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Quadratic formula ax^2 + bx + c = 0: x = (-b pm sqrt(b^2 - 4ac))/2a

This quadratic has a real root when b^2 - 4ac >= 0.

x^2 + y^2 = 4 + 12x + 6y

x^2 + y^2 = k + 4x + 12y

Graphs intersect when -67 <= k <= 43.

Area of a cube =6S^2, where S = Side length.
Area of cube =6 x 12^2
 [(a/2)-5] * [(b/3)-8] =a*b - [6*12^2 + 7] , solve for a, b
b = -(6 (5 a - 1312))/(8 a + 4), and (6 a + 4) cannot =0, 
a =42 and b=30

Here is the image to the lest question:

helpppppppppppppp and idk

The horizontal separation is tripled, the vertical separation is doubled, so the area will increase by a factor of 3*2 = 6.  i.e. new area = 75*6.

Add the two equations together to get

10x + 10y + 10z = 50,

and divide by 10.

Thanks Oldtimer,

I am fond of you too and I admire your perseverance.

(I assume this is a retirement learning activity for you)

Maybe you had an easier method?

Anyway, I wish you a great Christmas and the best of new years.

Let  two of the equal  sides  = x      Let the other side  = y

So

2x +  y  =  400   →     y = 400 - 2x

Area  = xy

A  =  x ( 400 - 2x)

A  = 400x  - 2x^2        take the  derivative

A' =  400 - 4x

Set  the  derivative  = 0 

400  - 4x  =  0

400  = 4x

x = 100 

y = 400  -2x  =   200

Max area =   100 * 200  =   20000 ft^2

See the graph  of the Area  function here.....https://www.desmos.com/calculator/pvjvhyciaz

EP  and FP  could have the  same  value  (in this case  EP  = FP  = 4)

See here  :

This problem  would  have a definite answer if  AD  or BC    lengths were  specified

I made a mistake!!! I used the wrong angle.

Instead of √2 and √8  should be  1 and 3

Phill's answer is correct!!!

Please don't trust me. I'm not Nikola Tesla.

g(a)  = 5 -10a

p (g(a))     =        4  + 2 sqrt [  ( 5 -10a)  + 2  ]   =  4 + sqrt [ 7 - 10a ] 

The function in the radical cannot  evaluate to <  0....therefore...

a ≤  7/10

Answered here :  https://web2.0calc.com/questions/can-someone-explain-induction-to-me

(a (10)   +  b )   / 2   =   ab

10a  + b  = 2ab

10a =    2ab - b

10a  =   b ( 2a - 1)

b =   10a  / ( 2a  -1)

b will be the smallest possible   integer, 6,  when a = 3

So....the  number  = 36.....just as the Guest  found  !!!

DE + DF = 6√3 + 12  

Point G is the centroid of 4ABC (∠C = 90◦ ). Find CG is AB = 24 cm

CG = 2/3 (24 / 2)   ==>    CG = 8 cm

2x + 8               2 ( x + 4)          2          (x + 4)                (1/4) (x + 1) / (2x + 1)

_______  =   __________  =  ___ *   _______  =       

16 x + 8           8 ( 2x + 1)          8        (2x + 1)

Base  6  number =  m(6^2)  + 1(6)  + 5  =   36m + 11

Let  m  = 1...  =  115₆  =    47₁₀  

Base  9 number

n(9)  + 5 

Let n = 5  .... =  55₉  =     50₁₀

50₁₀ - 47₁₀  =   3₁₀   =   10₃

The last two digits are  1 and 0

6x  +  5y    - 2z =   -4

4x  +  5y   + 12z  = 54

We have infinite  soutions for  x, y and z

Subtract the  second equation from the  first

2x  -14z  =  -58     →  x - 7z =  -29   →  x = -29 + 7z

6(-29 + 7z)  + 5y  - 2z =  -4   →   -174 + 42z + 5y  - 2z =  -4  →  5y  + 40z = 170  →  y + 8z  = 34  → y =34 - 8z

Letting  z   = 1     x = -22    and y =  26

In this case ......x + y +  z  =    5

tysm! i got it :>

There  are  5 + 2  =  7 equal parts of work

Ben  does  2/7  of them

So

(2/7)  (1400)  =   2 ( 1400/7)  =  2 (200)  = 400......just as the Guest  found  !!!

3x^2  + 4x  + 12   =  0

You're  on  the right track....we  just need to use a little Algebra

r + s  = -4/3      square both sides

r^2  + 2rs  +  s^2 =  16/9      (1)

And

rs = 4  .....   so.....2rs  = 8      (2)

Sub (2)  into (1)    and you will find what you need

Primes are  2,3 and  5 =    1/2  chance of showing a prime

Nonprimes are  1, 4, 6 =  1/2  chance of showing a non-prime

P (that three show primes )  =  C(6,3)  ( 1/2)^3  ( 1/2)^3    =  20  * (1/8)^2  =  20/64  =  5/16