CPhill

Here's my attempt at this one .....

Note

P = C  ......    total distances walked  = 9 + 4 + 2 + 5  =  20

P =  midpoint of    BC    .....total distances  walked  = 7 + 2 + 2 +  4 +  7 =  22

P = one unit to the right of B.....total distances walked  =  6 + 1 + 3 + 5 + 8  =  23

P = one unit to the left of C  ....total distances walked  = 8 + 3 + 1 + 3 + 6  = 21

P =  one unit to the right of C  .....total distances walked  = 10 + 5 + 1 + 1 + 4  = 21

Let P   be somewhere to the left of C

With a little help from WolframAlpha

Min

sqrt(9-x)2 +  sqrt (4-x)^2  +  sqrt (x^2) + sqrt (2 + x)^2 + sqrt (5 + x)^2   = 20 and   x = 0 ....P =  (9,0) ,  AP  = 9  

Now....let P  be somewhere to the right of C

Min

sqrt (9 + x)^2  + sqrt  (4 + x)^2  + sqrt (x^2)  + sqrt (2 -x)^2 + sqrt (5 - x)^2   = 20  and  x = 0....P =(9,0), AP  = 9

So...it appears that the walking distances are minimized if all meet at C  =  P

10                                       10

 ∫     3x  dx  =     3 (x^2 / 2) ]               =  [  3 (10^2)/2  - 3 (4)^2 / 2 ] =    [ 150  - 24 ]  =  126

4                                         4

If repeats are allowed

26^5 * 5^2    =    297,034,400

When it hits the ground,  the  height =   0    ......so......

-16t^2  + 96t  + 880  =  0      divide through by -16

t^2  - 6t  - 55  =  0

(t - 11) (t + 4) = 0

The first factor set =  0  gives us what we want

t - 11  = 0

t =  11   sec  to  hit the  ground

Here :    https://web2.0calc.com/questions/polygons_3

M + W =   1143       →    1143 - W  = M       (1)

(3/7)M + (5/6)W =  672      (2)

Sub   (1) into (2)

(3/7) ( 1143 - W)  + (5/6)W =  672        

Solving for W  produces    450 = W

So  (b)  =      450 /1143 =  50 / 127

And

a)   M =   1143 - 450 =   693

So  (3/7) (  693)  =    297  M   volunteered   

After running for 2 minutes at hyperturbo speed, the battery has

1  - 2/20 =  1 - 1/10  =  .9   of its  battery life left

And after running for another 18  minutes at turbo speed, the battery has

.9  - 18/40  =   .9  - 9/20  =   .9  - .45 =  .45  of its battery life left

So....after switching to normal speed

The car  can run for     60  * .45  =    27   more minutes 

Rearrange as

x =  2y^2 -14y  + 13

This is a parabola that opens to the  right

The  y coordinate of the vertex =    -(-14) / [2*2]  =   14/4  =  7/2

And the x coordinate  is     2 (7/2)^2  - 14 (7/2)  + 13   =  49/2 - 49 + 13  =  -11.5

Vertex =   (-11.5 , 7/2)  =   (-11.5 , 3.5)

Rearrange  as

x^2   - 6x   +  y^2 + 14y =   6           complete the square on  x, y

(x^2 - 6x + 9)  + ( y^2 + 14y +  49 )  =   6 + 9 + 49

(x -3)^2  +  ( y + 7)^2  =  64

This is a  circle centered at  ( 3 , -7)   with a radius of 8

The area enclosed =    pi * 8^2  =     64 pi    units^2

x + 3 < x^2 + 2x + 14       rearrange as

x^2 + x - 11  >  0

Let's solve  this

x^2  + x - 11 =  0

x^2 + x  =  11                complete the square on x

x^2  + x + 1/4=    11 + 1/4

(x + 1/2)^2  =  45/4       take both roots

x + 1/2  =  sqrt (45) / 2      or   x  + 1/2  = -sqrt (45)  / 2             {  sqrt (45)  =  3sqrt (5)  }

x = [   3 sqrt (5) - 1  ]  / 2     or       x = [  -3sqrt (5)  - 1 ]  / 2

These two answers are the x intercepts  of  a parabola that turns upward

Every real number less than the second answer will be in  the  solution and every  real number greater than the first answer will be in the  solutin ....so.....

x =  ( -inf ,  [-3 sqrt 5 - 1 ] / 2 )  U    (  [ 3sqrt (5) - 1 ]  / 2  , inf )