CPhill

y = x^2  - 7

This is  the parabola y = x^2 shifted down by 7 units

Thus.....the minimum value for y  = -7

Here's a graph, Logic  : https://www.desmos.com/calculator/torephbrbq

First term simplified    =  ( b^4 + a^4 - 2a^2b^2)             (a^2 - b^2)^2

                                       ___________________   =    ___________

                                                 a^2b^2                              a^2b^2

Second term simplified  =  [(a + b)^2 + (b - a)^2 ]        2a^2 + 2b^2       2 (a^2 + b^2)

                                         ___________________ = ____________ = __________

                                                 b^2 - a^2                     - (a^2 - b^2)        - (a^2 - b^2)

Third term simplified   =    b^2 + a^2           a^2  - b^2             

                                        _________  -  ___________   =  

                                          a^2 - b^2          a^2 + b^2               

(a^2 + b^2)^2 - (a^2 - b^2)^2

________________________  =

  (a^2 - b^2) (a^2 + b^2)

[ (a^2 + b^2) + (a^2 - b^2] *  [ (a^2 + b^2) - ( a^2 - b^2) ]           ( 2a^2) ( 2b^2)

____________________________________________  =       __________________ = 

                       (a^2 - b^2) ( a^2 + b^2)                                        (a^2 - b^2)(a^2 + b^2)

        4a^2b^2

__________________

(a^2 - b^2) (a^2 + b^2)

Mutiplying the first term by the third we have

(a^2 - b^2)^2                4a^2b^2                                         4(a^2 - b^2)

____________ *  _____________________   =          _____________       

  a^2b^2                   (a^2 - b^2) (a^2 + b^2)                       (a^2 + b^2)

Multiply this result by the second term and we have

4(a^2 - b^2)          2(a^2 + b^2)                4  *  2

___________  *  ____________  =        ______   =     - 8  

(a^2 + b^2)           - (a^2 - b^2)                    -1

Note that  angle CAD  = 30°

Since this an inscribed angle.....then the measure of  minor arc CD is twice this  = 60°

And angle ACB  = 40°

Then......by simiar reasoning.....minor arc AB  = 80°

Then the sum of  of the measures of minor arcs  AD  and BC  must be  360 - 60 - 80  =  360 - 140  = 220°

But these arcs are intercepted by angles ACD  and CAB

So.....the sum of their measures must = (1/2) of the sum of these arcs  = (1/2)(220°) = 110°

x^2 + 2y^2  = 16

2y^2  =  16 - x^2

y^2  =  (16 - x^2)/2      take the positive root on both sides

y  =  [ (1/2)(16 - x^2)]^(1/2)    =   [ 8 - x^2/2]^(1/2)

xy  =  x [ 8 - x^2/2]^(1/2)       take the derivative of this with respect to x   and set to 0

y '  = [ 8 - x^2/2]^(1/2)  - (x)(1/2) [ 8 - x^2/2] ^(-1/2) (x)  =

[8 - x^2/2]^(1/2) - (x^2/2) [ 8 - x^2/2]^(-1/2)  = 0      factor

[8 -x^2/2]^(-1/2)  [ [8-x^2/2  -x^2/2]  = 0     

[8 -x^2/2]^(-1/2)  [  8 - x^2]  = 0       divide both sides by the first factor and we have that

8 - x^2  = 0

x^2  =  8       take the positive root

x = √8  = 2√2

And y  =  √ [ 8 - (√8)^2 / 2 ]  =  √[ 8 -8/2] = √ 8 -4]  = √4  = 2

So.....the max value of xy on the graph is  (2√2) (2)  =  4√2

What is the radius, in inches, of a right circular cylinder if its lateral surface area is 3.5 square inches and its volume is 3.5 cubic inches?

The lateral surface area  is given by  2pi *r * height

The volume is given by  pi * r^2 * h

So  we have that

2pi * r * h  =  3.5   ⇒    h  =   [ 3.5] / [ 2 pi *r]    (1)

pi * r^2 * h  = 3.5  (2)

Sub ( 1) into (2) for h     and we have that

pi * r^2  *  [ 3.5] / [ 2 pi * r ]  =  3.5     simplify

r * 3.5/2  =  3.5     multiply both sides by  [ 2] / 3.5

r  = 2 inches

2x^2  - 8x + 3y^2  + 6y  + 5  = 0

2x^2 - 8x + 3y^2  + 6y  =  - 5       compete the square on x , y

2 ( x^2 - 4x + 4)  + 3(y^2 + 2y + 1)   =   -5  + 8 + 3

2 ( x - 2)^2  +  3 ( y + 1)^2   =  6           divide both sides by 6

(x - 2)^2          (y + 1)^2

_______  +     ________    =       1

      3                    2

We have the form

(x - h)^2          ( y - k)^2

________  +  _________  =  1

     a^2                  b^2

The center of this ellipse  is  ( 2, - 1)

And the major  axis lies along  x

And the max value of x  =   ( 2 + a)  =  (2 + √3)  ≈  3.732

Here is a graph that shows this :  https://www.desmos.com/calculator/qu9bzxtafp

105^4   = 121550625

We can solve this as follows:

Suppose that the first term is 1  and the last term is  15592

Then....the sum of these terms is    (15592) (15593) / 2  = 121563028

Then the difference in the sums is    121563028 - 121550625  =  12403

So....we need to find out how many of the first n positive integers sum to  12403

So  we have

n ( n + 1) / 2 =  12403

n^2 + n  = 24806

n^2 + n  - 24806  =   0

The square root of   24806  ≈  157....

So.... we might guess that the factoriztion of this is  ( n - 157) ( n + 158)  =0

Expanding this gives us  n^2  + n - 24806  = 0

Taking the positive value of n shows that n  = 157

So.....the first 157 positive integers sum to  12403

Which means that the sum of the integers from 158 to 15592   gives us 105^4  =  121550625

So....the first term is 158    the last is 15592   and the number of terms is   15592 - 158 + 1  = 15435

-89 *  (100x^2)  =

-89 * 100 * x^2  =

-8900x^2

Jo's sum is easy   = (100)(101)/2  =  50 * 101   =  5050

Kate  would first start with 5  [ since 1 - 4 ]  are rounded to 0

Starting with 5 thru 14  she will have ten 10s

15 - 24   she will have  ten 20s

25 - 34 she will have  ten 30s

.

.

.

85 - 94 she will have ten 90s

95 - 100   she will have six 100s

So.....her sum is

10  [10  + 20 + 30 + 40 + 50 + 60 + 70 + 80 + 90]  + 6 [100] =

10  [ (10) (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) ] + 6[100]  =

10 (10)  [  sum of the first nine positive integers ] + 6 [ 100]

100  [  (9)(10) / 2 ]  + 6[100]  =

100 [ 45]  + 6[100] =

100 [ 45 + 6 ]  =

100 [ 51 ] =

5100

So....the positive difference is    5100 - 5050  =  50

Here's one more way to solve this

Since AM    =  2√3   and is the diagonal of the square....then AN  must be  2√3 / √2  = √2 * √3  =  √6

And angle ANE  is right    and angle NAE = 15°

So angle AEN  =  75°

So....using the Law of Sines

EN / sin NAE  =  AN / sin AEN

EN / sin 15  =  √6 / sin 75

EN  =  √6 sin15 / sin 75  = √6 sin15/cos15  = √6 tan 15   = √6 [ 1 - cos 30] / sin (30)  =

√6 [ 1 - √3/2] / (1/2)  =  √6 [ 2 - √3]

So  the area of triangles AEN and ALF    = AN * EN  =  √6 * √6 [ 2 - √3]  =  6 [2 - √3]    (1)

So the area  common to the square and triangle =   [ AEMF]  =  area of square - (1)  =    6 - 6[2 -√3]  = [ 6√3 - 6]  units^2  ≈ 4.39 units^2