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Multiply top / bottom  by     1 + 3^(1/3)  + 3^(2/3)   

Note that the bottom becomes

(3^(1/3)  - 1)  ( 1 + 3^(1/3) + 3^(2/3) )  =

3^(1/3) + 3^(2/3)  + 3   - 1 - 3^(1/3) - 3^(2/3)  =     2

So  we have

1 + 3^(1/3)  + 3^(2/3)             ∛1  + ∛3 + ∛9

__________________  =     _____________

                 2                                    2

A + B + C + D   =   15

100 -  (17 + 43 + 28 + 12)   = 0%  have  no children

See alan's answer here :

https://web2.0calc.com/questions/pre-calc-problem

Note that only the inside 4 x 4 x 4 cube will not be painted. 

Can you take it from here?

g(x)  = f(x)  * f(x)   =

1  / (x- 3)^2

The x not in the domain of g(x)   =   3     (it makes the denominator =   0  )

answered

https://web2.0calc.com/questions/percent_93

Why don't you try and factor both quadratics? Can you cancel anything out?

Not much to do here

5 =  -4b  + 5      

5 - 5 =  4b

0  = 4b

b = 0

4u - 5v  =  23       (1)

2u + 4v =  -8 + 4u  →    u + 2v  = -4 + 2u  →    2v  + 4  = u     (2)  

Sub (2) into (1)

4 ( 2v + 4) - 5v =  23

8v + 16  - 5v = 23

3v  =  23 - 16

3v  = 7

v =  7/3              and      2 (7/3) + 4  =  u   =    26/3

u + v =     26/3 + 7/3   =    33/3  =   11

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

(1 * 2) * 3   =     -2 * 3  =    (-2 * 3) / ( -2 - 3)  =  -6 / -5  =   6 / 5

T(b + 1)  - T(b)  =  

(b + 1) (b + 2)  - (b) (b + 1)

______________________   =

                   2

(b + 1)  ( b + 2 - b)

_______________     =  

                 2

2 ( b + 1)

_______  =   

      2

b +  1

T(63)  = 2016

So

b + 1 =   T(63)

b + 1 =  2016

b  =  2015

Radius =  5 in

Area  =   pi *5^2 =  25pi in^2

Cost =   25 pi * 25 cents =      625 pi  cents ≈  1963 cents  ≈  $19.63

Note that if  m = 3   and n = 2    then

3^2 + 14(3) - 32  =   2(2)^3 + 3

                19        =      19

m + n =    5

I believe your answer is correct. But your solution is "wrong". First off, the first student gets 10 choices and not one. And the answer is not 10 * 2^8 because we'll have to divide 10. Remember two seatings are the same if each studnet in both seats hsa the same sttudent to the left and to the right. Since there are 10 seats, we divide by 10 to get 2^8 as our answer. 

Lets see,

The first stuent has only one place to chose from as they denot the 'begining of the circe'

The second student has 2 places to chose from

The third has 2 to shooe from 

The ninth has 2 to chose from and the last has only one.

1*2*2*2*2*2*2*2*2*1

2^8 = 256 

I am not certain that it is right though.

Please add a link to the previous question/answer

Expand the product 

Hello Guest!

\(4r(2r-4) \color{blue}=8r^2-16r\\ =8r(r-2)=r(8r-16)=8(r-1)^2-8\)

  !

Factor \(9x^7+6x^5\)

Hello Guest!

\(9x^7+6x^5\\ \color{blue}=3x^5(3x^2+2)=x^5(9x^2+6)=(\sqrt{6}-3ix)(\sqrt{6}+3ix)x^5\)

  !

The smallest m is 142.

Chris, in step 3 could you have divided both sides by (N – 6)? 

If the cube of a positive number is three times the square of the number times two, what is the number?    

What I noticed was that         3 • n² • 2  is the same as  6 • n² 

And since                                        n³   is the same as  n • n²  

Then you could say that             6 • n²  =  n • n²  

From which follows that                      n  =  6  

_.

Another solution! :D

P(x) = 2x^2 - 10x + 5 = 2(x - a)(x - b)

(2a - 3)(4b - 6) = 8(a - 3/2)(b - 3/2)

P(3/2) = 2(3/2 - a)(3/2 - b) = -11/2

-11/2 * 4 = -22. 

x * y * z = 630 x 14

3x^2 * z = 630 * 14

x^2 * z = 210 * 14

x^2 * z = 2^2 * 3 * 5 * 7^2

In order to minimize x + y + z, we get:

x = 7, z = 60... Therefore the minimum value of x + y + z = 88.

Hmm interesting. In order to find the last 3 digits of 101^{107}, then we have to take it to mod 1000.

101^{107} (mod 1000)

= 201^{53} * 101 (mod 1000)

= 401^{26} * 201 * 101 (mod 1000)

= 801^{13} * 201 * 101 (mod 1000)

= 601^6 * 801 * 201 * 101 (mod 1000)

= 210^3 * 801 * 201 * 101 (mod 1000)

=210^4 * 801 * 101 (mod 1000)

= 401^2 * 801 * 101 (mod 1000)

= 801 * 801 * 101 (mod 1000)

= 601 * 101 (mod 1000)

= 701 (mod 1000)

The last three digits are therefore 701.