CPhill

Nice, Melody !!!

We can prove that this will always be so  no matter the current rate or Andy's rate  [ as long as his rate is greater than the current rate ]

Let Andy's normal rate be  A   and the current rate, C

Call T  the time elapsed between when his hat is knocked off and he notices that it's gone

In this time the hat has traveled  C * T

And Andy  has traveled   (A - C)T

When he turns around his rate becomes A + C

And the distance that he travels to retrieve the hat is (A + C) T

And this equals the distance he was from the bridge when he turned around plus the distance the hat has traveled from the bridge  = C *2T  ....so we have......

(A + C ) T  =  (A - C) T   + C*2T

AT + CT  =  AT - CT + 2CT

AT + CT  =  AT + CT

Note to  GA....I verified this result with Lancelot Link......he gave it a "thumbs up" ... [ at least I think that was the digit he was using....]

I like that "table" trick, Melody.....that makes the counting way easier !!!

A = (3,5)   B  = (7, 11)   C  = ( 13,17)

Because of the nature of the points.....this graph will turn downward

The slope  of the line between (3,5) and ( 7,11)  is 6/4  =  3/2

And the equation of this line is

y  = (3/2) (x - 3) + 5    ⇒  (3/2)x + 1/2

And this line forms the "left " branch of the absolute graph

For the equation  of the line forming the other half of the graph...it will have a slope that is the negative of the first line and it will pass through ( 13, 17) 

So the equation of this line is just

y = (-3/2) ( x - 13) + 17    ⇒  (-3/2)x  +  73/2

The intersection of these two lines will form the "vertex" of the function....so we have....

(3/2)(x) + 1/2  =  (-3/2)x + 73/2

3x    =  72/2

3x  = 36     ⇒  x  =  12

And y  =  (3/2)(12) + 1/2  =  18.5

So   b =  12   and c =  18.5

To find "a"  we can use any of the points

So we have that

5  =  a l 3 - 12  l  + 18.5

-13.5  = a (9)

a  =  -13/5 / 9   =   -3/2  =  -1.5

Here's the graph of the lines and the absolute value graph itself :

https://www.desmos.com/calculator/1eyojys50i

3 * 5 * 6

There are a total of  90 cubes

The unpainted ones will be the ones in the middle of the prism

Let 3 be the height

There will be 2 * 6 * 5  =  60  painted cubes on the top and bottom

There will  2 * 6  +   2 * 3  =  18 more painted cubes "sandwiched" between the top and bottom surfaces

So...the total unpainted cubes  is  90  - 78   =   12

So their total surface area is just   [12 * 6 ] cm^2   =  72 cm^2

P.S.  -  I believe there is also a "formula" for this....we are cutting away 2 inches in every dimension...so  (3 - 2) (5 - 2) (6 - 2)  =  1 * 3 * 4    =   12 cubes

Call S  the radius of the smallest circle  and  M the radius of the intermediate circle

pi (12)^2  -  pi (M)^2  =  pi ( S)^2     (1)       and

pi (M)^2  - pi(S)^2   = pi (S)^2    ⇒  pi (M)^2  = 2 pi (S)^2   (2)

Sub (2)  into  (1)

pi (12)^2  - 2pi(S)^2  = pi (S)^2

12^2  - 2S^2 + S^2

12^2   =  3S^2

144  = 3S^2

48 = S^2

√48  = S

4√3  cm  = S   ≈  6.93 cm

She used trial and error....but..with the presence of the "5"  she knew that the base had to  be  6 or greater.......so...it didn't take her long to get to 7....LOL!!!!

This was answered by Melody a few days ago...I believe that she found the answer to be base 7 as follows:

[ 1b^2 + 3b  + 5 ]  / [ 2b + 1]     and when b  = 7 we have that

[ 7^2 + 3*7 + 5 ] / [ 2*7 + 1  ]  =

[ 49 + 21  + 5 ] / [ 15 ]

[ 70 + 5 ]  / [ 15]    =

[ 75] / [ 15]   =

5

p(x)   = √ [ -x ]    q(x)  = 8x^2 + 10x - 3

So   p(q(x) )    =     √  [ - ( 8x^2 + 10x - 3 ]   =  √ [  3 - 8x^2 - 10x ]

We want to find the x values that make

3 - 8x^2 -10x  =  0     multiply through by -1

8x^2  + 10x  -  3  =   0  factor

(4x  - 1) (2x + 3)  = 0

Setting both factors to 0 and solving for x we have that  x  = 1/4   and x  = -3/2

However......the quantity under the radical, 3 - 8x^2 - 10x   must be ≥ 0

So...we have three possible intervals that may solve this

(-inf, -3/2] U [ -3/2, 1/4}] U {1/4, inf )

We only need to test a point in  the middle interval..if it "works," this interval will solve the equation

3 - 8x^2 - 10x ≥  0

Testing  x  = 0  will make this true, so the interval that solves the equation is

-3/2  ≤ x  ≤ 1/4

So  b  -  a   =     1/4  - (-3/2)   =   1/4  - (-6/4)   =   7 / 4

In the form y   = mx + b   .......the slope is  "m"

So  in the line

y =  3x  -  2         the slope  =  3

Now........ line 2  is

3y - 9x + 5  = 0      ....we want to get y by itself

Add 9x to both sides, subtract 5 from both sides

3y   =  9x  - 5         divide both sides by 3

y  =  3x  -  5/3      and note that the slope is exactly the same as the first one

And two lines with the same slope are always parallel