CPhill

An arithmetic sequence 2,5,8, 11, 14,... is written in order in a book, one hundred numbers per page, beginning on page one. On which page will the number 11,111 appear?

The common difference is 3

So....we want to find what term 11,111 might be

11,111 = 2 + 3(n - 1)    subtract 2 from both sides

11,109 = 3(n - 1)    divide both sides by 3

3703 = n - 1         add 1 to both sides

3704 = n

So  11,111 will appear on page =   ceiling [ 3704/100]  =  page  38 

What is the largest three-digit number in the arithmetic sequence 1,4,7,10,13, 16, ...?

The common difference is  3

So....we'd like to solve this

1 + 3(n - 1) < 1000

3(n - 1) < 999

n - 1 < 333

n < 334

So....n = 333

So...the largest three-digit term is

1 + 3 ( 333 - 1)   =

1 + 3(332) =

997

D) Find all the real zeros of f(x) = 3x^5 - x^4 - 6x^3 + 2x^2

Factor out x^2

x^2 (3x^3 - x^2 - 6x + 2)

One zero = 0

Factors of 2 =  1 , 2

Factors of 3 = 1 . 3

So...The other possible rational roots are   ± [ 1,  2, 1/3 , 2/3 ]

A little trial and error shows that the only other rational root = 1/3

The other two roots are non-rational reals

Consider the function f(x) = 2x^2 + 5x^3 - 21x^2  - 36x

A) Explaing why the rational zero theorem cannot be directly applied to this function.

The last term must be a constant to apply the theorem

B) Factor out the common monomial factor of f.

 x (2x + 5x^2 - 21x - 36) = 0

x ( 5x^2 - 19x - 36) = 0

The possible rational zeoroes  are all the possible divisors of 36   ovet all the possibe divisors of 5

However....the only rational zero here is x =0

The other two solutions involve non-rational  (but real) answers

f(2)  = 2^2 - 8(2) + 6  =   -6

f(-5) = (-5)^2 - 8(-5) + 6  =  71

1.196 x 10^0   in   sci not

OK....thanks for catching my error....!!!!

Oops...I shouldn't have included 4   !!!!

See my edited answer

Not much on Science...but

1 mole = 6.022  x 10^23 molecules 

So

[7.2 x 10^23]  /  [ 6.022 x 10^23 ]  =    7.2 / 6.022  ≈ 1.196  moles

1/3 < 3/x < 3/4  

We have that

1/3 < 3/x            and      3/x < 3/4

             cross - multiply

x < 9           and                12 < 3x           divide both sides by 3

                                           4  < x

So....the integer solutions are 5, 6, 7, 8    

And their sum is   26  !!!!