CPhill

Whatever x's  make the denominator = 0  are the asymptotes.....these are also the roots of the polynomial

So

x^2 + 5x + 2  =  0

By Vieta, the sum of the  roots =  -5

So

a + b =   -5

First, find the slant height   = sqrt [ 20^2 - 3^2 ] =    sqrt [ 391]

So   the area of  one triangle   = (1/2)(6)sqrt [391]  =  3sqrt [391]

So  the surface  area  = 4 * 3sqrt [391]  + surface area of the base  =

12sqrt [391] + 6^2  =     12 sqrt [ 391 ] + 36  =   12 [ sqrt (391) + 3 ] units^2

To find the volume  find the height  

The diagonal across the base =  6sqrt 2

So (1/2)  of this =  3sqrt (2) =  sqrt (18)

So the height  =   sqrt [ 20^2 - (sqrt 18)^2 ]  =  sqrt  (382)

So the volume =  (1/3) area of base * height  =  

(1/3) 6^2 * sqrt (382)  =

12 sqrt (382)  units^3

Let the roots be m, n

The product of the roots =   5

Therefore     m * n =  5

But

1/m * 1/n  =  1/5

So

a*b  = 1/5

We can use the tangent

tan 57°  =   height  /   21

21*tan 57°   =   height  ≈   32.3 ft

Let C =  (0,0)   A =  (4,0)  and B = (0,3)

We have a 3-4-5 right triangle so BA =  5

We can find the coordinates of the incenter as  (  [ 4(3)  ] / 12  , [  3(4) / 12 ] =  (1 , 1)

Equation of line through   A and I   ....  y = (-1/3)(x - 4)

When x =  0    D = ( 0, 4/3)

Equation of line  through B and I  ......    y = (-2 / 1) x + 3

When y = 0,  E  =  (0, 3/2)

DE =   sqrt [ (4/3)^2  + (3/2)^2]   =   sqrt  [ 16/9 + 9/4 ]  =  sqrt  [  145 ] / 6   

S =  (7 + 8 + 9 ] / 2   =  12

Area = 

sqrt  [ S * (S  - 7) (S - 8) (S - 9) ]  = 

sqrt [ 12 * 5 * 4 * 3  ]   =

sqrt [ 12 * 12 * 5 ]  =

12sqrt  5  

1/3 + 2/3^2  + 1/3^3 + 2/3^4 + 1/3^5 + 2/3^6  + 1/ 3^7 + 2/3^8 .........

We can write this as

(3 + 2)/3^2   + (3 + 2)/3^4  + (3 + 2) / 3^6  + (3 + 2) / 3^8   .....   =

5 [  1/3^2  + 1/3^4  +  1/ 3^6  + 1/3^8  .....  ]    =

5 [ the sum of an infinite series  with  the first term = 1/3^2  and a  common ratio of 1/3^2 ]  = 

5 (1/3^2)  / ( 1 - 1/3^2)  =

(5/9) / ( 8/9)  =  

5 /  8

Not sure this is possible

To see why

Consider that no one takes both

Then 48 take Spanish and 49 take Chinese.....but we're told that 5 more take Chinese than Spanish

Consider that only  1 takes both

Then  48 + 1  =  49 take Spanish

So  53  take only Chinese and  1 takes both  =  54

But this gives us     53 + 1 + 48    =   102 total students (too  many)

Conslder that  2 take both

Then 50 take Spanish   and  55 would take Chinese   (53 taking only Chinese and 2  taking both)

But this gives us 53 + 2 + 48  = 103

The same thing happens if 3 take both

51 take Spanish   and 56 take Chinese  (53 taking only Chinese and 3 taking both)   =  53 + 3 + 48 = 104

So.....it appears impossible that 48 take only Spanish

THX !!!

Nice answer  and Welcome  !!!

B

                sqrt 14              

C       3             D

BC =  sqrt [ (sqrt 14)^2 - 3^2 ]   = sqrt  5

tan B  =   CD  /   BC  =     3 /  sqrt 5  =   3sqrt (5)  /  3