CPhill

Thanks, Melody....

y=-x^3+x^2-2.....the tangent line will be horizontal wherever the derivative = 0

So, the derivative is given by

y ' = -3x^2 + 2x

Set this to 0...so we have.....

-3x^2 + 2x = 0    factor

x(-3x+ 2 ) = 0     set each factor to  0

x = 0   and   x = 2/3

Let's see

-1/2  = - .50

And

-1/3  = - .25

How about

-1 / 2.7 =  -0.3703703703703704   ???

Yeah....that looks pretty good....(of course, it's only one possibilty.....there are many others)

7^20/7^4  ... remember the rule .....  a^n / a ^m  = a^(n - m)......so we have...

7^(20 - 4)  =

7^16

(5-2)^2-(6-5)^3 =

3^2  - 1^3  =

9 - 1  =

8

(x^4+2x^3-7)÷(x^2-1) =

                x^2  + 2x  + 1

             --------------------------------------

x^2 -1   [  x^4  + 2x^3  + 0x^2   + 0x  - 7 ]

                x^4               -  x^2

------------------------------------------------

                           2x^3  +   x^2

                            2x^3               -2x

---------------------------------------------------

                                          x^2  + 2x - 7

                                          x^2          - 1

                                        ----------------

                                                    2x - 6

So, our answer is  x^2 + 2x + 1   +  [2x -6] / [x^2 -1]

First, let us show that it is true for n=1

So....a^(2(1) + 1) + b^(2(1) + 1)  =  a^3 + b^3  =  (a +b)(a^2 -ab + b^2)...and (a + b) is a factor

Now, let us assume it is true for  k=1, that is, a^(2(k) + 1) + b^(2(k) + 1) is true

Now, let's prove that it is true for k + 1

So we have

a^(2(k+1)+ 1) + b^(2(k+1) + 1) =

a^(2k + 3) + b^(2k + 3) =

a^3*a^(2k)  + b^3*b^(2k)......and using   a^3  =  (a +b)(a^2 -ab + b^2) - b^3 , we can write

[(a+b)(a^2 -ab + b^2) - b^3]a^(2k)  + b^3*b^(2k)  =

(a+b)(a^2 -ab + b^2)a^(2k) - b^3[ a^(2k) -  b^(2k)]

Notice that the last term is a difference of two even powers.........and it can be shown that the difference of two even powers can be factored with (a + b) as a factor thusly...

(a^(2n) - b^(2n)) = (a + b)[a^(2n-1) - a^(2n-2)b + a^(2n-3)b^2 - a^(2n-4)b^3 +.....+ ab^(2n-2) - b^(2n-1) ]

Thus, (a+b) is a factor of both terms, so (a +b) is a factor of a^(2k + 3) + b^(2k + 3)

Using Calculus, we can find the derivatve of the function.... which is the velocity at any time t..so we have

h' =-32t + v₀

And at the max ht, this is = 0, so we have

-32t + v₀ = 0     →  v₀ = 32t  ...and, substituting this back into the position function at the time of the max ht, we can find the time it takes to reach that point...so we have

1081 = -16t^2 + (32t)t + 921    simplify

1081 = -16t^2 + 32t^2 + 921

160 = 16t^2    divide both sides by 32

10 = t^2

t = √10 sec  ..and this is the time it takes to reach the max ht

So, using the first derivative to solve for v₀ when t= √10  , we have ... v₀  = 32(√10)ft/s

And our original position function becomes

h = -16t^2 + 32√10t + 921

Let the base be representd by h-8

And the area is given by  b*h /2 = (h-8)*h / 2

And this is "at least" 60cm² ....and "at least" is analogous to "≥" ... so we have...

(h-8)*h / 2  ≥  60 cm²   .... which we could also express as....

h² - 8h  ≥ 120 cm²