−r)4the fraction with numerator 6 r and denominator open paren 1 minus r close paren to the fourth power end-fraction

6𝑟(1−𝑟)4

\( \frac{1 + 3 + 5 + ... + 1999 + 2001 + 2003}{2 + 4 + 6 + ... + 2000 + 2002 + 2004 + 2006 + 2008 + 2010} \)

Number of odd  terms in numerator   =   [2003 +1] / 2  = 1002

Sum of  1st n odd terms =  n^2

Number of  even terms in  denominator =  [2010 - 2] / 2  + 1  = 1006

Sum of 1st n even terms  = (n)(n+1)

Evaluation

1002^2 / [ 1006 * 1007]  =   502002 /  506521

The sum of the series is 10424329

To simplify a fraction, you can find the HCF of the numerator and denominator. 

Divide both the numerator and denominator by the HCF. 

If you get a large fraction, you can find the HCF to ensure a fully simplified fraction. 

A common factor is any number that divides both numbers evenly. 

The sum of the series is  22704/117649

.

The answer is \(\frac{47}{95}\)

We start with 2. 2, 7, 12, 17, 22, 27, 32, 37, 42, 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, 97. So the answer is \(\boxed{97}\)

What is a scale factor -1 ?

As far as I know, a scaling factor is a real number > 0.

 !

The simplified expression for A + B is 

2x4+2x2+2

The smallest possible value of

a1+a2+a3a sub 1 plus a sub 2 plus a sub 3

𝑎1+𝑎2+𝑎3

is 30

We have these two  equations

a / ( 1 - r)  = 9     →  a = 9(1 - r)   → a^3  = 729( 1 -r)^3        (1)

a^3 / (1 - r^3)  = 36    →  a^3  = 36 * (1 -r^3)          (2)

Set (1) = (2)

729 ( 1 -r)^3  =  36 (1 - r^3)

729 [ ( 1 -r) (1- r)^2 ]  =  36 / [ (1-r) (1 + r + r^2)]

729 (1 - r)^2  = 36 ( 1 + r + r^2)

729 ( r^2 - 2r + 1)  =  36 ( 1 + r + r^2)

729r^2 - 1458r + 729  =  36 + 36r + 36r^2

693r^2  -1494r + 693  = 0

Solving for r produces

r  =  [83  - 8sqrt 15 ] / 77  ≈  .67553

and

r = [ 83 + 8sqrt 15 ] / 77  ≈  1.4803

Since, in an infinite series

-1 < r < 1....then   only the first solution is  good

We need to  work  with three equations 

7a1  =  a1 + a1r  + a1r^2  →    6a1  = a1r + a1r^2 →  6  = r + r^2      (1)

a1 + a1r  + a1r^2  + a1r^3  =  30  →  a1 ( 1 + r + r^2 + r^3)  =  30 →  1 +( r + r^2) + r^3  =  30/a1 → 

1 + 6 + r^3  = 30/a1  →  7 + r^3  = 30/a1      (2)

a1   ( 1 - r^4) / (1 - r)  =  30   →   (1-r^4) / (1 - r)  = 30/a1      (3)

Set (3)  =  (2)   and solve for  r

(1 - r^4) / (1 - r)  =  7 + r^3

1 -r^4  = (1 -r) (7 + r^3)

1 -r^4  =  7 + r^3 - 7r - r^4

1  = 7 + r^3 -7r

r^3 - 7r  + 6  =  0

Possible  solutions  for  r

r =1   (reject...it make equation 3  undefined)

r = 2 , -3 

If  r =  2

Then 

a1   ( 1 - 2^4) / (1 - 2)  = 30

a1 (-15)/(-1)  = 30

a1 * 15  =  30

a1  = 30 / 15  =  2 

If r =  -3

a1 [1 - (-3)^4 ]  / ( 1 + 3)   =30

a1 [ -80] / 4  = 30

a1 * -20  = 30

a1 = 30 / -20  =  -3/2

Simplify (x^4 + x^3 + x^2 + x + 1) + (x^4 - x^3 + x^2 - x + 1).       

stack them for addition    

  x⁴  +  x³  +  x²  +  x  +  1    

  x⁴  –  x³  +  x²  –  x  +  1    

 2x⁴           + 2x²         + 2   =   2(x⁴ + x² + 1)    

_.    

The roots are 3, -5, and 12.

\(x^3 - 4x^2 - 7x + 12=0\\ The\ equation\ is\ solved\ with\ \color{blue}WolframAlpha.\\ \color{blue}r=-2.0912\\ \color{blue}s=1.1648\\ \color{blue}t=4.9265\)

\(\dfrac{rs}{t^2} + \dfrac{rt}{s^2} + \dfrac{st}{r^2}\\ =\dfrac{-2.0912\cdot 1.1648}{4.9265^2} + \dfrac{-2.0912\cdot 4.9265}{1.1648^2} + \dfrac{1.1648\cdot 4.9265}{(-2.0912)^2}\\ \color{blue}=-6.38147970131\)

 !

yz / (y + z)  = 2                    xz / (x + z)  =  1

yz  = 2y + 2z                       xz =  x + z

yz - 2y  = 2z                        xz - x  = z

y ( z -2) = 2z                       x ( z -1)  = z

y  = 2z / ( z - 2)                   x  =  z/ (z -1)

xy / (x + y)  = 1

xy  = x + y

z / (z -1)  * 2z / (z -2)  =  z/(z-1) + 2z / (z -2)

2z^2 / [ ( z -1) (z -2)] =  [ z (z -2) + 2z (z -1) ] / [ (z -1) ( z -2) ]

2z^2  =  [ z ( z -2) + 2z ( z -1)

2z^2  = 3z^2  - 4z

z^2 - 4z  =  0

z ( z -4)  =  0

z = 0 (  reject  )     and        z =   4

r + s =  2m    square both sides

r^2  + 2rs + s^2  =  4m^2

r^2 + s^2  = 4m^2  - 2rs

rs  = m^2 - 6m + 11

2rs  = 2m^2 -12m + 22

So

r^2 + s^2  = 4m^2 - 2rs

r^2 + s^2  = 4m^2 - (2m^2 - 12m + 22)

r^2 + s^2  = 2m^2 + 12m - 22

(r - s)^2  =

r^2 + s^2  -  2rs   =

2m^2 + 12m - 22  - (2m^2 - 12m + 22)  =

24m - 44

This is a linear expression.....there is  no real number minimum

z +  1/z  =  1

z^2  + 1  =  z

z^2  - z  =    -1             complete the square on  z

z^2 - z + 1/4  =  -1  + 1/4

(z - 1/2)^2    =   -  3/4

z - 1/2 =   +/-  sqrt [ -3] /2

z  =   [1 - sqrt (3) i ] / 2                 or              z = [ 1 + sqrt (3) i ] / 2

z^2  =  [ 1 - 2sqrt(3)i - 3] / 4

z^2  =  [ -1/2  -sqrt(3)i / 2]

z^2  =  - [ 1 + sqrt (3) i ] / 2

z^3 = z * z^2

z^3 =  - [ 1 - sqrt (3) i] / 2 *  [1 +sqrt(3)i] / 2  =  - [1 + 3] / 4  =   -1

And

z^2  =   [ 1 + 2sqrt (3)i  -3 ] / 4  =  [ -1 + sqrt (3)i ] / 2

z^3  = z * z^2   =   [ 1 + sqrt (3) i ] / 2   *  [ -1 + sqrt (3) i ]  /  2   =

[-1  - 3 ] / 4   =    -1

Simplify as

-mn - 3m  + 15n  +   ____

- m ( n + 3)   +  15 (n  + 3)  =

( 15 - m) ( n + 3)

-mn - 3m  + 15n  +  45

[a + 2 ]  / [ a + 1]  = b / 3

3[a + 2] = b[a + 1]

3a + 6  = ab  + b

ab - 3a + b  =  6      multiply  the coefficients on a,b = (-3)(1)  = -3   and add to  both sides

ab - 3a + b - 3  =  6 - 3      factor the left side, simplify the right

(a + 1) (b - 3)  =  3

   a            b

  -4           2

  -2           0

   0           6

   2           4

(x+2) / (x)  •  (x+3) / (x+1)  •  (x+4) / (x+2)  =  10    

cancel the (x+2) ' s   

(1) / (x) • (x+3) / (x+1) • (x+4) / (1) = 10    

[ (1) • (x+3) • (x+4) ] / [ (x) • (x+1) • (1) ] = 10    

(x² + 7x + 12) / (x² + x) = 10    

x² + 7x + 12 = 10x² + 10x    

0 = 9x² + 3x – 12              this will factor (it would reduce first, but I'm skipping that step)   

(3x + 4) • (3x – 3) = 0    

                      3x + 4 = 0     this x will not produce integer numerators and denominators, so disregard    

                      3x – 3 = 0    

                            3x = 3    

                              x = 1     therefore (x+2) / (x) = 3 / 1 = 3 is the value of the first fraction   

_.