CPhill

OK!!!!

16x^2  + 24x   + 9       we have a perfect square trinomial.....take the square root of the first and last terms  = 4x  and 3

Write them once separated with a plus and  repeat  and we have

(4x + 3) ( 4x + 3)        

And that is the factorization  !!!

a + b + c  = 2    (1)    square both sides  ⇒  (a^2 + b^2 + c^^2) + 2(ab + ac + bc)  =  4    (2)     

a^2 + b^2 + c^2  = 4       (3)

Sub (3)  into (2)  and we have that 

 (4) + 2(ab + ac + bc)  =  4

2(ab + ac + bc)  = 0

ab + ac + bc  = 0   

 (ab + ac + bc)^2 = 0    square both sides

(ab)^2 + (ac)^2 + (bc)^2  + 2[a^2bc + ab^2c + abc^2]  = 0  ⇒

(ab)^2 + (ac)^2 + (bc)^2  + 2abc ( a + b + c)   = 0  ⇒

(ab)^2  + (ac)^2 + (bc)^2   =  -2abc ( a + b + c)      (4) 

Find      ab       ac      bc

           ___ +   __  +  ___   =

              c        b         a

ab*ab  + ac*ac + bc*bc

___________________  =

            abc

(ab)^2 + (ac)^2 + (bc)^2

___________________        sub ( 4)  for the numerator  and we have that

           abc

-2abc ( a + b + c)

_______________   =

      abc   

-2 (a + b + c)        sub (1)  into this

-2 (2)  = 

-4 

This is a bimomial probability problem

We flip the coin  3 times and we need to choose 2 of them to be heads....

So we have

C(3,2) (.30)^2 (.70)  = 189/1000  = .189  =  18.9%

Not great at probability, but we could either draw 3 red marbles, 3 blue marbles or 3 white marbles

P(red)   = (3/12) (2/11) (1/10)  =  6 / 1320

P( blue)  = (5/12) (4/11) (3/10)  = 60/1320

P(white) = (4/12) (3/11)(2/10)  = 24 / 1320

So.....the total probability =  [ 6 + 60 + 24 ] /1320   =  90 / 1320  =  9 /132   =   3 / 44

Good job, CU       [ At least we got the same answer  !!! ]

f(x)  = 3x^2 - 2x + 4

g(x) = x^2 - kx - 6

f(10)  =  3(10)^2 - 2(10) + 4  =  284

g(10)  =  (10)*2 - 10k - 6  =  100 - 6 - 10k  =  94 - 10k

f(10) - g(10)    = 10    ....so....

284  - [94 - 10k]  =  10

190 + 10k  =  10

10k  = -180

k = -18

x + y + z  = 6       (1)     

1/x + 1/y + 1/z = 2     ⇒  [ yz +  xz + xy] / xyz  = 2  ⇒  [ xy + xz + yz]  = 2xyz      (2)

                x + y          y + z            x + z

Find        ____  +   ______   +  ______    =

                 z               x                   y

xy ( x + y)   + yz( y + z)  +  xz ( x + z)

________________________________  =

                     xyz

xy  ( 6 - z)   + yz ( 6 - x)   +  xz ( 6 -y)

______________________________      =

                   xyz

6xy - xyz  + 6yz - xyz  +  6xz - xyz

_______________________________  =

                    xyz

6 [ xy  + xz +  yz ] -  3 [ xxz ]

_______________________   =

                   xyz

6  [ 2xyz ]   - 3 [xyz]

________________   =

          xyz

xyz [ 6*2  - 3 ]

____________  =

     xyz

6*2  - 3  =

12 - 3  =

9

Using dimensional analysis, we have

34 in  x  [ 1 ft] / [12 in]  x [ 30.5 cm] / [1 ft ]      cancel units and we are left with

34 x  [30.5 /12] cm  =

34 * 30.5  /12   cm   ≈  86.42 cm

You're on the right track  !!!!

10b+5v+5z=55   ⇒  5v + 5z  =  55 - 10b    (1)
3b+5v+5z=34      ⇒  5v + 5z  = 34  - 3b    (2)
5b+6v+3z=42         (3)

Note that   (1)  must = (2)....therefore

55 - 10b  =  34 - 3b   rearrange as

55-34  = 7b

21 = 7b 

b = 3

Knowing this  (1)  becomes 

10(3) + 5v + 5z  = 55

30 + 5v + 5z = 55

5v + 5z  = 25    [divide through by 5 ]

v + z =  5     →    z = 5 - v     (4)

Sub (4)  into (3)  and sub 3 for b   and we have that

5(3) + 6v + 3(5 - v )  = 42

15 + 6v + 15 - 3v  = 42

30 + 3v  = 42

3v = 12

v = 4

So  z = 5 - v  =  5 - 4   = 1

So    { b, v , z}   =  ( 3, 4 , 1}

OK, man  !!!