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If arc LM  = 90    then angle  LOM  =90  because it is a  central agle in the larger circle

Then LOM is a right triangle with legs OM , LM  =1

[ LOM ]   =  (1/2) (OM) (LM)   =   (1/2) (1)(1)   = 1/2

Area of large square   =  (1 + 1.5)^2  = (2.5)^2  = 6.25

2k - 3(-3)^2  + 7  =  -10  +  4 ( 2k -3(-5)^2  + 7)

2k - 27 + 7  = -10   + 4 ( 2k - 75 + 7)

2k -27 + 7 =  -10 + 8k - 300 + 28

2k -20  =  8k - 282

262  = 6k

262 / 6  =  k  =  131 / 3

radius =  (9 + 16 + 9) /  2  =   17

Let O  be the center of the circle

QO =  8  OD  = 17

Triangle QOD is right  with OD=  the hypotenuse

So

DQ = sqrt ( 17^2 - 8^2) =  sqrt ( 225) =  15

So ABCD is a trapezoid  with height  =  15    and bases of 34  and 16

[ ABCD ] =  (1/2) ( 15) ( 34 + 16)  =   375

4 feet  =  48  inches = width

Each can has a diameter=  2 in

So

48/2 =   24 cans can  be  packed in one row

And 23  can be packed in the next row

15 in / 5 =    3  =  number of  cans in a  stack

Each two rows   of cans occupy   [10 + 5sqrt (3)] in  of the length

So     integer ( 24  / [ 10 + 5sqrt 3 ]  )  =     1  two row unit

This means only  two rows of cans can  be fitted  lengthwise   

Total number of cans  =  3 ( 24 + 23)  =  147

Surface area =

pi  [ L (R + r)   ]  +   pi ( R^2 + r^2 )             where  L = sqrt [H^2 + (R - r)^2 ]

So

pi [ sqrt ( 4^2 + (12 - 9)^2 ) ( 12 + 9 )]  +  pi (12^2 + 9^2) ] =

pi [ sqrt ( 16 + 9)  *  (21)  ]  +  pi ( 225) =

pi [ 5 * 21 + 225 ] =

330 pi

Volume  =

pi  (H /  3 )   *   ( R^2  + Rr + r^2)

So

pi (4/3) * (12^2 + 12*9 + 9^2)   =

pi (4/3) * ( 144 + 108 + 81)  =

pi (4/3) ( 333) = 

444 pi

A + V =    (330 + 444) pi  =   (774) pi

h(x)  is a polynomial and is  continuous for al real  numbers x

f(x)  is a rational  function and is continuous everywhere except at x  =  7

This is AoPS homework.  Do not post AoPS homework.

y = 2x^2

y = x + c

Set the functions equal 

2x^2   = x + c

2x^2 - x -  c =  0

To find one  point  of intersection set the discriminant  = 0  and solve for  c

(1)^2  - 4(2)(-c)  = 0

1 + 8c =  0

1 = -8c

c = - 1/8

Let the number of "8" slips we need to  add = N

So

3 ( 8 / (10 + N) )   +   8 ( 2+ N) / (10 + N )  =  4       simplify

[ 24 + 16 + 8N ]  / (10 + N)  = 4

[ 40 + 8N ] =  4 (10 + N)

40 + 8N  =  40 + 4N

8N = 4N

(8- 4) N = 0

4N  = 0

N =  0         (we don't need to  add any !!! )

Proof   :   

Expected value =  Outcome * Probability  =   3 (8/10)  + 8(2/10)  =  [ 24 + 16 ] / 10  =  40 / 10    = 4

Expected value  = Outcome * probability

So   (3) (8/17)  +( 8 )(9/17)  =    [24 + 72 ] / 17  =  96 / 17 ≈   5.6

The ways to get to  (1,1)  from (0,0)  =  C(2,1)  = 2

Ways to get to *2,2) from (0,0) =    C(4,2)  = 6

So .....ways to  get from  (2,2) from (0,0)  without going through (1,1) =  6 - 2 =  4

(This is assuming you know modular arithmetic already)

Looking at the first equation, x would have to be the values 2,7,12,17,22,27,...,etc. Looking at the second equation, x would have to be the values 1,8,15,22,29,...,etc. When you compare both sequences, you can see that 22 is the only value common in both of them. Therefore, the only value of x that satisfies both equations is 22.

EF  = radius = EH =  9 -sqrt (27)  =  9 -3 sqrt (3)  = r

FD = HD

Angles EFD = EHD = 90

So  by SAS, triangle EFD  congruent to  triangle EHD

Angle HDF  = 30

ED bisects this angle

Angle EDF =15    angle FED =75

Let FD = x

By the Law of Sines

r / sin (15)  = x / sin (75)

 [ 9 -sqrt (27) ] / sin (15) = x / sin (75)

[ 9 -sqrt (27) ] * sin 75 / sin 15  =  x                     (sin 75  = cos 15)

[ 9 -sqrt (27) ] * (cos 15) / (sin 15) =  x

[ 9 - sqrt (27) ] * cot (15) =  x

[9 - sqrt (27) ] * cot (30/2)  = x

[ 9 -sqrt (27)] *  [ 1 + cos 30 ] / sin 30 =  x

[9 -sqrt (27) ] * [ 1 + sqrt (3) / 2 ] / [ 1/2 ]  = x

[9 - 3 sqrt (3)] * [ 2 + sqrt (3)] = x

[ 18 - 6sqrt (3) + 9sqrt (3)  - 9   =  x

[ 9 + 3sqrt (3) ]  = x =  FD

Side of square  =  CF + FD =     radius of circle + x   =    [ 9 - 3sqrt (3) ] +  [ 9 + 3sqrt (3) ]   =   18

\( A(t) = 3 - 5t^3*srqt(t) + 4 + 2^t. Find A(2) - A(1) \)

A(2)  = 3 - 5(2)^3 * sqrt (2) + 4 + 2^2  =   11 - 40sqrt (2)

A (1) =  3 - 5(1)^2*sqrt (1) + 4 + 2^1   =     4

A(2) - A(1)  =   7 - 40sqrt (2)  

We can write

2 ( 1 + 2 + 3 + 4 +......+ 670 + 671)                2

____________________________   =       ____

3 ( 1 + 2 + 3 + 4 +.......+670 + 671)                3

No circle shown  but I assume that  P,Q,R,S are in order  ????

We just need to   convert  radians to  degrees  using conversion  factor    180 /  pi

Note   (x, y)  =  (r * cos (angle measure)  , r * sin (angle measure) )   ,   r = 1 in this  case

2 *  180 / pi ≈  114.6°        point   ≈  ( cos 114.6° , sin 114.6°)  ≈  (-.42 , .91)       =     P

-.5 * 180 / pi  ≈ -28.6°  ≈  360 - 28.6  ≈  331.4°        point ≈ (cos 331.4°, sin 331.4°) ≈ (.88 , -.48)     =  S

10 * 180 / pi  ≈  573° ≈  (573 - 360) ≈   213°         point  ≈  (cos 213° , sin 213°) ≈ (-.84,-.54)  =   R

-3 * 180 / pi  ≈  -171.9° ≈ (360 - 171.9) ≈  188.1°        point  ≈ (cos 188.1° , sin 188.1°) ≈ (-.99, -.14)  =  Q

\( \frac{4x - 1}{5x + 2} + \frac{2x + 3}{x} \)

3 = (4x -1) / (5x + 2)  +  (2x + 3) / x

3=   [(4x -1)(x) ]  + (5x + 2)(2x + 3) ]  /  [ (5x + 2) (x) ]

3 [ (5x +2) x ]=   14x^2 + 18x + 6

15x^2 +6x  =  14x^2 + 18x + 6

x^2 - 12x - 6  =  0

x^2 - 12x  + 36 =  6 + 36

(x - 6)^2  = 42

x - 6 =    ± sqrt (42)

x = 6 ± sqrt (42)

Call the total muffins sold =  T

Cheese Muffins sold = (3/5)T

Apple sold = (3/10)T - 6

Banana sold =  T - (3/5)T  - (3/10T - 6)  =    (1/10) T + 6

Quantity sold each * price each  = Revenue

(3/5)T * (1.50) + (3/10T - 6) * (1.20) + (1/10 T + 6) *(1.40) = 561.20   simplify

.90T  + .36T - 7.20  +  .14T  + 8.40  = 561.20          multiply through by 100

90T + 36T - 720 + 14T + 840 = 56120

140T  + 120  = 56120

140T =  56000

T = 56000 / 140  = 400

Number of Banana sold =      (1/10)(400) + 6  =    46

A

 N              

C         M            B

We have the equations

AC^2 + CM^2  =  AM^2

NC^2 + BC^2 = BN^2                            CM = (BC/2)       CN  =  (AC/2)

AC^2  +(BC/2)^2  = 5     (1)

(AC/2)^2 + BC^2  = 5     (2)              subtract (2)  from (1)

(3/4)AC^2  - (3/4)BC^2   =  0

(3/4)AC^2 = (3/4)BC^2

AC^2 = BC^2

AC = BC

Therefore using (1)

AC^2 + (AC/2)^2 =  5

(5/4)AC^2  = 5

AC^2  = 5(4/5)

AC^2 = 4

AC = 2 =  BC

hypotenuse AB = sqrt ( AC^2 + BC^2)  = sqrt (2^2 + 2^2)  =  sqrt (8) = 2sqrt (2)

\(\frac{2x + 3}{x - 7} + \sqrt{-1/x} \)

Domain  of  first function  =   (-inf, 7) U ( 7 , inf)

Domain of second  function  =  (-inf , 0)

Most restrictive interval  is   ( -inf , 0) =  domain