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f (g (x))   just  means we  are putting    "g"  into  any "x"    of    "f"    and  evaluating

So  3  ( x^2 - 5x - 1)   +  2   =

3x^2  - 15x  - 3  + 2   =

3x^2  - 15x  - 1        ........ and that's  it   !!!

Let's  use  a couple of  examples  and  see if  we  can generate an answer

Let one  series be

1, 3, 5 , 7 , 9

The  mean is     [ 1 + 3  +  5 + 7 + 9 ]  /  5  =  25 /  5  =  5  =  y

Sum of largest and smallest values =   10  =  2y

Let  another  series  be

11, 13,  15, 17, 19

The mean  is    15

So....it  appears that  the mean  =   the   median value

Sum of largest and smallest value =   30   =  2y

So...... it appears  that  the  sum of the  largest  and smallest  value  =  2 * average  =    2y

Note  that  x  cannot   be   any perfect  square  from  x =  1  to  x = 100  (i.e.,  x= 1^2  to x = 10^2)  because  any  of  these  make  the  denominator  = 0

So......we  only  have   90   possible   x values  for  the  numerator 

(360 - 76 * 2) ÷ 2 

= (360 - 152) ÷ 2

= 208 ÷ 2

= 104^o

5x - 24 = 3 * 2 

5x - 24 = 6

5x = 30

  x = 6

The roots are -1 - sqrt(5/2) and -1 + sqrt(5/2), and the sum of their squares is 7.

I used a computer program, and found 120 possible ways.

The set of all possible a is [-2,2].

The total number of students at the school is 360.

The sum telescopes to x.

This question has been posted before:

\(3(2-0.9h) + (-1.3h - 4) = (6-2.7h) -1.3h - 4 = \boxed{-4h+2}\)

Let the height be h. Then, we have \(h(h+8)(h-2) = 96.\) (volume is height x length x width)

Solving, we get h = 4 inches.

\(\text{slope} = m = \frac{3-1}{3+2} = \frac{2}{5}\)

Plug in (3, 3) into the equation y = mx + b.

\(3 = \frac{2}{5}(3) + b \quad \rightarrow \quad b = \frac{9}{5}\)

The equation is \(\boxed{y = \frac{2}{5}x + \frac{9}{5}.}\)

\(f(x) - g(x) + h(x) = (2x^2 + 3x - 9) - (5x + 11) + (-3x^2+1)\)

\(\boxed{-x^2-2x-19}\)

Let the side length of each triangle be 1. The total perimeter of the corrals is 12, so the side length of the square corral is 3.

Area of triangles: \(4(\frac{\sqrt{3} \cdot 1^2}{4}) = \sqrt{3}\)

Area of square: \(3^2 = 9\)

Ratio: \(\boxed{\frac{\sqrt{3}}{9}}\)

Thanks Cubey

\(s(x) = (4x-7) + (x+1)^2 = x^2 +6x - 6\)

\(s(3) = 3^2 + 6(3) - 6 = \boxed{21}\)

\(5x \geq -85\)

\(\boxed{x \geq -17}\)

There are a total of \({7 \choose 2} = 21\) different pairs. We will count the pairs where the difference is 1 or 2. Note that a difference of 0 is not possible since the numbers have to be distinct.

Case 1: difference = 1

7-6, 6-5, 5-4, 4-3, 3-2, 2-1

6 pairs

Case 2: difference = 2

7-5, 6-4, 5-3, 4-2, 3-1

5 pairs

\(\text{probability} = 1 - \frac{11}{21} = \boxed{\frac{10}{21}}\)

So can someone else help?

If an integer is odd, then it is of the form \(2n+1.\)

Square it: \((2n+1)^2 = 4n^2 + 4n +1 = 4(n^2 + n) + 1 \equiv 1 \quad (\mod 4)\)

I don't think it does.

I'm pretty sure the constant is squared and not the whole thing.

So like \((x-1)(x-4)(x-9)...(x-10000)\)

Show that when the square of an odd integer is divided by 4, the remainder is always  =  1

Let  the  odd integer  =   2n -  1

(2n  - 1)^2 /  4  =   

( 4n^2 - 4n  +  1)  /   4  =

4n^2         4n              1

____  -  ______   +   ___   =

  4             4                4

n^2   -  n     +   1/4      →      remainder  = 1