CPhill

Get a common  denominator  between  3 5 and 7  = 105

So we have

70 / 105  <  21x / 105  <  90/105

Note that if x = 4   we have that

70 /105  <  84/ 105  < 90 /105

The top left gragh

The second function is the tip-off.....

l x  l  -  1     if - 1 ≤  x  ≤  1

When x = -1  . the function exvalutes to 0

When x = 0 , the function  evaluates to - 1

When x = 1, the function evaluates back to 0 

Consider  the set   ( A, B , C, D, E, F, G ,H , I ,J )....let A,B  be the indistinguishable  chairs and the  rest  distinguishable stools

We can chose any  8 of 10  places  for the distinguishable chairs to be placed and for each of these they  can be  arranged  in 8! ways.....note that once we  do  this, the indistinguishable stools don't matter because their places are "set" and the placement of A....B  looks just  like the placement of  B ....A

So

C (10,8) * 8!  =  1,814,400  arrangements

Parallel lines ( or their segments) have the same slope

So  slope of   AB =  [ -4 - 0 ] / [ 0 - - 4 ] =  -4 / 4 =  -1

So ....using the same idea

[ k - 18  ] / [  20 - 0 ]  =  -1

[ k - 18 ] / 20  = -1     multiply both sides by 20

k -18  = -20    add 18 to both sides

k = -2

V = √2 * (side)^3  / 3

243  = √2  (side)^3 / 3     multiply both sides by 3 / √2

243 * 3 / √2  =  side^3     take the cube root of both sides

side ≈  8.018

Good job, Imnotamaster  .....!!!!

The circumference of the  inside circle can  be expressed as 2pi R₁

The circumfernce of the outside circle can be expressed as  2piR₂

So

20pi  = 2pi R₂ - 2pi R₁

20pi  =  2pi  (R2 - R1)     divide both sides by 2 pi

10  = R₂ - R₁    [ the track is 10 ft wide  ]

I'll let you fill in the  details

H= distance separating bases

R1 = 12

R2 = 9

V = (pi/3) * H *  [ (R1)^2 + R1* R2  + (R2)^2 ]

A =  pi   [ ( R1 + R2)  √ [ (R1 - R2)^2 + H^2 ] + (R1)^2 + (R2)^2 ]

I/m using a technique that I "borrowed" from a really good  mathematician on  here, Alan  !!!!

pq  + 2p + q  =  348

qr  + 4q  + 3r  = 208

rs  + 8r  + 6s =  722

Manipulating the first equation

p (q + 2)  =  (348 - 1q)

p = (348 - 1q) / (q + 2)

p = [ (350 - 1( q + 2) ] / (q + 2)

p = (350) / ( q + 2) -  1 

Manipulating the  second equation

qr + 4q + 3r = 208

q ( r + 4) =  208 - 3r

q = ( 208 - 3r) ( r + 4)

q =  [ (220 -3(r + 4) ] / (r + 4)

q = (220) / (r + 4) - 3

Manipulating the  third  equation

rs + 8r + 6s  = 722

r (s + 8)  =  722 - 6s

r = (722 - 6s) / ( s + 8)

r = [722 - 6(s + 8)] / (s + 8)

r = [ 770 - 6(s +8) ] / (s + 8)

r = [ 770 ] / (s + 8) -  6

Note  that  770  factors  as   2 * 5 * 7  * 11 =  2 * (35) * 11 = 2 * (27 + 8) * 11

If we let s = 27     then we have that

r = 770 / 35 - 6  =   22 - 6  =  16

q=  (220) / (16 + 4)  - 3  =  11 - 3 = 8

p = (350) / ( 8 + 2) - 1  =  35 - 1  =  34

Nice, hectictar  !!!!