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Thanks for that info CPhill.  Maybe I should hunt down some of these essays.  

Thanks Chris  :)

This is yet another that I would really like to find the time to examine properly.

there just is not enough hours in a day.  :(

Here's the prime factorization from Wolfram Alpha.......

(I'll take their word for it.....!!!!)

If you read Isaac Asimov's essay, "The Imaginary That Isn't," he demonstrates that an "imaginary" number isn't any more "imaginary" than any other number!!!  (He does so with great humor, too!!).....all of Asimov's essays on mathematics are excellent...and easy to understand....I highly recommend any of them !!!!

(Asimov even demonstrates that there is no such thing as a "half a piece of chalk!!"

A trapezoid is a quadrilateral.  That is it is a 4 sided polygon.

The angle sum of all quadriaterals is 360 degrees.

so the missing angle = 360-(sum of the other 3 angles)

Aw shucks, folks.....it was nothing....really......!!!

Geno had logic out the ying-yang....me???....I was just throwing something against the wall and hoping that some part of it would stick !!!!

Yes he is!!

Hi anon,

unless I am very much mistaken I do not think that this is your only good answer today.

why don't you join up and become a part of out team?

So you have this function, which for your convenience I'm going to rewrite as $$y = 4^x$$.

Now make a plot of the x- and y-axis. 

Now calculate some values for x such as;

x=-2, x=-1, x=0, x=1 and x=2 which give you respectively, y = 1/16, y = 1/4, y = 1, y = 4 and y = 16.

Now mark these points in your graph and then you probably have an idea of what the graph looks like. 

If you want to check your answer, visit this website;

V =(4/3)pi.r³

dV/dt = 4.pi.r².dr/dt so

dr/dt = -10^-4

r = r0 - 10^-4t

where (4/3)pi.r0³ = 10^-6 m³

r0 = ³√(3*10^-6/4pi) m

We want the time when r = 0:

t = r0/10^-4

(the units of 10^-4 aren't given!).

.

I really like Gino's and Chris's logic.  Thanks guys  

f(x) = x^4 - 4x^2 - 3

f'(x) = 4x^3 - 8x

Setting the first derivative to 0 to find the critical points, we have

4x^3 - 8x = 0    factor

4x(x^2 - 2) = 0      and setting each factor to 0 we have    x = 0,  x =±√(2)

And finding the second derivative, f"(x),  we have ....  12x^2 - 8

And "plugging the critical points into this we have

f"(0) = -8    so the curve is concave down at x = 0

And

f"(±√(2)) = 12(2) - 8   = 24-8 = 16, so the curve is concave up at x=  ±√(2)

So, the curve is concave down on  (-√2, √(2 ) = (-1.414, 1.414)

I'll just do part ii) since you can already do part i) 

If you start with the original parabola  $${{\mathtt{x}}}^{{\mathtt{2}}} = {\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{ay}}$$

$$(2ap,ap^2)$$    This has a focal length of $$a$$ and a directrix of   $${\mathtt{y}} = {\mathtt{\,-\,}}{\mathtt{a}}$$

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

If you reflect this about the x axis it will turn upside down and you will have a parabola with a focal length of $${\left|{\mathtt{\,-\,}}{\mathtt{a}}\right|} = {\mathtt{a}}$$         and a directrix of    $${\mathtt{y}} = {\mathtt{a}}$$   and the equation will be

The equation will be     $$x^2=4(-a)y\;\;\rightarrow \;\;
x^2=-4ay$$

The parametric equation will be       $$(2ap,-ap^2)$$

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

If this is now translated down 2a units  

The directix will translate down 2a units to        $$y=a-2a\quad\rightarrow\quad y=-a$$

The focal length will still be $$a$$

The equation will be      $$x^2=-4a(y+2a)$$

The parametric equation will be       $$(2ap,-ap^2-2a)$$

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

Therefore     $$G(2ap,-ap^2-2a)$$

has the same focal length and directrix as the original parabola.

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

Here is a graph to show what is happening

what is a thirteen over five times b two over 2 divided by a negative seven over ten times b negative one over three?

(13/5)  x (2b/2) / (-7/10) x (-b/3) =

(13/5)  x b / (-7/10) x (-b/3) =

(13b/5) x (-10/7) x (-b/3)  =

(-130b/35) x (-b/3) =

(130b^2  /  105)  =

(26/21)b^2

Thanks, I thought there was something wrong. Wonder how the luctured did not notice the problem.....

If b = 4   then  (b -2^2) = (4 -2^2) = (4-4)  =  0

But, we'd be dividing 100 by 0 and that's undefined....thus.....there is no "answer" here!!!

There is no sloution to this.

If we add three consecutive odd integers we get an odd result..........and subtracting 21 from this is an even result.

And that can't be equal to "the smallest odd'.....or any other odd !!!!!

6^(x-3)=2^x  take the log of both sides

log6^(x-3) = log 2^x      and by a log property we have

(x-3)log6= xlog2   simplify

xlog6 - 3log6 = xlog2      rearrange

xlog6 - xlog2 = 3log6      factor

x(log6 - log2) = 3log6      divide both sides by (log6 - log2)

x =  (3log6) /(log6 - log2)  = about 4.89

The jacket requires [3 + 1/4] - [5/6] =  [13/4] - [5/6]....the common denominator is 12, so the fractions become:

[39/12 - 10/12]   =  [29/12]  = [2 + 5/12]

So the total requirement is:

[3 + 1/4] + [2 + 5/12]    =   [3 + 3/12] + [ 2 + 5/12] = [5 +8/12] = [ 5 + 2/3]

Well...if she repaid 3/7, then there remains 4/7 to pay

So....just multiply (4/7) x 4200....and that will be your answer....

45 1/5 , 126 3/4 , and 88 3/8

Let's forget the whole numbers and just concentrate on adding 1/5 + 3/4 + 3/8

The common denominator is 40, so the fractions become 8/40 + 30/40 + 15/40....and adding all these together gives us  53/40  = 1 + 13/40

And adding back the whole numbers, we get  ( 45 + 126 + 88) + (1 + 13/40) = 260 +13/40

424,400 / 284,994 = about 1.49 times as large

Mmmmm...this looks like a "parts" problem

So we have

∫.03x dx + ∫.01xsin(54x) dx =

.03∫x dx   + .01∫x sin(54x) dx

The first is easy...it's just   .03[x^2/2} + C  =  .015x^2 + C

The second requires a bit more doing ....let's forget the .01 out in front....we'll tack it on at the end....

let u = x  and   du = dx

let   v = -(1/54)cos(54x)  and  dv = sin(54x)dx   

So we have

∫u dv =

u*v - ∫v du =

-(x)(1/54)[cos(54x) - ∫-(1/54)cos(54x) dx    and......evaluating the second integral, we have

+(1/54)∫cos(54x) dx = [(1/54)^2][sin(54x)] + C

So we end up with (hopefully!!)....

[.015x^2 + C ] + .01 [ -(1/54)(x)[cos(54x) + [(1/54)^2] sin(54x) + C]= (final answer) =

[.015x^2) + (1/100)(1/2916)sin(54x) - (1/5400)(x)cos(54x)] + C

(Yep...I verified this with WolframAlpha....it's definitely a strange one!!)