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12/100*3000     now use the calc on the home page  

The calc is on the home page.

What do you want us to do with this?  - it is already fully simplified  

I don't think that "Fermat's Last Theorem" got as much attention as this one did...!!!

{But..it took a lot longer to prove....over 350 years....}

Regardless...it was a good one to play around with, Mathematician.....many different approaches by our "professionals"

And even some bumbling and stumbling by me, too!!!

Mathematician, I bet you didn't think that your question would attract THIS much attention  LOL  

Thank you Chris  :D

More for me to think about :)))

See if this helps 

You can verify your answers easily with calculator: e^ia = cos(a) + i*sin(a)..

$$\\59-3|6-9^2|\\
=59-3*|6-81|\\
=59-3*|-75|\\
=59-3*+75\\
=59-225\\
=5-166\\$$

This boils down to finding the side length of an equiateral triangle inscribed in a circle. Of course, the area is maximized when the side lengths are maximized....!!!

Applying the Calculus to our given problem where r = 4.5 and x = s

We have

A = .5x(4.5 + √(20.25 - .25x^2))

A = 2.25x + .5x√(20.25 - .25x^2)

dA/dx = 2.25 +.5(20.25 - .25x^2)^(.5) - (.125x^2)(20.25 - .25x^2)^(-.5) = 0

18 + 4(20.25 - .25x^2)^(.5) - x^2(20.25 - .25x^2)^(-.5) = 0  rearrange

18 + 4(20.25 - .25x^2)^(.5) = x^2(20.25 - .25x^2)^(-.5)

18 (20.25 - .25x^2)^(.5) + 4 (20.25 - .25x^2) = x^2

18(20.25- .25x^2)^(.5) = x^2 - 4 (20.25 - .25x^2)     square both sides

324(20.25- .25x^2) = x^4 - 8 (20.25 - .25x^2)x^2 + 16(20.25-.25x^2)^2

6561 - 81x^2 = x^4 - 162x^2 + 2x^4 + 16(6561/16 - 61/8x^2 + 1/16x^4)

6561 - 81x^2 = 4x^4 - 324x^2  +6561

4x^4 - 243x^2 = 0   factor

x^2 (4x^2 - 243) = 0      reject x =0

4x^2 - 243 - 0

4x^2 = 243

x^2 = 243/4

x = √(81 * 3) / 2 =   (9/2)√3  = 4.5√3 = s

And this result can be generalized to r√3 for any given radius

There a answer, it's 2750

If you added 100 to f(x) then you would know.

$$\\11x^2+13x-28+100=0\\
11x^2+13x+72=0\\
\triangle=b^2-4ac\\
\triangle=13^2-4*11*72\\
\triangle<0\\
$therefore the graph will never reach as low as $ f(x)=-100$$

how many cubic meters is 1,669 trillion cubic feet

Let x be the number of dollars, rhen 2x will be the number of quarters....

So

x + .25(2x) = 96    simplify

x +.50x = 96

1.5x = 96

(3/2)x = 96

x = 64

And 2x = 128 = the number of quarters

The discriminant = b^2 - 4ac.... so we have

13^2 - 4(11)(-28) = > 0    so this tells us we have real roots

But...maybe I'm missing something.....the discriminant doesn't tell us anything about whether or not f(x) ever = -100....it just tells us whether or not we may have "real" zeroes

We'd need to find the vertex for this

The x coordinate for the vertex is given by -b/2a  = -13/2(11)  = -13/22

And putting this back into the function.....we have the y value at the vertex = about -31.84

And since this parabola turns "up," (-13/22, -31.84) is the lowest point on the graph.....so f(x) never = -100....

Don't worry, saseflower, I misread it too, at first....!!!

Suggestion:

x-3<1

x<4

the web2 calc handles this easily

[2nd][sec](1.2934292)

$${asec}{\left({\mathtt{1.293\: \!429\: \!2}}\right)} = {\mathtt{39.363\: \!427\: \!756\: \!281^{\circ}}}$$

on othe calcs you often have to do this

$$\\sec \theta = 1.2934292\\
cos\theta = 1/1.2934292\\
\theta= cos^{-1}(1/1.2934292)\\$$

arc cos is the same as inverse cos

I haven't tried that yet either.   

Thank you Heureka   

21*sqrt(31.4)

$${\mathtt{21}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{31.4}}}} = {\mathtt{117.674\: \!976\: \!099\: \!423\: \!959\: \!1}}$$

Using the Gauss method to add 1+2+3+...+98+99+100= 50(1+100)=5050,

Find the sum of  1x+2+3x+4+5x+...+98+99x+100.

Can you please show how you got the answer as well, thank you.

$$\small{\text{
$
1x+2+3x+4+5x+ \dots +98+99x+100 =
$
}}\\
\small{\text{
$
\underbrace{
[\ 1x+3x+5x+7x+ \dots + 97x+99x \ ]
}_{\text{Part}\ 1}
+
\underbrace{
[ 2+4+6+8+\dots +98+100\ ]
}_{\text{Part}\ 2}
$
}}$$

Part 1:

$$\small{\text{
$[\ 1x+3x+5x+7x+ \dots + 97x+99x \ ]
=
2x[\ \underbrace{ 1+2+3+4+\dots + 49+ 50}_{=(1+50)*\frac{50}{2}}\ ] -50x $
}}\\
\small{\text{
$=(1+50) *50x-50x = 2550x-50x = 2500x
$
}}$$

Part 2:

$$\small{\text{
$
[ 2+4+6+8+\dots +98+100\ ] = 2*[\ \underbrace{ 1+2+3+4+\dots + 49+ 50}_{=(1+50)*\frac{50}{2}}\ ]
$
}}\\
\small{\text{
$=(1+50) *50 = 2550
$
}}$$

Part 1 + Part 2 = 2550 + 2500x