CPhill

Using l'Hopitals rule

lim     sin  θ  =    0

θ → 0   

\(\cfrac{1}{1 + \cfrac{1}{4 + \cfrac{1}{4 + \cfrac{1}{2}}}} \)

Working from  the bottom-up

4 + 1/2 = 9/2

1/ (9/2) =  2/9

4+ 2/9  = 38/9

1/ (38/9)  =9/38

1 + 9/38 = 47/38

1 / (47/38)  = 38 / 47

x^2 + y^2  =  1

(x - 1)^2 +y^2  =  1        subtract the second from  the first

x^2 - ( x -1)^2  =  0

x^2 - x^2 + 2x  - 1   =0

2x - 1  = 0

x = 1/2

y^2  =  1  - (1/2)^2

y^2  = 3/4

y = sqrt (3) / 2          and  y =   -sqrt (3) / 2

Pts of intersection  are   ( 1/2 , sqrt (3) / 2)    and ( 1/2 , -sqrt (3) / 2)

Since the x values are the same, the slope is undefined  

Construct perpendicullar line  to the given one passing through (7.1)

The equation of this line  is

y = (x - 7) + 1

y = x  -6

The x intersection of  these lines is

-x + 5 = x - 6

2x =11

x = 11/2  =  5.5

And the y coordinate of the intersection is

5.5 - 6  =   -.5

The intersection point is ( 5.5 , -.5)

The reflected point  is  given by  [ 5.5 - (7 - 5.5) , -.5 - ( 1 -  -.5) ]  = [( 5.5 - 1.5) ,( -.5 - 1.5) ] =

( 4 , -2)

Construct a circle with a radius of  (2)  at the origin

The  line l will be tangent to this  circle

The equation of this  circle is

x^2 + y^2  = 4     (1)

The slope of a tangent line to this  circle at any  point will be -x /y

So

-x/ y =  1

-x = y  (2)

Sub (2) into (1)

x^2 + (-x)^2 = 4

2x^2  = 4

x^2  = 2

x =  sqrt2

y =-sqrt 2

The equation of the line is

y = 1 (x - sqrt 2)  - sqrt 2

y = x - 2sqrt 2

y=  x -sqrt 8

When x = 0, y= - sqrt 8

When y = 0 x = sqrt 8

So points (0.-sqrt 8)  and (sqrt 8,0)  are on the line

The base of  the triangle is  sqrt [ (sqrt 8)^2 + (-sqrt 8)^2 ] ]=  sqrt (16)  = 4

Its area is  (1/2) base * height  =  ( 1/2)(4) (2)  =  4

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Just solve for each variable from the bottom-up

a = 6

b = -22/5

And so on

Borrowing from proyaop's answer

14 + 2sqrt (41)  ≈ 26.8

This is  the larger root  and all  integer x's > than this will make the polynomial > 0

So.....the greatest integer that makes  this le to 0   is 26

a + 2b = 3

5a + (2+ m)b = k 

This system will have infinite solutions when the equations are the  same

Multiply the first equation through by 5

5a + 10b  = 15

5a + (2 + m)b  =  k

This will  have infinite solutions when   m = 8  and  k  = 15

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