CPhill

6x^2 - x < 2       subtract  2 from both sides

6x^2 - x  - 2  <  0           factor the left side

(2x  + 1) (3x -  2)  <  0        (1)

Set both  of the linear  factors  to   0  and solve for x

2x + 1  = 0              3x - 2   = 0

2x  = -1                   3x  = 2

x = -1/2                     x  = 2/3

We  have three possible intervals to test

(-inf, -1/2)   ( -1/2, 2/3)     (2/3, inf)

Either the first and third intervals solve (1)   or   the middle interval does

Pick  an   x value   on  ( -1/2, 2/3)   [ I'll choose x = 0 ]  and  note that

(2(0) + 1) ( 3(0) -2)  <  2     ????

(1) ( -2)  =   <  2     ???

-2  < 2      is true

So.....the interval   (-1/2, 2/3)   is the solution....

You are going in the right direction.....we can split your answer  up into these  functions

         cosΔx - 1                             sinΔx 

(1/2)  _______    +    (√ 3/2)    _________ 

            Δx                                       Δx

We have two identities to remember  :

As   x →  0,     cos Δx  -  1

                       __________     approaches 0......

                                Δx 

So   the limit of the first function  just simplifies to  0

Also    as x   → 0 ,     sin Δx

                                  ______   approaches  1......

                                     Δx

So the limit of the second function just simplifies  to (√ 3/2) 

So  

lim               sin ( pi/6  +  Δx)  - 1/2           

 x→ 0         __________________   =   (√ 3/2) 

                            Δx

See the graph here : https://www.desmos.com/calculator/mnpxjiqitj

Here is the graph :   https://www.desmos.com/calculator/eyxfsgsa6y

The only x intercept  is   x  = 0

y intercepts   →  y  = 0

To find any possible holes in the graph......factor  the numerator  and denominator

x ( x - 3 )                          x

____________  =        ______       a "hole"  occurs when  x - 3  = 0  ⇒   x  =   3 

(2x + 3) (x - 3)              2x - 3

And the y coordinate  of this point  is

   3                       3                   1

______  =         _____   =      ____      so....the "hole"   is  at    (3, 1/3)

2(3) + 3                9                    3

We have a horizontal asymptote at    y  = 1/2

And we have vertical asymptotes at  x  =   -3/2  

Thanks, tertre and heureka.....those "formulas"  provide a nice shorthand way to solve this type of problem  !!!!

We actually  have the coordinates ( 1,3) (2,1) (3.5) (4,4)  (5,2)

The inverse reverses these  so we would have ( 3,1) (1,2) ( 5,3) (4,4) (2,5)

So  f^-1(1)  =   2

And   f^-1 ( f^-1(1))  =  f^-1(2) = 5

Finally

f^-1 ( f^-1 (f^-1 (1) ))   =    f^-1(5)  = 3

I like doing this in the following manner....construct a square with a side  = 4  [ any side length would work ]

Let A  = (0,0)  B  = (0,4)  C = (4,4)  and D  = (4,0)  E  = (2,4)  F  = (4,2)  M = (0, 2)

Here's a pic : 

Triangles BCF  and DAM are congruent right triangles

And the slope of BF  =  [ 4-2] / [ 0 - 4  ]=  2/-4  = -1/2

And the slope of EA =   [ 4-0]  / [2-0]  = 4/2  = 2

So....these segements have reciprocal slopes  so they are perpendicular

This means that triangles   BCF  and BGE  are similar by AA congruency

BC  = 4     CF  = 2     BE  = 2    BF  =  √ [ BF^2 + CF^2]  = √ [4^2 + 2^2] = √20  = 2√5

So.....by similar  triangles......  BG / BE   =  BC/ BF   →  BG/ 2  = 4/ [ 2√5 ]   →  BG  = 4/√5 

And GE  =   √[BE^2  - BG^2]  = √[2^2 - (16/5]  = √  4 - (16/5]  = √ [ 20 - 16] / √5  =  2/√5   

And the  area of triangle  BCF  =  (1/2) BC * CF  = 4

And since BG / BC  =  (  4 / [ √5] ) / 4   =  1/ √5.....the area  of triangle BGE  = 4 (1/√5)^2  =  4/5

Likewise  triangles MAD  and  MNA   are similar  and we can show that triangles BGE  and ANM are congruent

So....BG/MN  = BG / GE   =  [4/√5 ] / [2 /√5]    =   2

And FG  =  BF - BG   =     2√5  - 4/√5   =  [10 √5 - 4√5] / 5  =  6√5/5  =  6/√5

So....FG/ MN  =  6/√5 / [ 2/√5] =  6/2  =  3

And DN  = DM - MN  =   BF - GE =  2√5  - 2/√5  =  [ 10√5 - 2√5] / 5 = 8√5/5  = 8/√5

So DN / MN  = [8/√5]/ [2/√5]  = 8 /2  =  4 

Area  of BMDF  =  Area of the square  less areas  of BCF  and DAM  =  4^2  - 2 [4] = 16 - 8  = 8

Area of  GBMN  =  area of BEA - areas of BGE  and ANM  =  (1/2)BE * BA  - 2(4/5)  = 

(1/2)(2)(4) - 8/5   =

4- 8/5   =

[20 - 8] / 5  =

12/5

So the area of  GFDN  = area of BMDF  - area of GBMN  =   8 - 12/5 =  [40 - 12]/5 = 28/5

So  GFDN / GBMN  =  [ 28/5] / [ 12/5] =  28/12   =  7 / 3

If a set  has n elements, then it possesses  2^n   subsets   [including the empty set ]

To see this....consider that every element is either  in or out of a subset....so....each element has two " states of being"

So....for  n elements, we have 2*2*2*2.....  = 2^n  subsets

So.....we have 10  members of set A

So....it possesses  2^10   =   1024  subsets

The hour hand  will  move at  a rate  of 

360°  in 12 hrs

30° in  1 hr

30°/ 60  =  .5° in one minute

So  at   5:44.....the hour  hand has  moved   5 (30°) +  44(.5°)  =  [150 + 22]° = 172°   from 12 noon

The minute hand  will move  at a rate of 360°/ 60  = 6° per minute

So at 5:44...it will have moved  [44 * 6]°  = 264°  from the top of the hour

So...the smaller angle formed by the hands  =  [264 - 172]°  =  92°

You should remove 5......if it remains and is chosen, there is no possibility of of a sum of 10

sin (x)  = 2/3      ⇒  cos(x)   =  √[3^2 - 2^2]  / 3  =  √[5]/3

sec (y)  =  5/4  ⇒  cos (y)   =  4/5  ⇒  sin(y)   =  3/5

sin (x + y)   =  sin (x) cos (y)  + cos (x)  sin( y )  =

(2/3)(4/5)  + (√5)/3 (3/5)  =

[ 8   + 3√5 ]                       

_________               

     15