CPhill

Very nice, Melody  !!!

Thanks, alan.....here's a non-Calculus qpproach

Let the product   xy   =  a    

And  xy will be maximized where this function is tangent to the ellipse...so we need to find  the "a" that makes this true

So   xy  = a   ⇒  y =  a/x

Subbing this into   x^2 + 2y^2  = 16   we get

x^2  + 2(a/x)^2  =  16       multiply through by   x^2

x^4  +  2a^2  =  16x^2

Let  v = x^2

v^2  -  16v  +  2a^2  =  0

We will have  a double real (positive) vaue for v whenever

256 - 8a^2  =  0

256 =  8a^2

a^2  =  32    ⇒  2a^2  =  64

a = √32  =  4√2

So    

v^2 - 16v  -  64 =  0

(v - 8)^2  =  0   ⇒ v =  8  ⇒  x^2  =  8  ⇒  x = ±√8  ⇒  x = ±2√2

And  y  =  (a/x)  =  (4√2 / ( ±2√2) )  =  ±2

So.....the product xy is maximized whenever  (x, y)  =  ( ±2√2, ±2)

And   xy  is  maximized  when  a  =  4√2    as this graph shows :

https://www.desmos.com/calculator/vmykhikdjb

Wecome back   !!!!

James's original money  = M

Brother's original money  =  2M

Sister's original money  = M + 27

After the payoff

( [ M + 27 ] + 12)  =  2 (2M - 24)   simplify

M + 39  =  4M - 48     add 48 to both sides, subtract M from both sides

87  =  3M      divide both sides by 3

29  =   M   

So....James now has   29 +  12  =  $41

9.  Not as bad as it looks

these points form a trapezoid  with a top base = 5   a bottom base = 7   and a height = 2

So....the area  is

(height)/2  *  ( sum of the bases)  =

(2)/2  *  ( 5 + 7)  =     1 * 12   =    12 units^2 

Here's a pic  :  https://www.desmos.com/calculator/c3knzz9qii

10.   pi r^2

        pi r^2 h

        1/3

11. Volume of a sphere  =  (4/3) (3.14) (diameter / 2)^3  =  (4/3) (3.14) (4/2)^3  ≈ 33.5 cm^3 

I WOULD say the same thing...!!!

5. length  =  √  [ 48^2  + 61^2  ]  =  √ 6025   ≈  77.6 ft  

6. equal

    pi r^2 h

7.    [ 7x - 40 ] / 18   = 25 / 15

       [ 7 x - 40 ] / 18 =  5 / 3    multiply both sides by 18

        7x  - 40   =   (5/3) * 18

        7x  -  40   =  30      add 40 to both sides

        7x   =    70           divide both sides by  7

         x   = 10 

8.   √ [ 19^2 - 13^2  ]  =  √ 192  ≈  13.9 

( 7k - 3)  ( k^2 - 2k  +  7)  =

7k ( k^2  - 2k  +  7 )  =          7k^3  -   14k^2   +  49k

+                                       +

-3 ( k^2 - 2k  + 7)   =                          -3k^2    +    6k   -   21

_____________                  _________________________

                                             7k^3   -   17k^2  +  55k   -  21

I'll answer a few....but not all 33   !!!!

1.   XZ / YZ  =  DF / EF     ⇒   15 / 10  =  DF /  8     multiply both sides by  8

     (8 *15) / 10  =  DF

      120 / 10  =  DF  =   12 

2.   diameter

      circumference

      pi

      diameter

3.  SAS Similarity Theorem

4.   The center of  the circle  is  ( 2, - 1)

And the radius is  √  [  ( 3 - 2)^2  +  ( -1 - 1)^2 ]  =  √ 5

So...the area of the semi-circle is   5pi/ 2  =  5(3.14)/2  = 7.85 

And  one side of the quadrilateral is 2√5

And the other dimension is √ [ ( 1 - 3)^2  + ( 2 - 1)^2 ] =  √ 5

So...its area  is  2√5 *√ 5  =  10

So...the total area is   [ 7.85 + 10 ]  units^2  =  17.85 units^2  =  17.9 units^2

3t +  4  =  12 + 3t    subtract 3t from both sides

4  =   12

Since this is never true  there are no solutions