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Answered on the original.

Just doing the substitutions.

\(h(14)=72+10*14-0.7*14^2\\ h(14)=72+140-137.2\\ =74.8\)

\(27A=1(mod29)\\ 27A=-29N+1\\ 29N+27A=1\\~\\ 29=1(27)+2\\ 27=13(2)+1\\ 27-13(2)=1\\ 27-13(29-1(27))=1\\ 27-13(29)+13(27)=1\\ 14(27)-13(29)=1\\ 27*14=1(mod29)\\ \)

So the modular inverse of 27 mod 29 is 14

Happy birthday!

Don't eat it all at once!

{ 2,.....,14}   number of possible  subsets with these two members  = 2^(14 - 2 -1)  = 2^11

( 4,....,7}     number of possible subsets with these two members  = 2^(7 - 4 - 1)  =  2^2

Total  subsets =   2^11 + 2^2   =    2052

To see this, consider the second set  .....either the  integers  5 and 6 are in or out of each subset

So  2 states ^( number of integers)  =   = 2^2   = 4  subsets

And in the first subset   the integers  3,4,5,6,7,8,9,10,11,12,13    are  either in or out of each subset=

2states ^(number of integers) =  2^11

Slope =   ( 0 - 25)  / (160 - 0)  =  -25  /160  =   -5/32

y  changes by    -5  units  when x  changes by  32

The number of integer coordinates =  1 + (160/5)  =  1 + 32 =   33

  ABCD

+  BCD

+    CD

+       D

______

----  4D

Note that this is impossible

4D   must  be   an even  number for any positive digit D     so    it  cannot  end with a   "1"

4^1   = 4

4^2  =  16

4^3  =  64

4^4 = 256

4^(2n)   ends in 6

4^(2n -1)  ends in 4

4 ^(2 * 18   - 1)   =   4^35     ends in  4

A counter-example is given by f(x) = x*sin(1/x) and g(x) = x^2.

See my answer under your original post

30 (A +.10)  + 35P  =  33.50    →   30A + 3 + 35P  =  33.50  →  30A + 35P  = 30.50    (1)  

24A + 24P  =  22.80     (2)

Multiply (1) by 24  and (2)  by  - 35

720A + 840P =  732

-840A - 840P  =  -798        add these

-120 A  =  -66

A =  66/120  =  .55 =  price of apple in store Y

Let the side of the cube be \(s\). We have the equation \(s^3 = 4 \times 12s\)

Solving the equation gives us \(s = \sqrt{48} = 4 \sqrt{3}\)

So, the volume of this is \((4 \sqrt 3)^3 = \color{brown}\boxed{192 \sqrt 3}\)

This simplifies to \(i^{14} = \color{brown}\boxed{-1}\)

The smallest vallue occurs at 100, or \(1 \times 15^2 + 0 \times 15 + 0 = \color{brown}\boxed{225}\)

Because the median is 20, we know that \(b = 20\). 

Now, we want to minimize a, so we can maximize c. 

Can you take it from here?

See my answer under your original post

In vertex form we have

y = a ( x -  h)^2  + k             (h,k)  is the vertex  and (x,y)  is a point on the graph

So

-8  = a( 0 - 1)^2  + (-9)

1  =  a1^2

a =  1 

So we have

y = ( x - 1)^2 - 9 

y  =x^2 - 2x + 1   - 9

y  = x^2  - 2x  -  8

b = -2       c =  -8

a + 2b  - c   =

1 + (2)(-2)  - (-8)  = 

1   - 4  +  8   =

5

What diagram? 

(5/15) (660) (6)  =  

(1/3) (660) (6)  =

2 * 660   =

1320  are sick

Let the number of cows = 30

Let the number of chickens = x

Let the number of horses =  50 - 30 - x =  20 - x

When he trades away  as many cows as horses he has   30 - (20- x)  =  10 + x cows left

So 

10 + x =  x + (20 - x)

10 + x =  20

x =10

So   he now has  10 +  x = 10 + 10 =  20 cows left

And the number of  chickens and horses =    x  + 20 - x  =   10 + 20 - 10  =   20 

And the number of  cows =  the number of  chickens and horses

(18, 19, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 46, 48)==28 unique sums.