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Assuming  f(x) = ax² + bx + c,

f(0)  =  a(0²) + b(0) + c  =  c     and so   c  =  0

f(1)  =  a(1²) + b(1) + c  =  a + b + c  =  a + b + 0  =  a + b     and so   a + b  =  2

f(2)  =  a(2²) + b(2) + c  =  4a + 2b + c  =  4a + 2b + 0  =  4a + 2b     and so     4a + 2b  =  9

Now since  a + b = 2  ,  we can subtract  b  from both sides to get   a = 2 - b

Now we can substitute  2 - b  in for  a  into the equation  4a + 2b = 9  to get:

4(2 - b) + 2b  =  9          distribute  4  to both terms in parenthesees

8 - 4b + 2b  =  9          combine like terms

8 - 2b  =  9          subtract  8  from both sides of the equation

-2b  =  1          divide both sides of the equation by  -2

b  =  -1/2

Now we can substitute  -1/2  in for  b  into the equation  a + b = 2  to get:

a + b  =  2

a - 1/2  =  2

a  =  2 + 1/2

a  =  5/2

And so  f(x)  =  \(\frac52\)x²  - \(\frac12\)x  +  0

Now to find  f(3) ,  plug in  3  for  x  into the function.

f(3)  =  \(\frac52\)(3)²  - \(\frac12\)(3)  +  0   =   \(\frac{45}2\)  -  \(\frac32\)   =   \(\frac{42}{2}\)   =   21

p/q = pi / 3

(3+2√2)^2n-1 + (3-2√2)^2n-1 - 2

\((3+2\sqrt2)^2*n-1 + (3-2\sqrt2)^2*n-1 - 2\\ =n(3+2\sqrt2)^2 + n(3-2\sqrt2)^2-4\\ \)

Is this what you intended to write or do you need brackets anywhere?

Have you tried it?

Here is the fromula

Using the letters A, B, and C and D, how many different four-letter code words can be formed?    

This would be the permutation of 4 things taken 4 at a time, abbreviated 4P4 in math. 

There are 4 choices for the 1st letter, 

There are 3 choices for the 2nd letter, 

There are 2 choices for the 3rd letter, 

There remains 1 choice for the 4th.           The total number is 24.  

_.

(3 + x)/(5 + x) = (1 - x)/(2 + x)     cross multiply to get

6 + 5x + x^2    =   5 -4x - x^2

2x^2 + 9x +1 = 0        Use Quadratic Formula to find x =  - 9/4 +- sqrt(73)/4

First position   4 choices   ( A M P S)

  then 5 digits     10000 choices   (00000 through 9999)

      4 * 10 000 = 40 000 licenses available

Tens digit, sorry, my mistake

There are: 4! ==24 such numbers

1234  1243  1324  1342  1423  1432  2134  2143  2314  2341  2413  2431  3124  3142  3214  3241  3412  3421  4123  4132  4213  4231  4312  4321  Total =  24

You connect the points (0, 0) and (9, 6) with a line segment. Starting at (0, 0), you move 1/2 of the way along the segment. What is the sum of the coordinates of the point you land upon?

Hello Guest!

The point is P(4.5, 3), the sum is 7.5.

  !

I = V/R

Initially    .300 = 50/R     R = 166.667 ohms      Cutting the wire in half also cuts R in half to 83,33 ohms

I = 300/83.33 = 3.6 amps     = 3600 mAmps

Oh finally you answered 

Subtract  4x  from both sides

x^3  >  5x^2          rearrange  as

x^3  - 5x^2  >  0        factor  as

x^2  ( x - 5) >  0

Set  as  an equality

x^2 ( x - 5)  =  0

The  possible  solution intervals  are   x  =     ( -inf , 0)    or  ( 0, 5)  or   ( 5 , inf)

If  we let x < 5 , the  first two intervals do not solve the original inequality

So......the  solution is   ( 5, inf)

"Bi - monthly"   can either  mean  "twice a month"   or  "once  every two  months"

If  the first  definition is meant,  we   have   5 * 12 * 2  =  120 compounding periods  and  the  amt is

24000  * ( 1.01)^120  = $79209

If the second definition,  we  have   5 * 6  =   30 compunding periods  and the amt is

24000 *  ( 1.01)^30   = $32348

x ( x + 6)  >   16  +  2x

x^2  +  6x  >  16  +  2x               rearrange    as

x^2 + 4x  -  16   >  0

Solve  this as an equality

x^2 +  4x  -  16  =  0               complete  the square on  x

x^2  +  4x  + 4   =  16  + 4

(x  + 2)^2  = 20            take  both roots

x + 2  =     ±sqrt 20

x  =  sqrt 20  - 2 ≈   2sqrt (5) - 2  ≈ 2.47          or          x  =  -sqrt 20  -  2  =  -2sqrt (5) - 2   ≈  -6.47

Note  that   the  intervals  that  possibly  solve  the original inequality  are  x  =

(-inf ,  ≈ -6.47)      or    ( ≈ -6.47 ,  ≈ 2.47)    or    ( ≈ 2,47 , inf)

If we let  x  =  0  ,the   middle interval  does not  solve  the original inequality

-

So.....the   solutions    are   ( -inf , -2sqrt (2) - 2 )   and  ( 2sqrt (5)   -2   ,  inf)

x>-2+2sqrt(5)

Muluples  of    89                     Multiples  of 23             Diff

1  = 89                                      4 =  92                         3

2 = 178                                     8 = 184                        6

3  = 267                                    12  = 276                     9

.                                                    .

.                                                    .

,                                                    .

9 = 801                                    36 = 828                     27

So

35 (23)  -  9(89)   =   805  -  801   =    4

thanks dude

3.65 /  5.87   x     10^8  /  10^12  =

0.6218057922  x     10^-4  = 

6.218057922  x  10^-5  =

6.218057922E-5

(2y  + 1)  (  4y^10  +  2y^9  + 4y^8 + 2y^7  +  2y^6  + 4y^5)  =

(2y) ( 4y^10  + 2y^9  + 4y^8 + 2y^7  + 2y^6  + 4y^5)  =

8y*11  +  4y^10  + 8y^9  + 4y^8 + 4y^7  + 8y^6          (A)

And

(1)  (  4y^10  +  2y^9  + 4y^8 + 2y^7  +  2y^6  + 4y^5)  =     4y^10  +  2y^9  + 4y^8 + 2y^7  +  2y^4  + 4y^5    (B)

So....combining A  and  B we get

8y^11  + 8y^10  + 10y^9  + 8y^8  + 6y^7  + 10y^6 + 4y^5 

C =    [14 + 3 ( 14 - 2)  ,   4  + 3 ( 4  - - 2)  ]  =   

[  14   +  3(12)   ,   4  +  3(6)   ]   =

( 50   ,  22)

The  number of total  subsets =   2^10  =  1024  (including the empty set)

The number   that  DON'T  contain  "5 "  =   2^9  =  512  (including the empty set)

So....the  number that DO  contain  "5"  =    1024  - 512  = 512

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