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It was 2. I figured it out. Take the derivitive of (1/4)a^2+2a+6. This is a/2+2=0. a=-4. Now plug it in.

Thank you so much!!

This solution is completely nonsensical.

"4" should be the number typed in the  small box outside to  represent those  dogs that got no ribbons

B

7                 7sqrt (3)  =  sqrt (147)

A           D            C

AC =  sqrt  [ 147 - 49 ]  = sqrt [98 ] =  7sqrt (2)

DC = 3.5sqrt (2) = AD  =  sqrt [24.5]

BD  =  sqrt [ 7^2 + 24.5]  = sqrt [ 73.5] 

Using Law of Cosines

BC^2  = BD^2  + DC^2  -  2(BD* DC) cos (BDC)

147  = 73.5 + 24.5 - 2 [ sqrt [ 73.5] * sqrt [ 24.5] ] cos (BDC)

49 / [- 2 sqrt (7203 / 4 )  ] =    cos (BDC)

-49 / sqrt(7203)  = cos (BDC)

sin(BDC) =   sqrt [  1 - (49)^2 / 7203 ]  =  sqrt (6) /  3

tan BDC =  [sqrt (6) / 3 ]  / [ -49 / sqrt (7203) ] =    [sqrt(6)*sqrt (7203) / [ -49 * 3]  =  

(sqrt 2) (sqrt (3) * 49 * sqrt  (3) )  / [ -147]  = 

-sqrt (2) * 147  /  147  = 

   -sqrt (2) 

x + y = (a/2)     square  both sides

x^2 + 2xy + y^2 =  a^2/4

And

xy = -(a + 3)

2xy  = -2(a + 3)

So

x^2  - 2(a + 3) + y^2 = a^2/4

x^2 + y^2 =  a^2/4  - 2(a + 3)  =   (1/4)a^2 - 2a - 6        (1)

Take the derivative  of  (1)  and set    =  0 

(1/2)a  - 2  =  0

(1/2)a =  2

a =  4

The min value  of x^2 + y^2  =  (1/4)(4)^2  - 2(4)  - 6 =    4 - 8 - 6 =   -10

1 + 3 +5 + 7 + 9 + 11  = 36         SIX cosecutive odd POSITIVE integers before the sum exceeds 40 

1

2 + 1

3 + 1

4 + 2

5 + 1

I'll let you finish

      30   A

12        X

            4

      90  C               B  60

The triangle is 30 - 60 - 90

AC = 12    BC =  12/ sqrt (3) = 4sqrt (3)     AB  = 8 sqrt (3)

Area of triangle ABC  =  (1/2) BC * AC  =  (1/2) (4sqrt (3) * 12 =  24 sqrt (3)

XC =  4

Area of triangle  BXC =  (1/2) ( BC) (XC)  = (1/2)(4sqrt (3)) ( 4)  = 8 sqrt (3)

Area of  triangle BXA =  [24sqrt (3)  - 8 sqrt (3) ]  =  16sqrt (3)

x ( 3x - 4)  ≤ [ 6x^2 - 3x + 5 ] /  5      multiply both sides  by 5

5x ( 3x - 4) ≤ 6x^2 - 3x + 5

15x^2 - 20x ≤ 6x^2 - 3x + 5

9x^2 -17x - 5 ≤ 0

Set up as an equality......this  is a  parabola that turns upward......the solution will be found on the  interval  that includes  the  roots

9x^2 -17x  - 5 =  0

The roots are

[ 17 - sqrt (289 + 160) ] / 18 =    [ 17 - sqrt (469) ] / 18  

And   [ 17 + sqrt (469) ]  /18

The solution is

x =  [  (17 - sqrt (469) )   /18 ,  (17 + sqrt (469))  / 18 ] 

Re-arrange to 

x^2 -x -35 > 0      this is a  BOWL SHAPED parabola .....the values of x on either side of the roots are > 0 

Use quadratic formula to find   the roots as x = 1/2 + sqrt(141) /2     and   1/2 - sqrt(141)/2 

so the answer would be 

(- inf ,1/2 - sqrt(141)/2 )   U   (1/2 + sqrt(141) /2  ,  + inf) 

Here is a graph:

Here is one possibility 

  your mom gives birth to your sister when your mom is 30    .... you are 7 at the time 

      10 years later, your sister is 10, your mom is 40   and you are 17 

Integers are   x     x+1     x+ 2    and  x+3 

summed they equal 18 

x  +    x+1   +   x+2   + x+3  = 18 

4x + 6 = 18 

4x = 12 

x = 3 

Since ∠A=30∘ and ∠B=60∘, then ∠C=90∘. Also, ∠BXA=21​⋅90∘=45∘.

Triangle ABX is a 45-45-90 triangle, so BX=AX=22​12​=32​. Therefore, the area of triangle BXA is [\frac{1}{2} \cdot BX \cdot AX = \frac{1}{2} \cdot 3\sqrt{2} \cdot 3\sqrt{2} = \boxed{9}.]