All Answers 358468 Answers

1/2      +  11/12   

12/24  +   22/24

           =

        34/24

OR    1.416666

Notice that if we have a unit circle, any point on that circle can be represened as (rcosx, rsinx)

And using the Pythagorean Theorem, (rcosx)^2 + (rsinx)^2  = 1

But, since this is a unit circle, r = 1 ..... so

cos^2(x) + sin^(2)x = 1

no. multiplaying a number by a number more then 1 its bigger but is its smaler then 1 its smaller

$${\mathtt{50}}{\mathtt{\,\times\,}}{\mathtt{0.5}} = {\mathtt{25}}$$50 * 0.5 = 25

and

50 * 2 = 100

Yes because you are duplicating a number a certain number of times.

Ex:

5*10=50

the original number was 5 though it was duplicated 10 times, and now it is a 50 wich is bigger that a 5.

10 es el answer

10 es el answer

Oops! Two errors on line 2 there, Sasini.

1/2 + 11/12

6/12

1/2

0.5

that is the answer

32/-2

=-16

that is the answer

1.4166666666666667

The correct answer is 4.

I think I have heard of binomial expansion, but that is as far as I have got with it. I think I shall save that for another day to save frying my brain too much now.

Is that other reply after yours also an answer for this? 

with the calculator on the homepage

Simple. You turn the plus symbol to a minus one :-) LOL

as in this ^? you type it by pushing shift and 6

The examples that you gave can be generalized to:  sin(x)  =  cos(90° - x).

-6(4-t) = 12t

-24 -4t = 12t

-24      =  16t

-24/16  =  t

-6/4     =  t

-3/2      =  t

t           =  -1.5//

e^(1/x)-x=0

Root-Finding Algorithm with Newton's method: $$\small{\text{
$
\boxed{ x_{i+1} = x_i - \dfrac{f(x_i)}{f'(x_i)} } \quad f{(x)} = e^{\frac{1}{x}}-x $ and $ f'(x) = -\frac{ e^{\frac{1}{x}}}{x^2}-1 $ we have $ x_{i+1} = x_i + \dfrac{ e^{\dfrac{1}{x_i}}-x_i }
{\frac{ e^{ \left(\dfrac{1}{x_i} \right) } } {x_i^2}+1}
$
}$$

We start with:  

$$\small{\text{
$x_0 = 1$
}} \\
\small{\text{
$ x_1 = 1+\frac{e-1}{e+1}= 1.46211715726
$
}} \\
\small{\text{
$
x_2= 1.46211715726 + \frac{ 0.51955244051 }{1.92697260563 }= 1.73173824009
$
}} \\
\small{\text{
$
x_3= 1.73173824009 + \frac{ 0.04975957061}{1.59404698875}= 1.76295411450
$
}} \\
\small{\text{
$
x_4= 1.76295411450+ \frac{ 0.00042115233 }{1.56736524331}= 1.76322281532
$
}} \\
\small{\text{
$
x_5= 1.76322281532+ \frac{ 0.00000002982}{1.56714330612}= 1.76322283435
$
}} \\
\small{\text{
$
x_6= 1.76322283435+ \frac{ 0.000000000000000149544303}{1.56714329041}= 1.76322283435
$
}}$$

$$x\approx 1.76322283435$$

@@ End of Day Wrap   Tues 20/1/15     Sydney, Australia      Time 11:55pm

Hi all,

Great answers were delivered today by Alan, Heureka, CPhill, Tetration, geno3141, Rochelle, Will85237, numbers, Derhammer, Saseflower, angel07182007, and Tenacious.  Thank you all.    

Interest posts:

1) This tricky AP question had Chris and I stumped.  Heureka came to our rescue   Thanks Heureka.  

$$\frac{\left(\sqrt[3]{x}-1\right)}{2}$$

Thanks Tenacious   

You can certainly do it the way Tenacious did.

Her is another approach (some of you won't know what i am talking about yet)

I have used a binomial expansion.

$$\\(2a+1)^3=(2a)^3+3*(2a)^2+3*(2a)+1\\
(2a+1)^3=8a^3+3*4a^2+3*(2a)+1\\
(2a+1)^3=8a^3+12a^2+6a+1\\$$